Divisibility Rules
Key Concepts
Solved Examples: 36
Practice Questions: 3
Solved Examples
Question 1: Without performing long division, determine all the single-digit numbers (2–9) by which 18,720 is divisible.
Divisibility by 2: The last digit is 0, so the number is divisible by 2. ✓
Divisibility by 3: The sum of the digits is 1 + 8 + 7 + 2 + 0 = 18. Since 18 is divisible by 3, the number is divisible by 3. ✓
Divisibility by 4: The last two digits are 20. Since 20 is divisible by 4, the number is divisible by 4. ✓
Divisibility by 5: The last digit is 0, so the number is divisible by 5. ✓
Divisibility by 6: Since the number is divisible by both 2 and 3, it is divisible by 6. ✓
Divisibility by 7: Apply the divisibility rule for 7:
1872 − (2 × 0) = 1872
187 − (2 × 2) = 183
18 − (2 × 3) = 12
Since 12 is not divisible by 7, 18,720 is not divisible by 7. ✗
Divisibility by 8: The last three digits are 720. Since 720 ÷ 8 = 90, the number is divisible by 8. ✓
Divisibility by 9: The sum of the digits is 18. Since 18 is divisible by 9, the number is divisible by 9. ✓
Question 2: Find the value(s) of k such that 47k182 is divisible by 9.
For divisibility by 9, the sum of the digits must be divisible by 9.
Sum of digits
= 4 + 7 + k + 1 + 8 + 2
= 22 + k
Now test possible values of k.
Only when k = 5,
22 + 5 = 27, which is divisible by 9.
Answer: k = 5
Sum of digits
= 4 + 7 + k + 1 + 8 + 2
= 22 + k
Now test possible values of k.
Only when k = 5,
22 + 5 = 27, which is divisible by 9.
Answer: k = 5
Question 3: Find the value(s) of k such that 63k48 is divisible by 4.
For divisibility by 4, only the last two digits need to be checked.
The last two digits are 48.
Since 48 is divisible by 4, the value of k has no effect.
Therefore, every digit from 0 to 9 satisfies the condition.
Answer: k = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
The last two digits are 48.
Since 48 is divisible by 4, the value of k has no effect.
Therefore, every digit from 0 to 9 satisfies the condition.
Answer: k = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Question 4: Find the value(s) of k such that 58k324 is divisible by 12.
A number is divisible by 12 if it is divisible by 3 and 4.
Divisibility by 4: The last two digits are 24, which is divisible by 4.
Divisibility by 3: The sum of the digits is:
5 + 8 + k + 3 + 2 + 4 = 22 + k
For divisibility by 3,
22 + k must be divisible by 3.
Only k = 2, 5, 8 satisfy this condition.
Answer: k = 2, 5, 8
Divisibility by 4: The last two digits are 24, which is divisible by 4.
Divisibility by 3: The sum of the digits is:
5 + 8 + k + 3 + 2 + 4 = 22 + k
For divisibility by 3,
22 + k must be divisible by 3.
Only k = 2, 5, 8 satisfy this condition.
Answer: k = 2, 5, 8
Question 5: Which is the largest 6-digit number divisible by 81?
The largest 6-digit number is 999,999.
Divide it by 81:
999,999 = 81 × 12,345 + 54
The remainder is 54, so subtract it from 999,999:
999,999 − 54 = 999,945
Since 999,945 = 81 × 12,345, it is exactly divisible by 81.
Answer: 999,945
Divide it by 81:
999,999 = 81 × 12,345 + 54
The remainder is 54, so subtract it from 999,999:
999,999 − 54 = 999,945
Since 999,945 = 81 × 12,345, it is exactly divisible by 81.
Answer: 999,945
Question 6: Find the value(s) of k such that k53206k is divisible by 6.
Divisibility by 2: The last digit is k, so k must be an even digit. Hence, k = 0, 2, 4, 6, 8.
Divisibility by 3: The sum of the digits is: k + 5 + 3 + 2 + 0 + 6 + k = 16 + 2k which shall be divisible by 3
Testing the even values:
k = 0 → 16 ✗
k = 2 → 20 ✗
k = 4 → 24 ✓
k = 6 → 28 ✗
k = 8 → 32 ✗
Answer: k = 4
Divisibility by 3: The sum of the digits is: k + 5 + 3 + 2 + 0 + 6 + k = 16 + 2k which shall be divisible by 3
Testing the even values:
k = 0 → 16 ✗
k = 2 → 20 ✗
k = 4 → 24 ✓
k = 6 → 28 ✗
k = 8 → 32 ✗
Answer: k = 4
Question 7: The six-digit number 439xy5 is divisible by 125, where x and y are digits. How many different six-digit numbers can be formed?
A number is divisible by 125 if its last three digits are divisible by 125.
The last three digits are: xy5
Since the number ends in 5, the only multiples of 125 ending in 5 are: 125, 375, 625, 875
Therefore, the possible values of xy are: 12, 37, 62, 87
Hence, the possible six-digit numbers are:
439125, 439375, 439625, 439875
Therefore, the number of such six-digit numbers is 4.
The last three digits are: xy5
Since the number ends in 5, the only multiples of 125 ending in 5 are: 125, 375, 625, 875
Therefore, the possible values of xy are: 12, 37, 62, 87
Hence, the possible six-digit numbers are:
439125, 439375, 439625, 439875
Therefore, the number of such six-digit numbers is 4.
Question 8: For what value of Y is the seven-digit number 46393Y8 divisible by 11?
For divisibility by 11, the difference between the sum of the digits in alternate positions must be 0 or a multiple of 11.
Sum of digits in odd positions: = 4 + 3 + 3 + 8 = 18
Sum of digits in even positions:
= 6 + 9 + Y
= 15 + Y
Now, because: The maximum possible difference here is between 18 and 24 (since Y is at most 9), the difference can only lie between −6 and 3.
It can never be ±11.
Therefore, the only possible value is 0.
Therefore, 18 − (15 + Y) = 0
3 − Y = 0
∴ Y = 3
Sum of digits in odd positions: = 4 + 3 + 3 + 8 = 18
Sum of digits in even positions:
= 6 + 9 + Y
= 15 + Y
Now, because: The maximum possible difference here is between 18 and 24 (since Y is at most 9), the difference can only lie between −6 and 3.
It can never be ±11.
Therefore, the only possible value is 0.
Therefore, 18 − (15 + Y) = 0
3 − Y = 0
∴ Y = 3
Question 9: How many three-digit numbers are divisible by 6 but not by 9?
Three-digit numbers range from 100 to 999.
The three-digit multiples of 6 begin with 102 and end with 996.
Number of multiples of 6:
(996 − 102) ÷ 6 + 1 = 150
Every number divisible by both 6 and 9 must be divisible by:
LCM(6, 9) = 18
The three-digit multiples of 18 begin with 108 and end with 990.
Number of multiples of 18 = (990 − 108) ÷ 18 + 1 = 50
Therefore, the number of three-digit numbers divisible by 6 but not by 9 is 150 − 50 = 100
The three-digit multiples of 6 begin with 102 and end with 996.
Number of multiples of 6:
(996 − 102) ÷ 6 + 1 = 150
Every number divisible by both 6 and 9 must be divisible by:
LCM(6, 9) = 18
The three-digit multiples of 18 begin with 108 and end with 990.
Number of multiples of 18 = (990 − 108) ÷ 18 + 1 = 50
Therefore, the number of three-digit numbers divisible by 6 but not by 9 is 150 − 50 = 100
Question 10: Find the largest five-digit number that is divisible by both 24 and 35.
First find the LCM.
24 = 2³ × 3
35 = 5 × 7
Therefore
LCM = 2³ × 3 × 5 × 7 = 840
The largest five-digit number is 99,999
Divide by 840:
99,999 = 840 × 119 + 39
Subtract the remainder
99,999 − 39 = 99,960
So the largest five-digit number that is divisible by 24 and 35 is 99,960
Question 11: Find the smallest five-digit number that is divisible by 8, 9 and 11.
The number must be divisible by the LCM of 8, 9 and 11.
Since these numbers are pairwise coprime
LCM = 8 × 9 × 11
= 792
The smallest five-digit number is 10,000.
Now 10,000 ÷ 792 = 12 remainder 496
Therefore, the next multiple is:
792 × 13 = 10,296
Check:
10,296 ÷ 8 = 1,287
10,296 ÷ 9 = 1,144
10,296 ÷ 11 = 936
Therefore, it satisfies all three conditions.
So the smallest number is 10,296
Question 12: Find the sum of the smallest and the largest number which may replace k in the number 3281k6 to make the number divisible by 6.
A number is divisible by 6 if it is divisible by both 2 and 3.
Divisibility by 2:
The last digit is 6, so the number is automatically divisible by 2.
Divisibility by 3:
The sum of the digits is = 3 + 2 + 8 + 1 + k + 6 = 20 + k
Therefore, 20 + k must be divisible by 3.
Possible digits are:
k = 1 → 20 + 1 = 21 ✓
k = 4 → 20 + 4 = 24 ✓
k = 7 → 20 + 7 = 27 ✓
So the smallest value of k = 1 and the largest value of k = 7.
Therefore the sum is 1 + 7 = 8
Divisibility by 2:
The last digit is 6, so the number is automatically divisible by 2.
Divisibility by 3:
The sum of the digits is = 3 + 2 + 8 + 1 + k + 6 = 20 + k
Therefore, 20 + k must be divisible by 3.
Possible digits are:
k = 1 → 20 + 1 = 21 ✓
k = 4 → 20 + 4 = 24 ✓
k = 7 → 20 + 7 = 27 ✓
So the smallest value of k = 1 and the largest value of k = 7.
Therefore the sum is 1 + 7 = 8
Question 13: A 4-digit number is written as 3ab2. If the number is divisible by 12 and a ≠ b, how many such 4-digit numbers are possible?
For a number to be divisible by 12, it must be divisible by both 3 and 4.
Since the last two digits are b2, they must form a number divisible by 4.
The possible two-digit numbers ending in 2 that are divisible by 4 are:
12, 32, 52, 72, 92
Therefore, b = 1, 3, 5, 7 or 9
Now apply the divisibility rule for 3.
The sum of the digits is 3 + a + b + 2 = a + b + 5
This must be divisible by 3.
Case 1: b = 1
a + 6 must be divisible by 3.
Thus, a = 0, 3, 6, 9
None of these equals b = 1, so there are 4 possibilities.
Case 2: b = 3
a + 8 must be divisible by 3.
Thus, a = 1, 4, 7
None of these equals b = 3, so there are 3 possibilities.
Case 3: b = 5
a + 10 must be divisible by 3.
Thus, a = 2, 5, 8
Since a ≠ b and b = 5, a = 5 is excluded.
So, there are 2 possibilities.
Case 4: b = 7
a + 12 must be divisible by 3.
Thus, a = 0, 3, 6, 9
None of these equals b = 7, so there are 4 possibilities.
Case 5: b = 9
a + 14 must be divisible by 3.
Thus, a = 1, 4, 7
None of these equals b = 9, so there are 3 possibilities.
Therefore, the total number of possible 4-digit numbers is:
4 + 3 + 2 + 4 + 3 = 16
Since the last two digits are b2, they must form a number divisible by 4.
The possible two-digit numbers ending in 2 that are divisible by 4 are:
12, 32, 52, 72, 92
Therefore, b = 1, 3, 5, 7 or 9
Now apply the divisibility rule for 3.
The sum of the digits is 3 + a + b + 2 = a + b + 5
This must be divisible by 3.
Case 1: b = 1
a + 6 must be divisible by 3.
Thus, a = 0, 3, 6, 9
None of these equals b = 1, so there are 4 possibilities.
Case 2: b = 3
a + 8 must be divisible by 3.
Thus, a = 1, 4, 7
None of these equals b = 3, so there are 3 possibilities.
Case 3: b = 5
a + 10 must be divisible by 3.
Thus, a = 2, 5, 8
Since a ≠ b and b = 5, a = 5 is excluded.
So, there are 2 possibilities.
Case 4: b = 7
a + 12 must be divisible by 3.
Thus, a = 0, 3, 6, 9
None of these equals b = 7, so there are 4 possibilities.
Case 5: b = 9
a + 14 must be divisible by 3.
Thus, a = 1, 4, 7
None of these equals b = 9, so there are 3 possibilities.
Therefore, the total number of possible 4-digit numbers is:
4 + 3 + 2 + 4 + 3 = 16
Question 14: How many 3-digit numbers are divisible by 11 and have the sum of their digits equal to 11?
Let the 3-digit number be abc, where a ≠ 0.
We are given a + b + c = 11
For a number to be divisible by 11, the difference between the sum of digits in alternate positions must be 0 or a multiple of 11.
Thus,
a − b + c = 0 or 11 or −11
Since a + c = 11 − b, we get:
11 − b − b = 0, 11 or −11
So, 11 − 2b = 0, 11 or −11
Since b is a digit, the only possible value is b = 0
Therefore, a + c = 11
The possible pairs (a, c) are:
(2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2)
Hence, there are 8 such 3-digit numbers.
We are given a + b + c = 11
For a number to be divisible by 11, the difference between the sum of digits in alternate positions must be 0 or a multiple of 11.
Thus,
a − b + c = 0 or 11 or −11
Since a + c = 11 − b, we get:
11 − b − b = 0, 11 or −11
So, 11 − 2b = 0, 11 or −11
Since b is a digit, the only possible value is b = 0
Therefore, a + c = 11
The possible pairs (a, c) are:
(2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2)
Hence, there are 8 such 3-digit numbers.
Question 15: Which are the two numbers nearest to 19,596 that are divisible by 9?
Sum of its digits
1 + 9 + 5 + 9 + 6 = 30
Since 30 is not divisible by 9, 19,596 is not divisible by 9.
The nearest multiple of 9 below it is obtained by reducing the digit sum from 30 to 27.
So, 19,596 − 3 = 19,593
The nearest multiple of 9 above it is 19,596 + 6 = 19,602
Therefore, the two numbers nearest to 19,596 that are divisible by 9 are 19,593 and 19,602.
1 + 9 + 5 + 9 + 6 = 30
Since 30 is not divisible by 9, 19,596 is not divisible by 9.
The nearest multiple of 9 below it is obtained by reducing the digit sum from 30 to 27.
So, 19,596 − 3 = 19,593
The nearest multiple of 9 above it is 19,596 + 6 = 19,602
Therefore, the two numbers nearest to 19,596 that are divisible by 9 are 19,593 and 19,602.
Question 16: The number N = 111,111,111,111 (a 12-digit number made entirely of 1s) is not divisible by which of the following numbers?
A) 3
B) 11
C) 37
D) 101
A) 3
B) 11
C) 37
D) 101
Check divisibility by 3:
The sum of its digits is 12, which is divisible by 3.
Therefore, N is divisible by 3.
Check divisibility by 11:
The alternating sum of the digits is:
1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 = 0
Therefore, N is divisible by 11.
Check divisibility by 37:
Notice that 111 = 3 × 37
and N = 111 × 1,001,001
Therefore, N is divisible by 37.
Check divisibility by 101:
Since 11,111,111,111 ÷ 101
is not an integer, N is not divisible by 101.
Therefore, the correct option is (D) 101
The sum of its digits is 12, which is divisible by 3.
Therefore, N is divisible by 3.
Check divisibility by 11:
The alternating sum of the digits is:
1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 = 0
Therefore, N is divisible by 11.
Check divisibility by 37:
Notice that 111 = 3 × 37
and N = 111 × 1,001,001
Therefore, N is divisible by 37.
Check divisibility by 101:
Since 11,111,111,111 ÷ 101
is not an integer, N is not divisible by 101.
Therefore, the correct option is (D) 101
Question 17: A six-digit number of the form 537xy5 is divisible by 125. How many such six-digit numbers are possible?
A number is divisible by 125 if its last three digits are divisible by 125.
The last three digits of 537xy5 are xy5.
The three-digit multiples of 125 are 125, 250, 375, 500, 625, 750, 875
Among these, the numbers ending in 5 are 125, 375, 625, 875
Therefore, the possible values of xy5 are 125, 375, 625 and 875
Hence, there are 4 possible six-digit numbers are 537125, 537375, 537625 and 537875
Therefore there are 4 such numbers.
The last three digits of 537xy5 are xy5.
The three-digit multiples of 125 are 125, 250, 375, 500, 625, 750, 875
Among these, the numbers ending in 5 are 125, 375, 625, 875
Therefore, the possible values of xy5 are 125, 375, 625 and 875
Hence, there are 4 possible six-digit numbers are 537125, 537375, 537625 and 537875
Therefore there are 4 such numbers.
Question 18: For what value of m is the seven digit number 48393m8 divisible by 11?
For divisibility by 11, the difference between the sums of alternate digits must be 0 or a multiple of 11.
For the number 48393m8:
Sum of digits in alternate positions = 4 + 3 + 3 + 8 = 18
Sum of the remaining digits = 8 + 9 + m = 17 + m
Therefore, 18 − (17 + m) = 1 − m
Since m is a digit, 1 − m can be 0 or −11.
So,
1 − m = 0 ⟹ m = 1
or
1 − m = −11 ⟹ m = 12, which is not a digit.
Therefore, m = 1
For the number 48393m8:
Sum of digits in alternate positions = 4 + 3 + 3 + 8 = 18
Sum of the remaining digits = 8 + 9 + m = 17 + m
Therefore, 18 − (17 + m) = 1 − m
Since m is a digit, 1 − m can be 0 or −11.
So,
1 − m = 0 ⟹ m = 1
or
1 − m = −11 ⟹ m = 12, which is not a digit.
Therefore, m = 1
Question 19: Find the greatest possible value of (a + b) for which the 8-digit number 143b203a is divisible by 15.
Divisibility by 5
The last digit must be 0 or 5.
So, a = 0 or 5
Divisibility by 3
The sum of the digits is
1 + 4 + 3 + b + 2 + 0 + 3 + a = 13 + a + b
This must be divisible by 3.
We need to maximize a + b.
If a = 5, then:
13 + 5 + b = 18 + b
For this to be divisible by 3, b can be 0, 3, 6 or 9.
The greatest value is b = 9.
Thus,
a + b = 5 + 9 = 14
Therefore, the greatest possible value of (a + b) is 14
The last digit must be 0 or 5.
So, a = 0 or 5
Divisibility by 3
The sum of the digits is
1 + 4 + 3 + b + 2 + 0 + 3 + a = 13 + a + b
This must be divisible by 3.
We need to maximize a + b.
If a = 5, then:
13 + 5 + b = 18 + b
For this to be divisible by 3, b can be 0, 3, 6 or 9.
The greatest value is b = 9.
Thus,
a + b = 5 + 9 = 14
Therefore, the greatest possible value of (a + b) is 14
Question 20: If the seven-digit number 5X8146Y is divisible by 88, find the value of X.
Since 88 = 8 × 11, the number must be divisible by both 8 and 11.
For divisibility by 8, the last three digits 46Y must be divisible by 8.
Among the numbers from 460 to 469, only 464 is divisible by 8.
Therefore Y = 4
The number becomes 5X81464
Now apply the divisibility rule for 11.
Sum of digits in alternate positions = 5 + 8 + 4 + 4 = 21
Sum of the remaining digits X + 1 + 6 = X + 7
Therefore 21 − (X + 7) = 11
14 − X = 11
Hence, X = 3
For divisibility by 8, the last three digits 46Y must be divisible by 8.
Among the numbers from 460 to 469, only 464 is divisible by 8.
Therefore Y = 4
The number becomes 5X81464
Now apply the divisibility rule for 11.
Sum of digits in alternate positions = 5 + 8 + 4 + 4 = 21
Sum of the remaining digits X + 1 + 6 = X + 7
Therefore 21 − (X + 7) = 11
14 − X = 11
Hence, X = 3
Question 21: If the five-digit number 535ab is divisible by 3, 7 and 11, find the value of a² − b² + ab.
Since the number is divisible by 3, 7 and 11, it must be divisible by their LCM.
Since 3, 7 and 11 are pairwise coprime
LCM = 3 × 7 × 11 = 231
The number 535ab lies between 53,500 and 53,599.
Now, 231 × 231 = 53,361
and 231 × 232 = 53,592
Therefore, the only multiple of 231 in this range is 53,592
Comparing with 535ab a = 9 and b = 2
Hence, a² − b² + ab
= 9² − 2² + (9 × 2)
= 81 − 4 + 18
= 95
Since 3, 7 and 11 are pairwise coprime
LCM = 3 × 7 × 11 = 231
The number 535ab lies between 53,500 and 53,599.
Now, 231 × 231 = 53,361
and 231 × 232 = 53,592
Therefore, the only multiple of 231 in this range is 53,592
Comparing with 535ab a = 9 and b = 2
Hence, a² − b² + ab
= 9² − 2² + (9 × 2)
= 81 − 4 + 18
= 95
Question 22: What is the largest five-digit number that is divisible by 5, 8, 15, 16 and 21?
The required number must be divisible by the LCM of 5, 8, 15, 16 and 21.
Prime factors:
5 = 5
8 = 2³
15 = 3 × 5
16 = 2⁴
21 = 3 × 7
Therefore,
LCM = 2⁴ × 3 × 5 × 7
= 16 × 3 × 5 × 7
= 1,680
So, the required number must be a multiple of 1,680.
The largest five-digit number is 99,999.
Now,
99,999 ÷ 1,680 = 59 remainder 879
Therefore, the largest required multiple is 1,680 × 59 = 99,120
Prime factors:
5 = 5
8 = 2³
15 = 3 × 5
16 = 2⁴
21 = 3 × 7
Therefore,
LCM = 2⁴ × 3 × 5 × 7
= 16 × 3 × 5 × 7
= 1,680
So, the required number must be a multiple of 1,680.
The largest five-digit number is 99,999.
Now,
99,999 ÷ 1,680 = 59 remainder 879
Therefore, the largest required multiple is 1,680 × 59 = 99,120
Question 23: Which of the following single-digit numbers is an odd divisor of 2²⁵ + 2²⁶ + 2²⁷?
A) 5
B) 6
C) 7
D) 9
A) 5
B) 6
C) 7
D) 9
Take 2²⁵ common:
2²⁵ + 2²⁶ + 2²⁷ = 2²⁵(1 + 2 + 4)
= 2²⁵ × 7
Therefore, the expression is exactly divisible by 7.
2²⁵ + 2²⁶ + 2²⁷ = 2²⁵(1 + 2 + 4)
= 2²⁵ × 7
Therefore, the expression is exactly divisible by 7.
Question 24: Which of the following numbers completely divide 7^81 + 7^82 + 7^83?
A) 399
B) 389
C) 387
D) 397
A) 399
B) 389
C) 387
D) 397
7⁸¹ + 7⁸² + 7⁸³
Take 7⁸¹ common:
= 7⁸¹(1 + 7 + 7²)
= 7⁸¹(1 + 7 + 49)
= 7⁸¹ × 57
Therefore,
7⁸¹ + 7⁸² + 7⁸³ = 3 × 19 × 7⁸¹
= 3 × 7 × 19 × 7⁸⁰
= 399 × 7⁸⁰
Hence, the expression is completely divisible by 399
Take 7⁸¹ common:
= 7⁸¹(1 + 7 + 7²)
= 7⁸¹(1 + 7 + 49)
= 7⁸¹ × 57
Therefore,
7⁸¹ + 7⁸² + 7⁸³ = 3 × 19 × 7⁸¹
= 3 × 7 × 19 × 7⁸⁰
= 399 × 7⁸⁰
Hence, the expression is completely divisible by 399
Question 25: 5^71 + 5^72 + 5^73 + 5^74 + 5^75 is divisible by which of the following numbers
A) 71
B) 69
C) 89
D) 73
A) 71
B) 69
C) 89
D) 73
5⁷¹ + 5⁷² + 5⁷³ + 5⁷⁴ + 5⁷⁵
Take 5⁷¹ common
= 5⁷¹(1 + 5 + 5² + 5³ + 5⁴)
Now,
1 + 5 + 25 + 125 + 625 = 781
And:
781 = 11 × 71
Therefore,
5⁷¹ + 5⁷² + 5⁷³ + 5⁷⁴ + 5⁷⁵ = 5⁷¹ × 11 × 71
Hence, it is divisible by 71.
Take 5⁷¹ common
= 5⁷¹(1 + 5 + 5² + 5³ + 5⁴)
Now,
1 + 5 + 25 + 125 + 625 = 781
And:
781 = 11 × 71
Therefore,
5⁷¹ + 5⁷² + 5⁷³ + 5⁷⁴ + 5⁷⁵ = 5⁷¹ × 11 × 71
Hence, it is divisible by 71.
Question 26: Which of the following numbers will always divide a six-digit number of the form abcabc, where a, b and c are non-zero digits?
A) 1001
B) 1010
C) 1101
D) 1111
A) 1001
B) 1010
C) 1101
D) 1111
A number of the form abcabc can be written as abc × 10³ + abc
Taking abc common
= abc(10³ + 1)
= abc(1000 + 1)
= abc × 1001
Therefore, every six-digit number of the form abcabc is always divisible by 1001.
Taking abc common
= abc(10³ + 1)
= abc(1000 + 1)
= abc × 1001
Therefore, every six-digit number of the form abcabc is always divisible by 1001.
Question 27: Which of the following will always divide a six-digit number of the form xyxyxy where 1 < x < 9 and 1 < y < 9?
A) 1010
B) 10101
C) 11011
D) 11010
A) 1010
B) 10101
C) 11011
D) 11010
The six-digit number is of the form xyxyxy
This can be written as
xy × 10⁴ + xy × 10² + xy
Taking xy common
= xy(10⁴ + 10² + 1)
= xy(10,000 + 100 + 1)
= xy × 10,101
Therefore, every number of the form xyxyxy is always divisible by 10,101.
Hence, the correct option is B) 10101
This can be written as
xy × 10⁴ + xy × 10² + xy
Taking xy common
= xy(10⁴ + 10² + 1)
= xy(10,000 + 100 + 1)
= xy × 10,101
Therefore, every number of the form xyxyxy is always divisible by 10,101.
Hence, the correct option is B) 10101
Question 28: If a positive integer is divisible by 3, 5 and 7 then what is the next larger integer divisible by all these numbers?
The number is divisible by 3, 5 and 7.
Their LCM is
LCM(3, 5, 7) = 3 × 5 × 7 = 105
Therefore, all numbers divisible by 3, 5 and 7 occur at intervals of 105.
If the given integer is N, the next larger integer divisible by all three numbers is N + 105
Their LCM is
LCM(3, 5, 7) = 3 × 5 × 7 = 105
Therefore, all numbers divisible by 3, 5 and 7 occur at intervals of 105.
If the given integer is N, the next larger integer divisible by all three numbers is N + 105
Question 29: The sum of 3 digit number abc, cab and bca is not divisible by which of the following numbers?
A) a + b + c
B) 37
C) 31
D) 3
A) a + b + c
B) 37
C) 31
D) 3
The three numbers are:
abc = 100a + 10b + c
cab = 100c + 10a + b
bca = 100b + 10c + a
Adding them abc + cab + bca
= (100a + 10b + c) + (100c + 10a + b) + (100b + 10c + a)
Collecting like terms
= 111a + 111b + 111c
= 111(a + b + c)
Now, 111 = 3 × 37
Therefore,
abc + cab + bca = 3 × 37 × (a + b + c)
So the sum is always divisible by:
a + b + c ✓
3 ✓
37 ✓
But 111 is not divisible by 31, since 111 ÷ 31 = 3 remainder 18 also since a + b + c ≤ 27, it cannot be divisible by 31.
Therefore, the expression is not always divisible by 31.
abc = 100a + 10b + c
cab = 100c + 10a + b
bca = 100b + 10c + a
Adding them abc + cab + bca
= (100a + 10b + c) + (100c + 10a + b) + (100b + 10c + a)
Collecting like terms
= 111a + 111b + 111c
= 111(a + b + c)
Now, 111 = 3 × 37
Therefore,
abc + cab + bca = 3 × 37 × (a + b + c)
So the sum is always divisible by:
a + b + c ✓
3 ✓
37 ✓
But 111 is not divisible by 31, since 111 ÷ 31 = 3 remainder 18 also since a + b + c ≤ 27, it cannot be divisible by 31.
Therefore, the expression is not always divisible by 31.
Question 30: Which is the greatest whole number that divides n⁴ + 6n³ + 11n² + 6n + 24 for every natural number n?
A) 6
B) 24
C) 132
D) 48
A) 6
B) 24
C) 132
D) 48
First, factor the first four terms:
n⁴ + 6n³ + 11n² + 6n
= n(n + 1)(n + 2)(n + 3)
Therefore, the expression becomes:
n(n + 1)(n + 2)(n + 3) + 24
The product of four consecutive integers is always divisible by 24.
Therefore, n(n + 1)(n + 2)(n + 3) + 24 is always divisible by 24.
Now we need to check whether a larger option, 48, always divides the expression.
Take n = 5:
5 × 6 × 7 × 8 + 24
= 1,680 + 24
= 1,704
But 1,704 ÷ 48 = 35.5
So 48 does not always divide the expression.
Since 24 > 6 we have not checked it as a divisor and 132 cannot divide for n = 1 and 2 so this also is not considered.
So the correct option is 24.
n⁴ + 6n³ + 11n² + 6n
= n(n + 1)(n + 2)(n + 3)
Therefore, the expression becomes:
n(n + 1)(n + 2)(n + 3) + 24
The product of four consecutive integers is always divisible by 24.
Therefore, n(n + 1)(n + 2)(n + 3) + 24 is always divisible by 24.
Now we need to check whether a larger option, 48, always divides the expression.
Take n = 5:
5 × 6 × 7 × 8 + 24
= 1,680 + 24
= 1,704
But 1,704 ÷ 48 = 35.5
So 48 does not always divide the expression.
Since 24 > 6 we have not checked it as a divisor and 132 cannot divide for n = 1 and 2 so this also is not considered.
So the correct option is 24.
Question 31: If n is a whole number, which of the following numbers always divides n²(n² − 1)?
A) 16
B) 2
C) 10
D) 8
A) 16
B) 2
C) 10
D) 8
We can write n²(n² − 1) = n²(n − 1)(n + 1)
Among the three consecutive integers n − 1, n, n + 1 at least one must be even.
Since n² is also involved, the expression is certainly divisible by 2.
To check the other options, take n = 2
n²(n² − 1) = 2²(2² − 1)
= 4 × 3 = 12
Since 12 is not divisible by 8, 10 or 16, none of these can always divide the expression.
Therefore the number will always be divisible by 2 and not the others.
Among the three consecutive integers n − 1, n, n + 1 at least one must be even.
Since n² is also involved, the expression is certainly divisible by 2.
To check the other options, take n = 2
n²(n² − 1) = 2²(2² − 1)
= 4 × 3 = 12
Since 12 is not divisible by 8, 10 or 16, none of these can always divide the expression.
Therefore the number will always be divisible by 2 and not the others.
Answer: B) 2
Question 32: How many numbers between 1 and 100 are neither divisible by 3 nor by 5?
There are 100 numbers from 1 to 100.
Numbers divisible by 3
100 ÷ 3 = 33 numbers.
Numbers divisible by 5:
100 ÷ 5 = 20 numbers.
But numbers divisible by both 3 and 5 have been counted twice. These are numbers divisible by 15
100 ÷ 15 = 6 numbers.
Therefore, the number divisible by 3 or 5 is 33 + 20 − 6 = 47
Hence, the number divisible by neither 3 nor 5 is 100 − 47 = 53
Numbers divisible by 3
100 ÷ 3 = 33 numbers.
Numbers divisible by 5:
100 ÷ 5 = 20 numbers.
But numbers divisible by both 3 and 5 have been counted twice. These are numbers divisible by 15
100 ÷ 15 = 6 numbers.
Therefore, the number divisible by 3 or 5 is 33 + 20 − 6 = 47
Hence, the number divisible by neither 3 nor 5 is 100 − 47 = 53
Question 33: How many numbers are there between 500 and 650 both numbers inclusive which are neither divisible by 3 nor by 7?
There are 650 − 500 + 1 = 151 numbers from 500 to 650, inclusive.
Numbers divisible by 3:
The first multiple of 3 is 501 and the last is 648.
Number of multiples:
(648 − 501) ÷ 3 + 1 = 50
Numbers divisible by 7:
The first multiple is 504 and the last is 644.
Number of multiples are (644 − 504) ÷ 7 + 1 = 21
Numbers divisible by both 3 and 7 must be divisible by 21.
The first multiple is 504 and the last is 630.
Number of multiples = (630 − 504) ÷ 21 + 1 = 7
Therefore, numbers divisible by 3 or 7 are 50 + 21 − 7 = 64
Hence, numbers divisible by neither 3 nor 7 151 − 64 = 87
Numbers divisible by 3:
The first multiple of 3 is 501 and the last is 648.
Number of multiples:
(648 − 501) ÷ 3 + 1 = 50
Numbers divisible by 7:
The first multiple is 504 and the last is 644.
Number of multiples are (644 − 504) ÷ 7 + 1 = 21
Numbers divisible by both 3 and 7 must be divisible by 21.
The first multiple is 504 and the last is 630.
Number of multiples = (630 − 504) ÷ 21 + 1 = 7
Therefore, numbers divisible by 3 or 7 are 50 + 21 − 7 = 64
Hence, numbers divisible by neither 3 nor 7 151 − 64 = 87
Question 34: What is the remainder when the 100-digit number 7777…77 (consisting of one hundred 7s) is divided by 16?
For divisibility by 16, we only need to consider the last four digits of the number.
The given number consists entirely of 7s, so its last four digits are 7777
Now divide 7777 ÷ 16 = 486 remainder 1
Therefore, the remainder is 1
The given number consists entirely of 7s, so its last four digits are 7777
Now divide 7777 ÷ 16 = 486 remainder 1
Therefore, the remainder is 1
Question 35: For every natural number n, which of the following numbers always divides 3²ⁿ⁺² − 8n − 9?
A) 16
B) 32
C) 64
D) 128
A) 16
B) 32
C) 64
D) 128
Let E = 3²ⁿ⁺² − 8n − 9
For n = 1:
E = 3⁴ − 8 − 9
= 81 − 17 = 64
So, 64 divides E.
Now take n = 2:
E = 3⁶ − 16 − 9
= 729 − 25 = 704
704 = 64 × 11
So, 64 still divides E.
To show that 64 always divides E, consider two consecutive values of n.
The difference is Eₙ₊₁ − Eₙ
= 3²ⁿ⁺⁴ − 8(n + 1) − 9 − (3²ⁿ⁺² − 8n − 9)
= 3²ⁿ⁺²(3² − 1) − 8
= 8(3²ⁿ⁺² − 1)
Since 3²ⁿ⁺² is an odd number, 3²ⁿ⁺² − 1 is divisible by 8.
Therefore, 8(3²ⁿ⁺² − 1) is divisible by 64.
Since E₁ = 64 is divisible by 64, and the difference between consecutive values is always divisible by 64, E is divisible by 64 for every natural number n.
For n = 1:
E = 3⁴ − 8 − 9
= 81 − 17 = 64
So, 64 divides E.
Now take n = 2:
E = 3⁶ − 16 − 9
= 729 − 25 = 704
704 = 64 × 11
So, 64 still divides E.
To show that 64 always divides E, consider two consecutive values of n.
The difference is Eₙ₊₁ − Eₙ
= 3²ⁿ⁺⁴ − 8(n + 1) − 9 − (3²ⁿ⁺² − 8n − 9)
= 3²ⁿ⁺²(3² − 1) − 8
= 8(3²ⁿ⁺² − 1)
Since 3²ⁿ⁺² is an odd number, 3²ⁿ⁺² − 1 is divisible by 8.
Therefore, 8(3²ⁿ⁺² − 1) is divisible by 64.
Since E₁ = 64 is divisible by 64, and the difference between consecutive values is always divisible by 64, E is divisible by 64 for every natural number n.
Question 36: How many whole numbers between 200 and 400 (inclusive) are divisible by 4 or 6, but not by 8?
We need numbers between 200 and 400, inclusive, that are divisible by 4 or 6, but not by 8.
Numbers divisible by 4
The first is 200 and the last is 400.
Number of multiples (400 − 200) ÷ 4 + 1 = 51
Numbers divisible by 6
The first is 204 and the last is 396.
Number of multiples (396 − 204) ÷ 6 + 1 = 33
Numbers divisible by both 4 and 6
The LCM of 4 and 6 is 12.
The first multiple of 12 is 204 and the last is 396.
Number of multiples (396 − 204) ÷ 12 + 1 = 17
Therefore, numbers divisible by 4 or 6:
51 + 33 − 17 = 67
Exclude numbers divisible by 8
Every number divisible by 8 is already divisible by 4.
The first multiple of 8 is 200 and the last is 400.
Number of multiples(400 − 200) ÷ 8 + 1 = 26
Therefore, the required number is 67 − 26 = 41
Numbers divisible by 4
The first is 200 and the last is 400.
Number of multiples (400 − 200) ÷ 4 + 1 = 51
Numbers divisible by 6
The first is 204 and the last is 396.
Number of multiples (396 − 204) ÷ 6 + 1 = 33
Numbers divisible by both 4 and 6
The LCM of 4 and 6 is 12.
The first multiple of 12 is 204 and the last is 396.
Number of multiples (396 − 204) ÷ 12 + 1 = 17
Therefore, numbers divisible by 4 or 6:
51 + 33 − 17 = 67
Exclude numbers divisible by 8
Every number divisible by 8 is already divisible by 4.
The first multiple of 8 is 200 and the last is 400.
Number of multiples(400 − 200) ÷ 8 + 1 = 26
Therefore, the required number is 67 − 26 = 41
What are Divisibility Rules?
Divisibility rules are quick tricks that help you check whether one number can be divided exactly by another, without doing the actual division. Once you become familiar with these rules, you can solve many mathematical problems faster and with greater confidence.
Divisibility Rule for 2
A number is divisible by 2 if its last digit is 0, 2, 4, 6, or 8. In other words, every even number is divisible by 2.
Examples: 18, 42, 150, 906 and 946874 are all divisible by 2 because they end in an even digit.
Divisibility Rule for 3
A number is divisible by 3 if the sum of its digits is divisible by 3.
Examples: 123 (1 + 2 + 3 = 6) and 954 (9 + 5 + 4 = 18) are divisible by 3 because the sum of their digits is a multiple of 3.
Examples: 123 (1 + 2 + 3 = 6) and 954 (9 + 5 + 4 = 18) are divisible by 3 because the sum of their digits is a multiple of 3.
Divisibility Rule for 4
A number is divisible by 4 if the number formed by its last two digits is divisible by 4.
Examples: 316 (last two digits: 16) and 1,024 (last two digits: 24) are divisible by 4 because 16 and 24 are both divisible by 4.
Examples: 316 (last two digits: 16) and 1,024 (last two digits: 24) are divisible by 4 because 16 and 24 are both divisible by 4.
Divisibility Rule for 5
A number is divisible by 5 if its last digit is 0 or 5.
Examples: 85, 320, and 1,745 are divisible by 5 because they end in 0 or 5.
Examples: 85, 320, and 1,745 are divisible by 5 because they end in 0 or 5.
Divisibility Rule for 6
A number is divisible by 6 if it is divisible by both 2 and 3. In other words, the number must be even and the sum of its digits must be divisible by 3.
Example: 540 has 0 in the one’s place, so it is divisible by 2. The sum of its digits (5 + 4 + 0 = 9) is divisible by 3. Therefore, 540 is divisible by 6.
Divisibility Rule for 7
To check whether a number is divisible by 7, double the last digit and subtract it from the remaining part of the number. If the result is 0 or a multiple of 7, then the original number is divisible by 7.
Example: Consider 301. Double the last digit: 1 × 2 = 2. Subtract it from the remaining number: 30 − 2 = 28. Since 28 is divisible by 7, 301 is also divisible by 7. Here is another one, consider the number 7865424.
786542 − 2 × 4 = 786534
78653 − 2 × 4 = 78645
7864 − 2 × 5 = 7854
785 − 2 × 4 = 777
77 − 2 × 7 = 63
The final two-digit number is 63, which is divisible by 7 (63 = 7 × 9). Therefore, 7865424 is divisible by 7.
Example: Consider 301. Double the last digit: 1 × 2 = 2. Subtract it from the remaining number: 30 − 2 = 28. Since 28 is divisible by 7, 301 is also divisible by 7. Here is another one, consider the number 7865424.
786542 − 2 × 4 = 786534
78653 − 2 × 4 = 78645
7864 − 2 × 5 = 7854
785 − 2 × 4 = 777
77 − 2 × 7 = 63
The final two-digit number is 63, which is divisible by 7 (63 = 7 × 9). Therefore, 7865424 is divisible by 7.
Divisibility Rule for 8
A number is divisible by 8 if the number formed by its last three digits is divisible by 8.
Example: Consider 5,192. The last three digits are 192, which is divisible by 8. Therefore, 5,192 is divisible by 8.
Example: Consider 5,192. The last three digits are 192, which is divisible by 8. Therefore, 5,192 is divisible by 8.
Divisibility Rule for 9
A number is divisible by 9 if the sum of its digits is divisible by 9.
Example: Consider 6,372. The sum of its digits is 6 + 3 + 7 + 2 = 18. Since 18 is divisible by 9, 6,372 is also divisible by 9.
Example: Consider 6,372. The sum of its digits is 6 + 3 + 7 + 2 = 18. Since 18 is divisible by 9, 6,372 is also divisible by 9.
Divisibility Rule for 10
A number is divisible by 10 if its last digit is 0.
Example: Consider 8,430. Since the last digit is 0, 8,430 is divisible by 10.
Example: Consider 8,430. Since the last digit is 0, 8,430 is divisible by 10.
Divisibility Rule for 11
A number is divisible by 11 if the difference between the sum of the digits in alternate positions is 0 or a multiple of 11.
Example: Consider 3,817. The sum of the digits in odd positions is 3 + 1 = 4, and the sum of the digits in even positions is 8 + 7 = 15. The difference is 15 − 4 = 11, which is divisible by 11. Therefore, 3,817 is divisible by 11.
Example: Consider 3,817. The sum of the digits in odd positions is 3 + 1 = 4, and the sum of the digits in even positions is 8 + 7 = 15. The difference is 15 − 4 = 11, which is divisible by 11. Therefore, 3,817 is divisible by 11.
Divisibility Rule for 12
A number is divisible by 12 if it is divisible by both 3 and 4. In other words, the sum of its digits must be divisible by 3, and the number formed by its last two digits must be divisible by 4.
Example: Consider 2,916. The sum of its digits is 2 + 9 + 1 + 6 = 18, which is divisible by 3. The last two digits are 16, which is divisible by 4. Therefore, 2,916 is divisible by 12.
Example: Consider 2,916. The sum of its digits is 2 + 9 + 1 + 6 = 18, which is divisible by 3. The last two digits are 16, which is divisible by 4. Therefore, 2,916 is divisible by 12.
Summary of Divisibility Rules
| Number | Divisibility Rule |
|---|---|
| 2 | Last digit is 0, 2, 4, 6 or 8. |
| 3 | Sum of the digits is divisible by 3. |
| 4 | The number formed by the last two digits is divisible by 4. |
| 5 | Last digit is 0 or 5. |
| 6 | The number is divisible by both 2 and 3. |
| 7 | Double the last digit and subtract it from the remaining number. Repeat until a two-digit number is obtained. |
| 8 | The number formed by the last three digits is divisible by 8. |
| 9 | Sum of the digits is divisible by 9. |
| 10 | Last digit is 0. |
| 11 | The difference between the sums of alternate digits is 0 or a multiple of 11. |
| 12 | The number is divisible by both 3 and 4. |
Common Mistakes
- Students often check the entire number instead of only the required digits. Example: For divisibility by 4, check only the last two digits, not the whole number.
- Confusing the rules for 3 and 9. Some students add the digits correctly but forget that
- The sum must be divisible by 3 for divisibility by 3.
- The sum must be divisible by 9 for divisibility by 9.
- Forgetting the rule for 6. A number is divisible by 6 only if it is divisible by both 2 and 3. Being divisible by just one of them is not enough.
- Using the wrong digits for 8. For divisibility by 8, check only the last three digits. Checking the last two digits gives the wrong result.
- Applying the rule for 11 incorrectly. When testing divisibility by 11, add the digits in alternate positions and find the difference between the two sums. Do not simply add all the digits together.
Practice Questions
Question 1: Find the greatest possible value of (a + b) for which the 8-digit number 143b203a is divisible by 15.
Question 2: If the number 59a44b is divisible by 36 then what is the maximum value of (a + b)?
Question 3: If a 5-digit number 535ab is divisible by 3, 7 and 11, then what is the value of (a – b)?
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