HCF and LCM
Key Concepts
Solved Examples: 60
Practice Questions: 7
Solved Examples
Question 1: Find the HCF and LCM of 72, 120 and 168
72 = 2³ × 3²
120 = 2³ × 3 × 5
168 = 2³ × 3 × 7
HCF = 2³ × 3 = 24
LCM = 2³ × 3² × 5 × 7 = 2520
120 = 2³ × 3 × 5
168 = 2³ × 3 × 7
HCF = 2³ × 3 = 24
LCM = 2³ × 3² × 5 × 7 = 2520
Question 2: Using the Prime Factorisation Method, find the HCF and LCM of 150, 225 and 300.
150 = 2 × 3 × 5²
225 = 3² × 5²
300 = 2² × 3 × 5²
HCF = 3 × 5² = 75
LCM = 2² × 3² × 5² = 900
225 = 3² × 5²
300 = 2² × 3 × 5²
HCF = 3 × 5² = 75
LCM = 2² × 3² × 5² = 900
Question 3: Using the Division Method, find the HCF of 252 and 198.
252 ÷ 198 = 1 remainder 54
198 ÷ 54 = 3 remainder 36
54 ÷ 36 = 1 remainder 18
36 ÷ 18 = 2 remainder 0
Therefore,
HCF = 18
198 ÷ 54 = 3 remainder 36
54 ÷ 36 = 1 remainder 18
36 ÷ 18 = 2 remainder 0
Therefore,
HCF = 18
Question 4: Find the HCF of 696 and 609 using the Division Method.
696 ÷ 609 = 1 remainder 87
609 ÷ 87 = 7 remainder 0
Therefore, HCF = 87
609 ÷ 87 = 7 remainder 0
Therefore, HCF = 87
Question 5: The HCF of two numbers is 18 and their LCM is 540. If one of the numbers is 90, find the other number.
Product of the two numbers = HCF × LCM
= 18 × 540
= 9720
Other number = 9720 ÷ 90 = 108
= 18 × 540
= 9720
Other number = 9720 ÷ 90 = 108
Question 6: The HCF of two numbers is 16 and their LCM is 960. If one number is 80, find the other number.
Product of the two numbers = 16 × 960 = 15360
Other number = 15360 ÷ 80 = 192
Other number = 15360 ÷ 80 = 192
Question 7: Find the greatest number that divides 455, 527 and 689, leaving the same remainder in each case.
If a number leaves the same remainder when dividing several numbers, it must divide their pairwise differences.
The three pairwise differences are:
527 − 455 = 72
689 − 527 = 162
689 − 455 = 234
Therefore, the required number is the HCF of these differences.
HCF(72, 162, 234)
Prime factorisation:
72 = 2³ × 3²
162 = 2 × 3⁴
234 = 2 × 3² × 13
Therefore, HCF = 2 × 3² = 18
Hence, the greatest number that divides all three numbers leaving the same remainder is 18
Verification
455 ÷ 18 = 25 remainder 5
527 ÷ 18 = 29 remainder 5
689 ÷ 18 = 38 remainder 5
The remainder is 5 in each case. ✓
The three pairwise differences are:
527 − 455 = 72
689 − 527 = 162
689 − 455 = 234
Therefore, the required number is the HCF of these differences.
HCF(72, 162, 234)
Prime factorisation:
72 = 2³ × 3²
162 = 2 × 3⁴
234 = 2 × 3² × 13
Therefore, HCF = 2 × 3² = 18
Hence, the greatest number that divides all three numbers leaving the same remainder is 18
Verification
455 ÷ 18 = 25 remainder 5
527 ÷ 18 = 29 remainder 5
689 ÷ 18 = 38 remainder 5
The remainder is 5 in each case. ✓
Question 8: Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
If a number leaves the same remainder when dividing 43, 91 and 183, then it must exactly divide the differences of these numbers.
Find the differences:
91 − 43 = 48
183 − 91 = 92
Now find the HCF of 48 and 92.
48 = 2⁴ × 3
92 = 2² × 23
HCF = 2² = 4
Verification:
43 ÷ 4 = 10 remainder 3 ✔
91 ÷ 4 = 22 remainder 3 ✔
183 ÷ 4 = 45 remainder 3 ✔
Since all three leave the same remainder, the required greatest number is 4
Question 9: Find the least number which when divided by 12, 15 and 20 leaves no remainder in each case.
The required number must be exactly divisible by all three numbers.
Therefore, find the LCM of 12, 15 and 20.
12 = 2² × 3
15 = 3 × 5
20 = 2² × 5
LCM = 2² × 3 × 5 = 60
Verification:
60 ÷ 12 = 5 remainder 0 ✔
60 ÷ 15 = 4 remainder 0 ✔
60 ÷ 20 = 3 remainder 0 ✔
Therefore, the least required number is: 60
Therefore, find the LCM of 12, 15 and 20.
12 = 2² × 3
15 = 3 × 5
20 = 2² × 5
LCM = 2² × 3 × 5 = 60
Verification:
60 ÷ 12 = 5 remainder 0 ✔
60 ÷ 15 = 4 remainder 0 ✔
60 ÷ 20 = 3 remainder 0 ✔
Therefore, the least required number is: 60
Question 10: Find the least number which when divided by 8, 10 and 12 leaves a remainder of 5 in each case.
Subtract the common remainder from the required number.
Required number − 5 must be exactly divisible by 8, 10 and 12.
Find the LCM of 8, 10 and 12.
8 = 2³
10 = 2 × 5
12 = 2² × 3
LCM = 2³ × 3 × 5 = 120
Therefore, required number = 120 + 5 = 125
Required number − 5 must be exactly divisible by 8, 10 and 12.
Find the LCM of 8, 10 and 12.
8 = 2³
10 = 2 × 5
12 = 2² × 3
LCM = 2³ × 3 × 5 = 120
Therefore, required number = 120 + 5 = 125
Question 11: Find the least number which when divided by 9, 15 and 18 leaves a remainder of 4 in each case.
Required number − 4 must be exactly divisible by 9, 15 and 18.
Find the LCM of 9, 15 and 18.
9 = 3²
15 = 3 × 5
18 = 2 × 3²
LCM = 2 × 3² × 5 = 90
Therefore, required number = 90 + 4 = 94
Find the LCM of 9, 15 and 18.
9 = 3²
15 = 3 × 5
18 = 2 × 3²
LCM = 2 × 3² × 5 = 90
Therefore, required number = 90 + 4 = 94
Question 12: Find the least number that must be added to 378 so that the resulting number is exactly divisible by 24.
Find the remainder when 378 is divided by 24.
378 ÷ 24 = 15 remainder 18
To make the number exactly divisible by 24, add:
24 − 18 = 6
Verification:
378 + 6 = 384
384 ÷ 24 = 16 remainder 0 ✔
Therefore, the least required number is: 6
378 ÷ 24 = 15 remainder 18
To make the number exactly divisible by 24, add:
24 − 18 = 6
Verification:
378 + 6 = 384
384 ÷ 24 = 16 remainder 0 ✔
Therefore, the least required number is: 6
Question 13: The HCF of 48 and 72 is also the HCF of 72 and x. Find the smallest possible value of x greater than 72.
HCF of 48 and 72 = 24
Therefore, HCF of 72 and x must also be 24.
The smallest multiple of 24 greater than 72 is: 24 × 4 = 96
HCF of 72 and 96 = 24 ✔
Therefore, the smallest possible value of x is: 96
Therefore, HCF of 72 and x must also be 24.
The smallest multiple of 24 greater than 72 is: 24 × 4 = 96
HCF of 72 and 96 = 24 ✔
Therefore, the smallest possible value of x is: 96
Question 14: The LCM of 15 and 20 is also the LCM of 20 and x. Find the smallest possible value of x.
LCM of 15 and 20 = 60
Therefore,
LCM of 20 and x must be 60.
Prime factorisation:
20 = 2² × 5
60 = 2² × 3 × 5
The smallest value of x that contributes the missing factor 3 is: x = 3
LCM of 20 and 3 = 60 ✔
Therefore, the smallest possible value of x is: 3
Therefore,
LCM of 20 and x must be 60.
Prime factorisation:
20 = 2² × 5
60 = 2² × 3 × 5
The smallest value of x that contributes the missing factor 3 is: x = 3
LCM of 20 and 3 = 60 ✔
Therefore, the smallest possible value of x is: 3
Question 15: Two numbers have an HCF of 14 and an LCM of 420. If one number is 70, find the other number.
Product of the two numbers = HCF × LCM
= 14 × 420
= 5880
Other number = 5880 ÷ 70
= 84
Verification:
HCF of 70 and 84 = 14 ✔
LCM of 70 and 84 = 420 ✔
Therefore, the other number is: 84
= 14 × 420
= 5880
Other number = 5880 ÷ 70
= 84
Verification:
HCF of 70 and 84 = 14 ✔
LCM of 70 and 84 = 420 ✔
Therefore, the other number is: 84
Question 16: The LCM of two numbers is 840. One number is 120. Find the least possible value of the other number.
Prime factorisation:
120 = 2³ × 3 × 5
840 = 2³ × 3 × 5 × 7
The missing prime factor is 7.
Therefore, the least possible value of the other number is: 7
120 = 2³ × 3 × 5
840 = 2³ × 3 × 5 × 7
The missing prime factor is 7.
Therefore, the least possible value of the other number is: 7
Question 17: Find the HCF of 0.84, 1.26 and 2.10.
Convert the decimals into whole numbers by multiplying each by 100.
0.84 = 84/100
1.26 = 126/100
2.10 = 210/100
Find the HCF of 84, 126 and 210.
84 = 2² × 3 × 7
126 = 2 × 3² × 7
210 = 2 × 3 × 5 × 7
HCF = 2 × 3 × 7 = 42
Now divide by 100.
HCF = 42/100 = 0.42
Verification:
0.84 ÷ 0.42 = 2 ✔
1.26 ÷ 0.42 = 3 ✔
2.10 ÷ 0.42 = 5 ✔
Therefore, the required HCF is: 0.42
0.84 = 84/100
1.26 = 126/100
2.10 = 210/100
Find the HCF of 84, 126 and 210.
84 = 2² × 3 × 7
126 = 2 × 3² × 7
210 = 2 × 3 × 5 × 7
HCF = 2 × 3 × 7 = 42
Now divide by 100.
HCF = 42/100 = 0.42
Verification:
0.84 ÷ 0.42 = 2 ✔
1.26 ÷ 0.42 = 3 ✔
2.10 ÷ 0.42 = 5 ✔
Therefore, the required HCF is: 0.42
Question 18: Find the LCM of 2/3, 5/6 and 7/9.
Use the formula:
LCM of fractions = (LCM of numerators) ÷ (HCF of denominators)
Numerators: 2, 5, 7
LCM = 70
Denominators: 3, 6, 9
HCF = 3
Therefore,
LCM = 70 ÷ 3 = 70/3
Verification:
70/3 ÷ 2/3 = 35 ✔
70/3 ÷ 5/6 = 28 ✔
70/3 ÷ 7/9 = 30 ✔
Therefore, the required LCM is: 70/3
LCM of fractions = (LCM of numerators) ÷ (HCF of denominators)
Numerators: 2, 5, 7
LCM = 70
Denominators: 3, 6, 9
HCF = 3
Therefore,
LCM = 70 ÷ 3 = 70/3
Verification:
70/3 ÷ 2/3 = 35 ✔
70/3 ÷ 5/6 = 28 ✔
70/3 ÷ 7/9 = 30 ✔
Therefore, the required LCM is: 70/3
Question 19: Three ropes measuring 36 m, 48 m and 60 m are to be cut into pieces of equal length. What is the greatest possible length of each piece?
The greatest possible length of each piece is the HCF of the three lengths.
36 = 2² × 3²
48 = 2⁴ × 3
60 = 2² × 3 × 5
HCF = 2² × 3 = 12
Verification:
36 ÷ 12 = 3 ✔
48 ÷ 12 = 4 ✔
60 ÷ 12 = 5 ✔
Therefore, the greatest possible length of each piece is: 12 m
36 = 2² × 3²
48 = 2⁴ × 3
60 = 2² × 3 × 5
HCF = 2² × 3 = 12
Verification:
36 ÷ 12 = 3 ✔
48 ÷ 12 = 4 ✔
60 ÷ 12 = 5 ✔
Therefore, the greatest possible length of each piece is: 12 m
Question 20: Three buses leave a bus stop every 24 minutes, 36 minutes and 54 minutes, respectively. If they leave together at 8:00 AM, after how much time will they next leave together?
The required time is the LCM of 24, 36 and 54.
24 = 2³ × 3
36 = 2² × 3²
54 = 2 × 3³
LCM = 2³ × 3³ = 216 minutes
216 minutes = 3 hours 36 minutes
Therefore, next common departure time = 11:36 AM
24 = 2³ × 3
36 = 2² × 3²
54 = 2 × 3³
LCM = 2³ × 3³ = 216 minutes
216 minutes = 3 hours 36 minutes
Therefore, next common departure time = 11:36 AM
Question 21: A teacher has 84 chocolates, 126 biscuits and 210 candies. She wants to distribute them into identical gift packs without leaving any item unused. What is the greatest number of gift packs that can be made?
The required number of gift packs is the HCF of 84, 126 and 210.
84 = 2² × 3 × 7
126 = 2 × 3² × 7
210 = 2 × 3 × 5 × 7
HCF = 2 × 3 × 7 = 42
Therefore, the greatest number of gift packs is: 42
84 = 2² × 3 × 7
126 = 2 × 3² × 7
210 = 2 × 3 × 5 × 7
HCF = 2 × 3 × 7 = 42
Therefore, the greatest number of gift packs is: 42
Question 22: A shopkeeper has 96 apples, 144 oranges and 240 mangoes. He wants to pack them into identical boxes so that each box contains the same number of each fruit and no fruit is left over. What is the greatest number of boxes that can be prepared?
The required number of boxes is the HCF of 96, 144 and 240.
96 = 2⁵ × 3
144 = 2⁴ × 3²
240 = 2⁴ × 3 × 5
HCF = 2⁴ × 3 = 48
Therefore, the greatest number of boxes is: 48
Question 23: The least number which when divided by 5, 6, 7 and 8 leaves a remainder of 3, but when divided by 9 leaves no remainder, is?
Since the required number leaves a remainder of 3 when divided by 5, 6, 7 and 8,
Required number − 3 must be exactly divisible by all four numbers.
Find the LCM of 5, 6, 7 and 8.
LCM = 2³ × 3 × 5 × 7 = 840
Therefore, required number = 840k + 3
It must also be divisible by 9.
Since: 840 ≡ 3 (mod 9)
We have: 840k + 3 ≡ 3k + 3 ≡ 0 (mod 9)
⇒ 3(k + 1) is divisible by 9
⇒ k + 1 is divisible by 3
The smallest positive value is: k = 2
Therefore, required number = 840 × 2 + 3
= 1683
Verification:
1683 ÷ 5 = 336 remainder 3 ✔
1683 ÷ 6 = 280 remainder 3 ✔
1683 ÷ 7 = 240 remainder 3 ✔
1683 ÷ 8 = 210 remainder 3 ✔
1683 ÷ 9 = 187 remainder 0 ✔
Therefore, the least required number is: 1683
Required number − 3 must be exactly divisible by all four numbers.
Find the LCM of 5, 6, 7 and 8.
LCM = 2³ × 3 × 5 × 7 = 840
Therefore, required number = 840k + 3
It must also be divisible by 9.
Since: 840 ≡ 3 (mod 9)
We have: 840k + 3 ≡ 3k + 3 ≡ 0 (mod 9)
⇒ 3(k + 1) is divisible by 9
⇒ k + 1 is divisible by 3
The smallest positive value is: k = 2
Therefore, required number = 840 × 2 + 3
= 1683
Verification:
1683 ÷ 5 = 336 remainder 3 ✔
1683 ÷ 6 = 280 remainder 3 ✔
1683 ÷ 7 = 240 remainder 3 ✔
1683 ÷ 8 = 210 remainder 3 ✔
1683 ÷ 9 = 187 remainder 0 ✔
Therefore, the least required number is: 1683
Question 24: What is the greatest possible length of a measuring rod that can exactly measure 5 m 40 cm, 8 m 10 cm and 11 m 70 cm?
Convert all the lengths into centimetres.
The required length of the measuring rod is the HCF of 540, 810 and 1170.
Prime factorisation:
540 = 2² × 3³ × 5
810 = 2 × 3⁴ × 5
1170 = 2 × 3² × 5 × 13
HCF = 2 × 3² × 5 = 90 cm
Therefore, the greatest possible length of the measuring rod is: 90 cm
The required length of the measuring rod is the HCF of 540, 810 and 1170.
Prime factorisation:
540 = 2² × 3³ × 5
810 = 2 × 3⁴ × 5
1170 = 2 × 3² × 5 × 13
HCF = 2 × 3² × 5 = 90 cm
Therefore, the greatest possible length of the measuring rod is: 90 cm
Question 25: Three numbers are pairwise co-prime. The product of the first two numbers is 299, and the product of the second and third numbers is 667. What is the sum of the three numbers?
Let the three numbers be a, b and c.
Since they are pairwise co-prime,
ab = 299
bc = 667
Prime factorise the given products:
299 = 13 × 23
667 = 23 × 29
The common factor is 23, so
b = 23
Therefore,
a = 299 ÷ 23 = 13
c = 667 ÷ 23 = 29
Hence, Sum = 13 + 23 + 29 = 65
Verification:
13 × 23 = 299 ✔
23 × 29 = 667 ✔
The numbers 13, 23 and 29 are pairwise co-prime ✔
Therefore, the required sum is: 65
Since they are pairwise co-prime,
ab = 299
bc = 667
Prime factorise the given products:
299 = 13 × 23
667 = 23 × 29
The common factor is 23, so
b = 23
Therefore,
a = 299 ÷ 23 = 13
c = 667 ÷ 23 = 29
Hence, Sum = 13 + 23 + 29 = 65
Verification:
13 × 23 = 299 ✔
23 × 29 = 667 ✔
The numbers 13, 23 and 29 are pairwise co-prime ✔
Therefore, the required sum is: 65
Question 26: Find the HCF of 2/3, 8/9 and 16/81.
To find the HCF of fractions, use:
HCF = HCF of numerators ÷ LCM of denominators
For:
2/3, 8/9 and 16/81
HCF of the numerators:
HCF(2, 8, 16) = 2
LCM of the denominators:
LCM(3, 9, 81) = 81
Therefore,
HCF = 2/81
HCF = HCF of numerators ÷ LCM of denominators
For:
2/3, 8/9 and 16/81
HCF of the numerators:
HCF(2, 8, 16) = 2
LCM of the denominators:
LCM(3, 9, 81) = 81
Therefore,
HCF = 2/81
Question 27: Find the HCF of 18/35, 24/49 and 30/77.
HCF of fractions = HCF of numerators ÷ LCM of denominators
HCF of the numerators:
HCF(18, 24, 30) = 6
LCM of the denominators:
LCM(35, 49, 77)
Prime factorisation:
35 = 5 × 7
49 = 7²
77 = 7 × 11
Therefore,
LCM = 5 × 7² × 11 = 2,695
Hence, HCF = 6/2,695
HCF of the numerators:
HCF(18, 24, 30) = 6
LCM of the denominators:
LCM(35, 49, 77)
Prime factorisation:
35 = 5 × 7
49 = 7²
77 = 7 × 11
Therefore,
LCM = 5 × 7² × 11 = 2,695
Hence, HCF = 6/2,695
Question 28: Find the HCF of 2.16, 3.24 and 4.32.
Multiply each number by 100 to remove the decimal points:
2.16 × 100 = 216
3.24 × 100 = 324
4.32 × 100 = 432
Now, HCF(216, 324, 432) = 108
Since the original numbers were multiplied by 100, divide the HCF by 100:
HCF = 108 ÷ 100 = 1.08
2.16 × 100 = 216
3.24 × 100 = 324
4.32 × 100 = 432
Now, HCF(216, 324, 432) = 108
Since the original numbers were multiplied by 100, divide the HCF by 100:
HCF = 108 ÷ 100 = 1.08
Question 29: Find the HCF of 12x²y³ and 18x³y.
We have 12x²y³ and 18x³y
First, find the HCF of the numerical coefficients HCF(12, 18) = 6
For the variables, take the lowest power of each common variable:
For x:
x² and x³ → x²
For y:
y³ and y → y
Therefore, HCF = 6x²y
First, find the HCF of the numerical coefficients HCF(12, 18) = 6
For the variables, take the lowest power of each common variable:
For x:
x² and x³ → x²
For y:
y³ and y → y
Therefore, HCF = 6x²y
Question 30: Find the greatest number that will divide 148, 246 and 623 leaving remainders of 4, 6 and 11, respectively.
Subtract the respective remainders from the given numbers:
148 − 4 = 144
246 − 6 = 240
623 − 11 = 612
Therefore, the required number must be the HCF of 144, 240 and 612.
Now, HCF(144, 240) = 48
and
HCF(48, 612) = 12
Therefore, HCF(144, 240, 612) = 12
Verification
148 ÷ 12 = 12 remainder 4
246 ÷ 12 = 20 remainder 6
623 ÷ 12 = 51 remainder 11
Thus, all three conditions are satisfied.
So the required number is 12
148 − 4 = 144
246 − 6 = 240
623 − 11 = 612
Therefore, the required number must be the HCF of 144, 240 and 612.
Now, HCF(144, 240) = 48
and
HCF(48, 612) = 12
Therefore, HCF(144, 240, 612) = 12
Verification
148 ÷ 12 = 12 remainder 4
246 ÷ 12 = 20 remainder 6
623 ÷ 12 = 51 remainder 11
Thus, all three conditions are satisfied.
So the required number is 12
Question 31: Find the least number which when divided by 20, 25 and 30 leaves remainders of 14, 19 and 24, respectively.
Compute the differences between each divisor and its corresponding remainder
20 − 14 = 6
25 − 19 = 6
30 − 24 = 6
Therefore, if the required number is N, then:
N + 6 must be exactly divisible by 20, 25 and 30.
Hence, N + 6 must be the LCM of 20, 25 and 30.
Prime factorisation:
20 = 2² × 5
25 = 5²
30 = 2 × 3 × 5
Therefore,
LCM = 2² × 3 × 5² = 300
Thus,
N + 6 = 300
N = 300 − 6 = 294
Verification
294 ÷ 20 = 14 remainder 14
294 ÷ 25 = 11 remainder 19
294 ÷ 30 = 9 remainder 24
Therefore, the least required number is: 294
20 − 14 = 6
25 − 19 = 6
30 − 24 = 6
Therefore, if the required number is N, then:
N + 6 must be exactly divisible by 20, 25 and 30.
Hence, N + 6 must be the LCM of 20, 25 and 30.
Prime factorisation:
20 = 2² × 5
25 = 5²
30 = 2 × 3 × 5
Therefore,
LCM = 2² × 3 × 5² = 300
Thus,
N + 6 = 300
N = 300 − 6 = 294
Verification
294 ÷ 20 = 14 remainder 14
294 ÷ 25 = 11 remainder 19
294 ÷ 30 = 9 remainder 24
Therefore, the least required number is: 294
Question 32: A courtyard is 15 m 17 cm long and 9 m 2 cm wide. Find the maximum size of square tiles needed to pave it without cutting.
Convert both dimensions into centimetres:
15 m 17 cm = 1,517 cm
9 m 2 cm = 902 cm
The maximum side length of the square tile is the HCF of 1,517 and 902.
Using the division method
1,517 = 902 × 1 + 615
902 = 615 × 1 + 287
615 = 287 × 2 + 41
287 = 41 × 7 + 0
Therefore, HCF = 41 cm
Hence, the maximum possible side length of each square tile is 41
15 m 17 cm = 1,517 cm
9 m 2 cm = 902 cm
The maximum side length of the square tile is the HCF of 1,517 and 902.
Using the division method
1,517 = 902 × 1 + 615
902 = 615 × 1 + 287
615 = 287 × 2 + 41
287 = 41 × 7 + 0
Therefore, HCF = 41 cm
Hence, the maximum possible side length of each square tile is 41
Question 33: Two numbers are in the ratio 3 : 4. If their HCF is 4, find their LCM.
Let the two numbers be 3x and 4x
Since 3 and 4 are co-prime, their HCF is x.
Given that the HCF is 4 x = 4
Therefore, the two numbers are
3 × 4 = 12
and
4 × 4 = 16
Now using HCF × LCM = Product of the two numbers
Therefore,
4 × LCM = 12 × 16
LCM = (12 × 16) ÷ 4
LCM = 48
Since 3 and 4 are co-prime, their HCF is x.
Given that the HCF is 4 x = 4
Therefore, the two numbers are
3 × 4 = 12
and
4 × 4 = 16
Now using HCF × LCM = Product of the two numbers
Therefore,
4 × LCM = 12 × 16
LCM = (12 × 16) ÷ 4
LCM = 48
Question 34: The sum of two numbers is 216 and their HCF is 27. How many such pairs of numbers can be formed?
Since the HCF is 27, let the two numbers be 27a and 27b
Their sum is 216, so:
27a + 27b = 216
27(a + b) = 216
Therefore, a + b = 8
Also, since 27 is the HCF, a and b must be co-prime.
The positive pairs whose sum is 8 are:
(1, 7) → co-prime ✓
(2, 6) → not co-prime ✗
(3, 5) → co-prime ✓
(4, 4) → not co-prime ✗
Thus, the possible pairs of numbers are
(27, 189) and (81, 135)
Therefore, the number of such pairs is 2
Their sum is 216, so:
27a + 27b = 216
27(a + b) = 216
Therefore, a + b = 8
Also, since 27 is the HCF, a and b must be co-prime.
The positive pairs whose sum is 8 are:
(1, 7) → co-prime ✓
(2, 6) → not co-prime ✗
(3, 5) → co-prime ✓
(4, 4) → not co-prime ✗
Thus, the possible pairs of numbers are
(27, 189) and (81, 135)
Therefore, the number of such pairs is 2
Question 35: The product of two numbers is 2028 and their HCF is 13. Find the number of such pairs.
Let the two numbers be 13a and 13b
Since their HCF is 13, the numbers a and b must be co-prime.
Given:
13a × 13b = 2,028
Therefore, 169ab = 2,028
ab = 12
Now find the factor pairs of 12:
(1, 12) → co-prime ✓
(2, 6) → not co-prime ✗
(3, 4) → co-prime ✓
Therefore, the possible pairs of numbers are (13, 156) and (39, 52)
Hence, the number of such pairs is 2
Since their HCF is 13, the numbers a and b must be co-prime.
Given:
13a × 13b = 2,028
Therefore, 169ab = 2,028
ab = 12
Now find the factor pairs of 12:
(1, 12) → co-prime ✓
(2, 6) → not co-prime ✗
(3, 4) → co-prime ✓
Therefore, the possible pairs of numbers are (13, 156) and (39, 52)
Hence, the number of such pairs is 2
Question 36: Find the greatest 3-digit number exactly divisible by 8, 12 and 15.
First, find the LCM of 8, 12 and 15.
8 = 2³
12 = 2² × 3
15 = 3 × 5
Therefore, LCM = 2³ × 3 × 5 = 120
So, the required number must be a multiple of 120.
The greatest 3-digit number is 999.
Divide 999 ÷ 120 = 8 remainder 39
Therefore, the greatest 3-digit multiple of 120 is
20 × 8 = 960
8 = 2³
12 = 2² × 3
15 = 3 × 5
Therefore, LCM = 2³ × 3 × 5 = 120
So, the required number must be a multiple of 120.
The greatest 3-digit number is 999.
Divide 999 ÷ 120 = 8 remainder 39
Therefore, the greatest 3-digit multiple of 120 is
20 × 8 = 960
Question 37: Find the smallest 4-digit number which is exactly divisible by 12, 18, 24 and 32.
LCM of 12, 18, 24 and 32.
Prime factorisation:
12 = 2² × 3
18 = 2 × 3²
24 = 2³ × 3
32 = 2⁵
Therefore,
LCM = 2⁵ × 3² = 288
So, the required number must be a multiple of 288.
The smallest 4-digit number is 1,000.
Now divide:
1,000 ÷ 288 = 3 remainder 136
Therefore, the next multiple of 288 is
288 × 4 = 1,152
Prime factorisation:
12 = 2² × 3
18 = 2 × 3²
24 = 2³ × 3
32 = 2⁵
Therefore,
LCM = 2⁵ × 3² = 288
So, the required number must be a multiple of 288.
The smallest 4-digit number is 1,000.
Now divide:
1,000 ÷ 288 = 3 remainder 136
Therefore, the next multiple of 288 is
288 × 4 = 1,152
Question 38: Five bells start tolling together and then toll at intervals of 2, 4, 6, 8 and 10 seconds, respectively. In 30 minutes, how many times will they toll together?
The bells will toll together at intervals equal to the LCM of 2, 4, 6, 8 and 10.
Prime factorisation:
2 = 2
4 = 2²
6 = 2 × 3
8 = 2³
10 = 2 × 5
Therefore, LCM = 2³ × 3 × 5 = 120 seconds
So, the bells toll together every 120 seconds.
Convert 30 minutes into seconds: 30 × 60 = 1,800 seconds
Number of complete 120-second intervals in 1,800 seconds:
1,800 ÷ 120 = 15
However, the bells also toll together at the starting time, before any interval has elapsed.
Therefore, total number of times they toll together: 15 + 1 = 16
Prime factorisation:
2 = 2
4 = 2²
6 = 2 × 3
8 = 2³
10 = 2 × 5
Therefore, LCM = 2³ × 3 × 5 = 120 seconds
So, the bells toll together every 120 seconds.
Convert 30 minutes into seconds: 30 × 60 = 1,800 seconds
Number of complete 120-second intervals in 1,800 seconds:
1,800 ÷ 120 = 15
However, the bells also toll together at the starting time, before any interval has elapsed.
Therefore, total number of times they toll together: 15 + 1 = 16
Question 39: A, B and C run around a circular track 1,200 m long at speeds of 2 m/s, 3 m/s and 5 m/s, respectively. If they start together from the same point and at the same time, after how long will they all meet again at the starting point?
First, find the time each person takes to complete one full round.
For A: Time = 1,200 ÷ 2 = 600 seconds
For B: Time = 1,200 ÷ 3 = 400 seconds
For C: Time = 1,200 ÷ 5 = 240 seconds
They will all be at the starting point together again after a time equal to the LCM of 600, 400 and 240.
Prime factorisation:
600 = 2³ × 3 × 5²
400 = 2⁴ × 5²
240 = 2⁴ × 3 × 5
Therefore, LCM = 2⁴ × 3 × 5² = 1,200 seconds
Convert into minutes: 1,200 ÷ 60 = 20 minutes
Therefore, they will meet at the staring point after 20 mins
For A: Time = 1,200 ÷ 2 = 600 seconds
For B: Time = 1,200 ÷ 3 = 400 seconds
For C: Time = 1,200 ÷ 5 = 240 seconds
They will all be at the starting point together again after a time equal to the LCM of 600, 400 and 240.
Prime factorisation:
600 = 2³ × 3 × 5²
400 = 2⁴ × 5²
240 = 2⁴ × 3 × 5
Therefore, LCM = 2⁴ × 3 × 5² = 1,200 seconds
Convert into minutes: 1,200 ÷ 60 = 20 minutes
Therefore, they will meet at the staring point after 20 mins
Question 40: Find the greatest 4-digit number which when divided by 10, 15 and 20 leaves a remainder of 4 in each case.
If a number leaves a remainder of 4 when divided by 10, 15 and 20, then subtracting 4 from the number must give a number exactly divisible by all three.
Therefore, find: LCM(10, 15, 20)
Prime factorisation:
10 = 2 × 5
15 = 3 × 5
20 = 2² × 5
Therefore, LCM = 2² × 3 × 5 = 60
So, the required number is of the form: 60k + 4
The greatest 4-digit number is 9,999.
Subtract the remainder 9,999 − 4 = 9,995
The greatest multiple of 60 not exceeding 9,995 is
60 × 166 = 9,960
Therefore, 9,960 + 4 = 9,964
Verification
9,964 ÷ 10 = 996 remainder 4
9,964 ÷ 15 = 664 remainder 4
9,964 ÷ 20 = 498 remainder 4
Therefore, find: LCM(10, 15, 20)
Prime factorisation:
10 = 2 × 5
15 = 3 × 5
20 = 2² × 5
Therefore, LCM = 2² × 3 × 5 = 60
So, the required number is of the form: 60k + 4
The greatest 4-digit number is 9,999.
Subtract the remainder 9,999 − 4 = 9,995
The greatest multiple of 60 not exceeding 9,995 is
60 × 166 = 9,960
Therefore, 9,960 + 4 = 9,964
Verification
9,964 ÷ 10 = 996 remainder 4
9,964 ÷ 15 = 664 remainder 4
9,964 ÷ 20 = 498 remainder 4
Question 41: Find the least number which, when divided by 3, 5, 6, 8, 10 and 12, leaves a remainder of 2 in each case, but is exactly divisible by 13.
If the required number leaves a remainder of 2 when divided by all the given numbers, then subtracting 2 from it must give a number exactly divisible by all of them.
First, find the LCM of: 3, 5, 6, 8, 10 and 12
Prime factorisation:
3 = 3
5 = 5
6 = 2 × 3
8 = 2³
10 = 2 × 5
12 = 2² × 3
Therefore, LCM = 2³ × 3 × 5 = 120
So, the required number is of the form is 120k + 2
It must also be divisible by 13.
We need the smallest value of k for which (120k + 2) is divisible by 13.
Since 120 = 13 × 9 + 3
we can write 120k + 2 = 13 × 9k + 3k + 2
Therefore, 3k + 2 must be divisible by 13.
Try successive values of k:
k = 1 → 3 + 2 = 5
k = 2 → 6 + 2 = 8
k = 3 → 9 + 2 = 11
k = 4 → 12 + 2 = 14
…
k = 8 → 24 + 2 = 26
Since 26 is divisible by 13, the smallest suitable value is k = 8
Therefore,
N = 120 × 8 + 2 = 962
Verification:
962 ÷ 3 = 320 remainder 2
962 ÷ 5 = 192 remainder 2
962 ÷ 6 = 160 remainder 2
962 ÷ 8 = 120 remainder 2
962 ÷ 10 = 96 remainder 2
962 ÷ 12 = 80 remainder 2
and
962 ÷ 13 = 74
Therefore, the number is 962
First, find the LCM of: 3, 5, 6, 8, 10 and 12
Prime factorisation:
3 = 3
5 = 5
6 = 2 × 3
8 = 2³
10 = 2 × 5
12 = 2² × 3
Therefore, LCM = 2³ × 3 × 5 = 120
So, the required number is of the form is 120k + 2
It must also be divisible by 13.
We need the smallest value of k for which (120k + 2) is divisible by 13.
Since 120 = 13 × 9 + 3
we can write 120k + 2 = 13 × 9k + 3k + 2
Therefore, 3k + 2 must be divisible by 13.
Try successive values of k:
k = 1 → 3 + 2 = 5
k = 2 → 6 + 2 = 8
k = 3 → 9 + 2 = 11
k = 4 → 12 + 2 = 14
…
k = 8 → 24 + 2 = 26
Since 26 is divisible by 13, the smallest suitable value is k = 8
Therefore,
N = 120 × 8 + 2 = 962
Verification:
962 ÷ 3 = 320 remainder 2
962 ÷ 5 = 192 remainder 2
962 ÷ 6 = 160 remainder 2
962 ÷ 8 = 120 remainder 2
962 ÷ 10 = 96 remainder 2
962 ÷ 12 = 80 remainder 2
and
962 ÷ 13 = 74
Therefore, the number is 962
Question 42: Find the least perfect square number that is exactly divisible by 21, 36 and 66.
First, find the LCM of 21, 36 and 66.
Prime factorisation:
21 = 3 × 7
36 = 2² × 3²
66 = 2 × 3 × 11
Therefore, LCM = 2² × 3² × 7 × 11 = 2,772
Now, 2,772 = 2² × 3² × 7 × 11
For a number to be a perfect square, all prime factors must have even powers.
The factors 7 and 11 each occur to an odd power. Therefore, multiply by 7 × 11 = 77
Thus, the least perfect square divisible by all three numbers is: 2,772 × 77 = 213,444
Also 213,444 = 462²
Therefore, the least perfect square number is 213,444
Prime factorisation:
21 = 3 × 7
36 = 2² × 3²
66 = 2 × 3 × 11
Therefore, LCM = 2² × 3² × 7 × 11 = 2,772
Now, 2,772 = 2² × 3² × 7 × 11
For a number to be a perfect square, all prime factors must have even powers.
The factors 7 and 11 each occur to an odd power. Therefore, multiply by 7 × 11 = 77
Thus, the least perfect square divisible by all three numbers is: 2,772 × 77 = 213,444
Also 213,444 = 462²
Therefore, the least perfect square number is 213,444
Question 43: Find the largest fraction that can divide 7/3 and 14/5 completely without leaving a fractional remainder.
For fractions,
HCF = HCF of numerators ÷ LCM of denominators
Therefore, HCF(7, 14) = 7 and LCM(3, 5) = 15
Hence, HCF = 7/15
Therefore, the greatest fraction that divides both 7/3 and 14/5 exactly is 7/15
Verification
(7/3) ÷ (7/15) = 5
(14/5) ÷ (7/15) = 6
Both quotients are whole numbers, so 7/15 divides both fractions exactly.
HCF = HCF of numerators ÷ LCM of denominators
Therefore, HCF(7, 14) = 7 and LCM(3, 5) = 15
Hence, HCF = 7/15
Therefore, the greatest fraction that divides both 7/3 and 14/5 exactly is 7/15
Verification
(7/3) ÷ (7/15) = 5
(14/5) ÷ (7/15) = 6
Both quotients are whole numbers, so 7/15 divides both fractions exactly.
Question 44: A and B start running from the same point on a circular track 600 m long at the same time, but in opposite directions. Their speeds are 3 m/s and 2 m/s, respectively. After how many minutes will they both meet again at the starting point?
The time each person takes to complete one full round.
For A: Time = 600 ÷ 3 = 200 seconds
For B: Time = 600 ÷ 2 = 300 seconds
They will both be at the starting point together when they have each completed a whole number of rounds. Therefore, we find the LCM of 200 and 300.
200 = 2³ × 5²
300 = 2² × 3 × 5²
Therefore, LCM = 2³ × 3 × 5² = 600 seconds
Convert into minutes = 600 ÷ 60 = 10 minutes
Therefore, they meet again at starting point after 10 minutes
For A: Time = 600 ÷ 3 = 200 seconds
For B: Time = 600 ÷ 2 = 300 seconds
They will both be at the starting point together when they have each completed a whole number of rounds. Therefore, we find the LCM of 200 and 300.
200 = 2³ × 5²
300 = 2² × 3 × 5²
Therefore, LCM = 2³ × 3 × 5² = 600 seconds
Convert into minutes = 600 ÷ 60 = 10 minutes
Therefore, they meet again at starting point after 10 minutes
Question 45: Find the Highest Common Factor (HCF) of (2¹² − 1) and (2¹⁸ − 1).
We use the algebraic property: HCF of (aⁿ − 1) and (aᵐ − 1) is equal to aᴴᶜᶠ⁽ⁿ⁺ᵐ⁾ − 1
Here, the base a = 2, exponent n = 12, and exponent m = 18.
Find the HCF of the exponents 12 and 18:
12 = 2² × 3
18 = 2 × 3²
HCF of 12 and 18 = 2 × 3 = 6
Substitute this back into the property:
Required HCF = 2⁶ − 1
Calculate the final value:
2⁶ − 1 = 64 − 1 = 63
Verification:
2¹² − 1 = 4095, which is exactly 63 × 65 ✔
2¹⁸ − 1 = 262,143, which is exactly 63 × 4161 ✔
Therefore, the required HCF is: 63
Question 46: A solid wooden block measures 12 cm × 15 cm × 18 cm. It is to be cut completely into identical cubes without any wastage. What is the minimum number of cubes that can be made?
The side of the largest possible cube must divide all three dimensions exactly.
Therefore, find HCF(12, 15, 18)
12 = 2² × 3
15 = 3 × 5
18 = 2 × 3²
Therefore, HCF = 3
So, each cube must have a side of 3 cm.
The number of cubes along each dimension is:
12 ÷ 3 = 4
15 ÷ 3 = 5
18 ÷ 3 = 6
Therefore, the total number of cubes is 4 × 5 × 6 = 120
Verification:
Cutting along the dimensions yields 4 segments, 5 segments, and 6 segments respectively, making 4 × 5 × 6 = 120 perfect cubes. ✔
Therefore, the minimum number of cubes is: 120
Therefore, find HCF(12, 15, 18)
12 = 2² × 3
15 = 3 × 5
18 = 2 × 3²
Therefore, HCF = 3
So, each cube must have a side of 3 cm.
The number of cubes along each dimension is:
12 ÷ 3 = 4
15 ÷ 3 = 5
18 ÷ 3 = 6
Therefore, the total number of cubes is 4 × 5 × 6 = 120
Verification:
Cutting along the dimensions yields 4 segments, 5 segments, and 6 segments respectively, making 4 × 5 × 6 = 120 perfect cubes. ✔
Therefore, the minimum number of cubes is: 120
Question 47: How many natural numbers between 200 and 600 are exactly divisible by both 12 and 15?
A number divisible by both 12 and 15 must be divisible by their LCM.
12 = 2² × 3
15 = 3 × 5
Therefore, LCM = 2² × 3 × 5 = 60
So, we need to count the multiples of 60 between 200 and 600.
The first multiple greater than 200 is
60 × 4 = 240
The last multiple less than 600 is 60 × 9 = 540
Therefore, the multiples are 240, 300, 360, 420, 480, 540
Number of multiples = 9 − 4 + 1 = 6
12 = 2² × 3
15 = 3 × 5
Therefore, LCM = 2² × 3 × 5 = 60
So, we need to count the multiples of 60 between 200 and 600.
The first multiple greater than 200 is
60 × 4 = 240
The last multiple less than 600 is 60 × 9 = 540
Therefore, the multiples are 240, 300, 360, 420, 480, 540
Number of multiples = 9 − 4 + 1 = 6
Question 48: Three electronic signals change color every 48 seconds, 72 seconds and 108 seconds respectively. If they all change simultaneously at 9:00 AM, at what time will they next change together?
They will all change together again after a time equal to the LCM of 48, 72 and 108 seconds.
Prime factorisation:
48 = 2⁴ × 3
72 = 2³ × 3²
108 = 2² × 3³
Therefore, LCM = 2⁴ × 3³ = 432 seconds
Convert 432 seconds into minutes and seconds = 7 minutes 12 seconds
Starting from 9:00 AM:
9:00 AM + 7 minutes 12 seconds = 9:07:12 AM
Prime factorisation:
48 = 2⁴ × 3
72 = 2³ × 3²
108 = 2² × 3³
Therefore, LCM = 2⁴ × 3³ = 432 seconds
Convert 432 seconds into minutes and seconds = 7 minutes 12 seconds
Starting from 9:00 AM:
9:00 AM + 7 minutes 12 seconds = 9:07:12 AM
Question 49: Find the smallest positive number that leaves a remainder of 3 when divided by 4 and a remainder of 1 when divided by 6.
Numbers that leave a remainder of 3 when divided by 4 are:
3, 7, 11, 15, 19, 23, …
Now check which of these leaves a remainder of 1 when divided by 6.
7 ÷ 6 = 1 remainder 1
Therefore, 7 satisfies both conditions.
Verification:
7 ÷ 4 = 1 remainder 3
7 ÷ 6 = 1 remainder 1
Hence, the smallest positive number is 7
3, 7, 11, 15, 19, 23, …
Now check which of these leaves a remainder of 1 when divided by 6.
7 ÷ 6 = 1 remainder 1
Therefore, 7 satisfies both conditions.
Verification:
7 ÷ 4 = 1 remainder 3
7 ÷ 6 = 1 remainder 1
Hence, the smallest positive number is 7
Question 50: The product of the HCF and LCM of two numbers is 924. If the difference between the two numbers is 10, find the two numbers.
Let the two numbers be x and y.
Property: Product of two numbers = HCF × LCM
Therefore, x × y = 924
We are also given their difference:
x − y = 10 → x = y + 10
Substitute this into the product equation:
(y + 10) × y = 924
y² + 10y − 924 = 0
Find two numbers that multiply to −924 and add up to 10. These numbers are 32 and −22:(y + 32)(y − 22) = 0
Since y must be a positive natural number, y = 22.
Find x:x = 22 + 10 = 32
Verification:
Difference: 32 − 22 = 10 ✔
Product (HCF×LCM): 32 × 22 = 924 ✔
Therefore, the two required numbers are 22 and 32
Property: Product of two numbers = HCF × LCM
Therefore, x × y = 924
We are also given their difference:
x − y = 10 → x = y + 10
Substitute this into the product equation:
(y + 10) × y = 924
y² + 10y − 924 = 0
Find two numbers that multiply to −924 and add up to 10. These numbers are 32 and −22:(y + 32)(y − 22) = 0
Since y must be a positive natural number, y = 22.
Find x:x = 22 + 10 = 32
Verification:
Difference: 32 − 22 = 10 ✔
Product (HCF×LCM): 32 × 22 = 924 ✔
Therefore, the two required numbers are 22 and 32
Question 51: A shopkeeper has 96 biscuits, 144 chocolates and 192 candies. He wants to make the greatest possible number of identical gift packs using all the items. Each pack must contain the same number of biscuits, chocolates and candies. How many gift packs can he make?
The greatest number of identical packs is HCF(96, 144, 192)
96 = 2⁵ × 3
144 = 2⁴ × 3²
192 = 2⁶ × 3
Therefore, HCF = 2⁴ × 3 = 48
So, the greatest number of gift packs is 48
Each pack contains:
96 ÷ 48 = 2 biscuits
144 ÷ 48 = 3 chocolates
192 ÷ 48 = 4 candies
96 = 2⁵ × 3
144 = 2⁴ × 3²
192 = 2⁶ × 3
Therefore, HCF = 2⁴ × 3 = 48
So, the greatest number of gift packs is 48
Each pack contains:
96 ÷ 48 = 2 biscuits
144 ÷ 48 = 3 chocolates
192 ÷ 48 = 4 candies
Question 52: The HCF of two numbers A and B is 12, and their LCM is 72. If A is the smaller number and A > 12, what is the value of A?
HCF × LCM = Product of the two numbers
Therefore, 12 × 72 = A × B
A × B = 864
Since the HCF is 12, both numbers must be multiples of 12.
The factor pairs of 864 in which both numbers are multiples of 12 are:
24 × 36 = 864
Also, HCF(24, 36) = 12 and LCM(24, 36) = 72
Since A is the smaller number A = 24
Therefore, 12 × 72 = A × B
A × B = 864
Since the HCF is 12, both numbers must be multiples of 12.
The factor pairs of 864 in which both numbers are multiples of 12 are:
24 × 36 = 864
Also, HCF(24, 36) = 12 and LCM(24, 36) = 72
Since A is the smaller number A = 24
Question 53: Three numbers are in the ratio 2 : 3 : 5. If their LCM is 900, what is their Highest Common Factor (HCF)?
Let the three numbers be 2x, 3x and 5x
Since 2, 3 and 5 are pairwise co-prime, their LCM is:
LCM = 2 × 3 × 5 × x = 30x
Given that their LCM is 900:
30x = 900
x = 30
Therefore, the three numbers are 60, 90 and 150
So HCF of the number is HCF(60, 90, 150) = 30
Since 2, 3 and 5 are pairwise co-prime, their LCM is:
LCM = 2 × 3 × 5 × x = 30x
Given that their LCM is 900:
30x = 900
x = 30
Therefore, the three numbers are 60, 90 and 150
So HCF of the number is HCF(60, 90, 150) = 30
Question 54: Find the smallest natural number n such that the HCF of n and 240 is 15, and the HCF of n and 560 is 35.
Since HCF(n, 240) = 15
So, n must be divisible by 15.
Similarly, since HCF(n, 560) = 35
n must be divisible by 35.
Therefore, n must be divisible by both 15 and 35.
Find their LCM:
15 = 3 × 5
35 = 5 × 7
Therefore, LCM(15, 35) = 3 × 5 × 7 = 105
So, the smallest possible value of n is 105.
Now verify:
HCF(105, 240) = 15
HCF(105, 560) = 35
Both conditions are satisfied.
Therefore, the smallest required number is: 105
So, n must be divisible by 15.
Similarly, since HCF(n, 560) = 35
n must be divisible by 35.
Therefore, n must be divisible by both 15 and 35.
Find their LCM:
15 = 3 × 5
35 = 5 × 7
Therefore, LCM(15, 35) = 3 × 5 × 7 = 105
So, the smallest possible value of n is 105.
Now verify:
HCF(105, 240) = 15
HCF(105, 560) = 35
Both conditions are satisfied.
Therefore, the smallest required number is: 105
Question 55: The sum of two positive integers x and y is 1,050. What is the maximum possible value of their HCF?
Let the HCF of x and y be d.
Since d is a common factor of both numbers, it must also divide their sum.
Therefore, d must be a divisor of 1,050.
To maximize the HCF, the two numbers should be equal:
x = y = 1,050 ÷ 2 = 525
Therefore, HCF(525, 525) = 525
So, maximum possible value of their HCF is 525
Since d is a common factor of both numbers, it must also divide their sum.
Therefore, d must be a divisor of 1,050.
To maximize the HCF, the two numbers should be equal:
x = y = 1,050 ÷ 2 = 525
Therefore, HCF(525, 525) = 525
So, maximum possible value of their HCF is 525
Question 56: Find the smallest positive number that leaves remainders of 4, 5, 6, 7 and 8 when divided by 5, 6, 7, 8 and 9, respectively.
The remainder in each case is 1 less than the divisor.
Therefore, if we add 1 to the required number, the result must be exactly divisible by:
5, 6, 7, 8 and 9
Find their LCM:
5 = 5
6 = 2 × 3
7 = 7
8 = 2³
9 = 3²
Therefore, LCM = 2³ × 3² × 5 × 7
= 2,520
So, the required number is 2,520 − 1 = 2,519
Verification
2,519 ÷ 5 = 503 remainder 4
2,519 ÷ 6 = 419 remainder 5
2,519 ÷ 7 = 359 remainder 6
2,519 ÷ 8 = 314 remainder 7
2,519 ÷ 9 = 279 remainder 8
Therefore, if we add 1 to the required number, the result must be exactly divisible by:
5, 6, 7, 8 and 9
Find their LCM:
5 = 5
6 = 2 × 3
7 = 7
8 = 2³
9 = 3²
Therefore, LCM = 2³ × 3² × 5 × 7
= 2,520
So, the required number is 2,520 − 1 = 2,519
Verification
2,519 ÷ 5 = 503 remainder 4
2,519 ÷ 6 = 419 remainder 5
2,519 ÷ 7 = 359 remainder 6
2,519 ÷ 8 = 314 remainder 7
2,519 ÷ 9 = 279 remainder 8
Question 57: In a large school auditorium, students are seated in rows. If 18 students are placed in each row, 14 students are left over. If 27 students are placed in each row, 23 are left over; if 32 students are placed in each row, 28 are left over; and if 40 students are placed in each row, 36 are left over. What is the minimum possible number of students in the auditorium?
In every case, the remainder is 4 less than the divisor:
18 − 14 = 4
27 − 23 = 4
32 − 28 = 4
40 − 36 = 4
Therefore, if we add 4 to the number of students, the resulting number must be exactly divisible by 18, 27, 32 and 40.
Find their LCM:
18 = 2 × 3²
27 = 3³
32 = 2⁵
40 = 2³ × 5
Therefore, LCM = 2⁵ × 3³ × 5 = 4,320
Hence, number of students + 4 = 4,320
Therefore, number of students = 4,320 − 4 = 4,316
Verification
4,316 ÷ 18 = 239 remainder 14
4,316 ÷ 27 = 159 remainder 23
4,316 ÷ 32 = 134 remainder 28
4,316 ÷ 40 = 107 remainder 36
18 − 14 = 4
27 − 23 = 4
32 − 28 = 4
40 − 36 = 4
Therefore, if we add 4 to the number of students, the resulting number must be exactly divisible by 18, 27, 32 and 40.
Find their LCM:
18 = 2 × 3²
27 = 3³
32 = 2⁵
40 = 2³ × 5
Therefore, LCM = 2⁵ × 3³ × 5 = 4,320
Hence, number of students + 4 = 4,320
Therefore, number of students = 4,320 − 4 = 4,316
Verification
4,316 ÷ 18 = 239 remainder 14
4,316 ÷ 27 = 159 remainder 23
4,316 ÷ 32 = 134 remainder 28
4,316 ÷ 40 = 107 remainder 36
Question 58: The LCM of two natural numbers p and q, where p > q, is 1,155. What is the maximum possible sum of the digits of q?
Prime factorise 1,155: 1,155 = 3 × 5 × 7 × 11
Since q < p and their LCM is 1,155, q must be a proper divisor of 1,155.
The relevant divisors of 1,155 are:
1, 3, 5, 7, 11, 15, 21, 33, 35, 55, 77, 105, 165, 231, 385
Their digit sums include:
231: 2 + 3 + 1 = 6
385: 3 + 8 + 5 = 16
The greatest digit sum is therefore obtained when q = 385
This is possible by taking p = 1,155 because
LCM(385, 1,155) = 1,155
Therefore, maximum possible sum of the digits of q
= 3 + 8 + 5
= 16
Since q < p and their LCM is 1,155, q must be a proper divisor of 1,155.
The relevant divisors of 1,155 are:
1, 3, 5, 7, 11, 15, 21, 33, 35, 55, 77, 105, 165, 231, 385
Their digit sums include:
231: 2 + 3 + 1 = 6
385: 3 + 8 + 5 = 16
The greatest digit sum is therefore obtained when q = 385
This is possible by taking p = 1,155 because
LCM(385, 1,155) = 1,155
Therefore, maximum possible sum of the digits of q
= 3 + 8 + 5
= 16
Question 59: Four logs of wood have lengths 5 ¼ m, 1 13⁄15 m, 3 ½ m and 4 9⁄10 m. They are to be cut into pieces of equal maximum length. Each piece is then given to a team of 2 carpenters. How many carpenters are needed altogether?
First, convert the lengths into improper fractions:
5 ¼ = 21⁄4 m
1 13⁄15 = 28⁄15 m
3 ½ = 7⁄2 m
4 9⁄10 = 49⁄10 m
The greatest possible length of each piece is the HCF of these fractions.
For fractions HCF = HCF of numerators ÷ LCM of denominators
Therefore, HCF(21, 28, 7, 49) = 7 and LCM(4, 15, 2, 10) = 60
Hence, HCF = 7⁄60 m
So, each piece will be 7⁄60 m long.
Now find the number of pieces obtained from each log:
For the first log: 21⁄4 ÷ 7⁄60 = 45 pieces
For the second: 28⁄15 ÷ 7⁄60 = 16 pieces
For the third: 7⁄2 ÷ 7⁄60 = 30 pieces
For the fourth: 49⁄10 ÷ 7⁄60 = 42 pieces
Total pieces: 45 + 16 + 30 + 42 = 133
Each piece is assigned to 2 carpenters.
Therefore, 133 × 2 = 266 carpenters
5 ¼ = 21⁄4 m
1 13⁄15 = 28⁄15 m
3 ½ = 7⁄2 m
4 9⁄10 = 49⁄10 m
The greatest possible length of each piece is the HCF of these fractions.
For fractions HCF = HCF of numerators ÷ LCM of denominators
Therefore, HCF(21, 28, 7, 49) = 7 and LCM(4, 15, 2, 10) = 60
Hence, HCF = 7⁄60 m
So, each piece will be 7⁄60 m long.
Now find the number of pieces obtained from each log:
For the first log: 21⁄4 ÷ 7⁄60 = 45 pieces
For the second: 28⁄15 ÷ 7⁄60 = 16 pieces
For the third: 7⁄2 ÷ 7⁄60 = 30 pieces
For the fourth: 49⁄10 ÷ 7⁄60 = 42 pieces
Total pieces: 45 + 16 + 30 + 42 = 133
Each piece is assigned to 2 carpenters.
Therefore, 133 × 2 = 266 carpenters
Question 60: Three storage tanks contain 2,520 L, 3,780 L and 5,040 L of water. The water from each tank is to be transferred into identical containers of the largest possible capacity, with no water left over. The containers are then to be distributed equally among 12 distribution centres, with every centre receiving the same number of containers from each tank. What is the largest possible capacity of each container, and how many containers will each distribution centre receive in total?
The container capacity must divide all three quantities exactly.
Therefore, first find HCF(2,520, 3,780, 5,040)
Prime factorisation:
2,520 = 2³ × 3² × 5 × 7
3,780 = 2² × 3³ × 5 × 7
5,040 = 2⁴ × 3² × 5 × 7
Therefore,
HCF = 2² × 3² × 5 × 7
= 1,260 L
However, using containers of 1,260 L gives:
2,520 ÷ 1,260 = 2
3,780 ÷ 1,260 = 3
5,040 ÷ 1,260 = 4
Total = 2 + 3 + 4 = 9 containers
Since 9 containers cannot be distributed equally among 12 centres, 1,260 L is not suitable.
The container capacity must therefore be a divisor of 1,260 such that the total number of containers is divisible by 12.
Total water = 2,520 + 3,780 + 5,040 = 11,340 L
For 12 centres to receive an equal number of containers:
11,340 ÷ container capacity
must be divisible by 12.
The largest suitable capacity is = 315 L
Now check:
2,520 ÷ 315 = 8 containers
3,780 ÷ 315 = 12 containers
5,040 ÷ 315 = 16 containers
Total containers 8 + 12 + 16 = 36
Therefore, each of the 12 centres receives 36 ÷ 12 = 3 containers
Therefore, first find HCF(2,520, 3,780, 5,040)
Prime factorisation:
2,520 = 2³ × 3² × 5 × 7
3,780 = 2² × 3³ × 5 × 7
5,040 = 2⁴ × 3² × 5 × 7
Therefore,
HCF = 2² × 3² × 5 × 7
= 1,260 L
However, using containers of 1,260 L gives:
2,520 ÷ 1,260 = 2
3,780 ÷ 1,260 = 3
5,040 ÷ 1,260 = 4
Total = 2 + 3 + 4 = 9 containers
Since 9 containers cannot be distributed equally among 12 centres, 1,260 L is not suitable.
The container capacity must therefore be a divisor of 1,260 such that the total number of containers is divisible by 12.
Total water = 2,520 + 3,780 + 5,040 = 11,340 L
For 12 centres to receive an equal number of containers:
11,340 ÷ container capacity
must be divisible by 12.
The largest suitable capacity is = 315 L
Now check:
2,520 ÷ 315 = 8 containers
3,780 ÷ 315 = 12 containers
5,040 ÷ 315 = 16 containers
Total containers 8 + 12 + 16 = 36
Therefore, each of the 12 centres receives 36 ÷ 12 = 3 containers
What are HCF and LCM?
Highest Common Factor (HCF), also known as the Greatest Common Divisor (GCD), is the largest number that divides two or more numbers exactly without leaving any remainder. It represents the greatest common factor shared by the given numbers.
Example: The HCF of 12 and 18 is 6, because 6 is the largest number that divides both numbers exactly.
Example: The HCF of 12 and 18 is 6, because 6 is the largest number that divides both numbers exactly.
Lowest Common Multiple (LCM) is the smallest positive number that is exactly divisible by two or more given numbers. It is the first common multiple shared by all the numbers.
Example: The LCM of 12 and 18 is 36, because 36 is the smallest number that is divisible by both 12 and 18.
HCF is commonly used to divide quantities into the largest possible equal groups, whereas LCM is used to determine when two or more repeating events occur together or to find a common denominator while working with fractions.
Summary of Divisibility Rules
| Concept | HCF | LCM |
|---|---|---|
| Full Form | Highest Common Factor | Lowest Common Multiple |
| Meaning | Largest number that divides all given numbers exactly. | Smallest number exactly divisible by all given numbers. |
| Also Known As | Greatest Common Divisor (GCD) | Least Common Multiple |
| Methods | Prime Factorisation or Division Method | Prime Factorisation or Division Method |
| Prime Factors Used | Common prime factors with the smallest powers. | All prime factors with the highest powers. |
| Applications | Equal grouping, largest possible size, simplifying ratios. | Repeating events, common denominators, finding the least common multiple. |
| Relation | For two numbers: HCF × LCM = Product of the two numbers | |
Common Mistakes
- Confusing HCF with LCM. Remember that HCF is the greatest common factor, whereas LCM is the smallest common multiple.
- Forgetting to convert units. Convert all quantities to the same unit (such as centimetres, metres, seconds or minutes) before finding the HCF or LCM.
- Using HCF when LCM is required. Problems involving repeated events usually require the LCM, while equal grouping or the greatest possible size usually require the HCF.
- Ignoring the phrase “same remainder”. When the same remainder is left after division, first subtract the common remainder before applying the HCF or LCM as required.
- Making errors in prime factorisation. An incorrect prime factorisation leads to an incorrect HCF or LCM.
- Missing a given number. When finding the HCF or LCM of three or more numbers, make sure every number is included in the calculation.
- Applying the formula HCF × LCM = Product of the numbers incorrectly. This formula is valid only for two positive integers.
Practice Questions
Question 1: Find the greatest number that divides 455, 527 and 671, leaving the same remainder in each case.
Question 2: Find the least number which when divided by 8, 12 and 15 leaves a remainder of 5 in each case.
Question 3: What is the greatest possible length of a measuring rod that can exactly measure 6 m 30 cm, 8 m 40 cm and 10 m 50 cm?
Question 4: Three numbers are pairwise co-prime. The product of the first two numbers is 437, and the product of the last two numbers is 713. What is the sum of the three numbers?
Question 5: Three athletes complete one round of a circular track in 45 seconds, 60 seconds and 72 seconds, respectively. If they start together from the same point, after how much time will they again meet at the starting point?
Question 6: Find the HCF of 0.72, 1.08 and 1.80.
Question 7: Find the LCM of 3/4, 5/6 and 7/8.
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