Remainders

Question 1: A number, when divided by 47, gives a quotient of 38 and a remainder of 19. Find the number.

Question 2: When a number is divided by 58, the quotient is 24 and the remainder is 17. If the same number is divided by 29, what will be the remainder?

Question 3: What is the largest possible remainder when a number is divided by 81?

Question 4: Find the smallest three-digit number that leaves a remainder of 27 when divided by 35.

Question 5: A number leaves a remainder of 63 when divided by 899. Find the remainder when the same number is divided by 29.

Question 6: When a number is divided by 144, the remainder is 53. What will be the remainder when it is divided by 12?

Question 7: Two numbers leave the same remainder when divided by 17. If one number is 358 and the other is 443, verify whether the statement is true.

Question 8: When two numbers are divided by the same divisor, they leave the same remainder. If the numbers are 786 and 1002, find the greatest possible divisor.

Question 9: Two numbers leave the same remainder when divided by a certain divisor. If the numbers are 845 and 1097, find all possible divisors greater than 10.

Question 10: What is the smallest number that should be added to 1000 to make it exactly divisible by 45?

Question 11: What is the greatest number that should be subtracted from 1250 so that the result is exactly divisible by 36?

Question 12: Find the greatest number that should be subtracted from 5321 so that the result is exactly divisible by 28.

Question 13: Find the remainder when 2¹⁰⁰ is divided by 5.

Question 14: Find the remainder when 3⁵⁰ is divided by 7.

Question 15: Find the remainder when 7⁷⁷ is divided by 10.

Question 16: Find the remainder when 11⁶⁵ is divided by 6.

Question 17: Find the remainder when 13¹²³ is divided by 5.

Question 18: Find the remainder when 99⁹⁹⁹ is divided by 10.

First, reduce the base.
99 leaves a remainder of 9 when divided by 10.
Now consider the powers of 9.
9¹ → 9
9² → 1
9³ → 9
9⁴ → 1
The cycle length is 2.
Since 999 is odd, the required remainder corresponds to the 1st term.
Therefore, remainder = 9

Question 19: Find the remainder when 26⁷⁵ is divided by 7.

Question 20: Find the remainder when 123⁴⁵⁶ is divided by 11.

Question 21: When a number is divided by 45, it leaves a remainder of 17. What remainder will be obtained when three times the number is divided by 45?

Let the number be
45k + 17
Three times the number is
= 3(45k + 17)
= 135k + 51
= 45(3k + 1) + 6
Therefore, the required remainder is 6.

Question 22: When a number is divided by 37, it leaves a remainder of 12. What remainder will be obtained when the number is increased by 25 and divided by 37?

Let the number be
37k + 12
After adding 25,
= 37k + 37
= 37(k + 1)
Hence, the number becomes exactly divisible by 37.
Therefore, the remainder is 0.

Question 23: When a number is divided by 52, it leaves a remainder of 41. What remainder will be obtained when twice the number is divided by 52?

Question 24: When a number is divided by 29, it leaves a remainder of 18. What remainder will be obtained when five times the number is divided by 29?

Let the number be
29k + 18
Five times the number is
= 145k + 90
Since 90 = 29 × 3 + 3
the expression becomes
= 29(5k + 3) + 3
Therefore, the required remainder is 3.

Question 25: A number leaves a remainder of 23 when divided by 31. What remainder will be obtained when the square of the number is divided by 31?

We need to find the remainder when 67⁶⁷ + 67 is divided by 68.
Since 67 = 68 − 1
we can write 67 ≡ −1 (mod 68)
Therefore 67⁶⁷ ≡ (−1)⁶⁷ (mod 68)
Since 67 is odd (−1)⁶⁷ = −1
Hence 67⁶⁷ ≡ −1 (mod 68)
Also 67 ≡ 67 (mod 68)
Therefore
67⁶⁷ + 67 ≡ −1 + 67 (mod 68)
≡ 66 (mod 68)
Since 66 is already less than 68, it is the required remainder.
∴ Remainder = 66

For the first term: 13² = 169 ≡ 19 (mod 25)
Now: 13⁴ ≡ 19² = 361 ≡ 11 (mod 25)
And 13²⁰ ≡ 1 (mod 25)
∴ 13¹⁰⁰ = (13²⁰)⁵ ≡ 1⁵ ≡ 1 (mod 25)
Similarly, for the second term:
17² = 289 ≡ 14 (mod 25)
17⁴ ≡ 14² = 196 ≡ 21 (mod 25)
Continuing the powers gives:
17²⁰ ≡ 1 (mod 25)
∴ 17¹⁰⁰ = (17²⁰)⁵ ≡ 1⁵ ≡ 1 (mod 25)
Hence 13¹⁰⁰ + 17¹⁰⁰ ≡ 1 + 1 (mod 25)
≡ 2 (mod 25)
∴ Remainder = 2

Let the number be 18q + 7, where q is an integer.
When divided by 12:
18q + 7 = 12q + 6q + 7
Since 18q contributes 6q to the remainder when divided by 12, we only need to consider whether q is even or odd.
Case 1: q is even
Then 6q is divisible by 12.
∴ N = 7
Case 2: q is odd
Then 6q leaves a remainder of 6 when divided by 12.
∴ N = 6 + 7 = 13
But a remainder must be less than the divisor 12, so
13 = 12 + 1
Hence N = 1
Thus, the possible values of N are 1 and 7
∴ Number of possible values of N = 2

Let the remainder when N is divided by 12 be r.
Then N = 12q + r
Squaring N² = (12q + r)²
= 144q² + 24qr + r²
The first two terms are exactly divisible by 24. Therefore, the remainder of N² when divided by 24 is determined by r².
Given that N² leaves remainder 1 when divided by 24.
∴ r² must leave remainder 1 when divided by 24.
Since the remainder when N is divided by 12 can be
0, 1, 2, …, 11
we check their squares modulo 24.
The values that satisfy the condition are
1² = 1
5² = 25 = 24 × 1 + 1
7² = 49 = 24 × 2 + 1
11² = 121 = 24 × 5 + 1
Thus, the possible remainders when N is divided by 12 are:
1, 5, 7, 11
∴ The possible remainders are 1, 5, 7 and 11.
Number of possible remainders = 4

When a number is divided by 30, the remainder can be any integer from 0 to 29.
Since p is a prime number greater than 150, it cannot be divisible by 2, 3, or 5.
Therefore, the remainder q cannot be divisible by 2, 3, or 5.
The possible remainders from 0 to 29 that are divisible by none of 2, 3, or 5 are:
1, 7, 11, 13, 17, 19, 23, 29
Thus, there are 8 possible values of q.
∴ Number of possible values of q = 8

Let the number be N.
Since N leaves a remainder of 4 when divided by 9
N = 9k + 4
It also leaves a remainder of 3 when divided by 5
N ≡ 3 (mod 5)
Substituting
9k + 4 ≡ 3 (mod 5)
Since 9 ≡ 4 (mod 5)
4k + 4 ≡ 3 (mod 5)
∴ 4k ≡ −1 (mod 5)
or 4k ≡ 4 (mod 5)
Hence k ≡ 1 (mod 5)
So k = 5m + 1
∴ N = 9(5m + 1) + 4
N = 45m + 13
Thus, the numbers are 13, 58, 103, 148, …
The common difference is 45.
For the largest value not exceeding 1500
45m + 13 ≤ 1500
45m ≤ 1487
m ≤ 33.04…
∴ m = 0, 1, 2, …, 33
Number of values 33 − 0 + 1 = 34

Let the three numbers be A, B and C.
Their remainders when divided by N are 27, 35 and 41.
Therefore, their sum can be written as:
A + B + C = Nq + 27 + 35 + 41
A + B + C = Nq + 103
But their sum leaves a remainder of 7 when divided by N.
∴ 103 − 7 = 96
must be exactly divisible by N.
Hence, N must be a factor of 96.
The factors of 96 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96
Since 41 is one of the remainders, the divisor must be greater than 41.
Therefore, the possible values of N are 48 and 96
∴ The possible values of N are 48 and 96.

Step 1: Count the digits
The numbers 1 to 9 contribute 9 digits
The numbers 10 to 54 contribute 45 × 2 = 90 digits
∴ 9 + 90 = 99 digits
So, the first 100 digits consist of 1, 2, 3, …, 54, followed by the first digit of 55, which is 5.
Step 2: Find the digit-sum remainder
The digit sum of 1 to 9 is 1 + 2 + … + 9 = 45 which is divisible by 9.
For the numbers 10 to 54, we don’t even need to calculate their individual digit sums. Each number has the same remainder modulo 9 as its digit sum.
So consider 10 + 11 + 12 + … + 54
The sum is (10 + 54) × 45 ÷ 2 = 1440
Since 1440 = 9 × 160 this also leaves remainder 0.
Finally, the 100th digit is the first digit of 55, which is 5.
Therefore, the entire 100-digit number leaves the same remainder as
45 + 1440 + 5
= 1490
Now 1490 = 9 × 165 + 5
∴ Remainder = 5

Let N = 31³ + 32³ + 33³ + 34³
Pair the terms N = (31³ + 34³) + (32³ + 33³)
Notice that 31 + 34 = 65 and 32 + 33 = 65
Using the identity
a³ + b³ = (a + b)(a² − ab + b²)
both pairs are therefore divisible by 65.
Hence, N is divisible by 65.
Now 130 = 2 × 65
We need to determine whether N is also divisible by 2.
The four terms have the parity 1³ + 32³ + 33³ + 34³
= odd + even + odd + even
= even.
Therefore, N is divisible by 2 as well as 65.
So N is divisible by 2 × 65 = 130.
∴ Remainder = 0

First, find x.
The powers of 3 when divided by 5 follow the cycle:
3¹ → 3
3² → 4
3³ → 2
3⁴ → 1
The cycle repeats every 4 powers.
Since 61284 ÷ 4 leaves remainder 0, 3⁶¹²⁸⁴ corresponds to the 4th term of the cycle.
Therefore x = 1
Now find y.
When 4 is divided by 6, the powers follow:
4¹ → 4
4² → 16 → 4
4³ → 4
Thus, every positive power of 4 leaves remainder 4 when divided by 6.
∴ y = 4
Hence 2x − y
= 2(1) − 4
= 2 − 4
= −2

The remainders are each 1 less than the corresponding divisor:
9 = 10 − 1
8 = 9 − 1
7 = 8 − 1
Therefore, if N is the required number, then N + 1 must be exactly divisible by 10, 9 and 8.
So, N + 1 must be a common multiple of 10, 9 and 8.
Find their LCM:
10 = 2 × 5
9 = 3²
8 = 2³
∴ LCM(10, 9, 8) = 2³ × 3² × 5
= 360
Hence N + 1 = 360
N = 359
Check:
359 ÷ 10 = 35 remainder 9
359 ÷ 9 = 39 remainder 8
359 ÷ 8 = 44 remainder 7
All three conditions are satisfied.
∴ Smallest positive integer = 359

Only the units digits of the bases matter. Therefore, we need to find the units digits of 2²²² and 8⁸⁸⁸.
For powers of 2, the units digits repeat in a cycle of 4:
2, 4, 8, 6
Since 222 ÷ 4 leaves remainder 2
the units digit of 2²²² is the 2nd digit in the cycle 4
For powers of 8, the units digits also repeat in a cycle of 4: 8, 4, 2, 6
Since 888 ÷ 4 leaves remainder 0 the units digit of 8⁸⁸⁸ is the 4th digit in the cycle 6
∴ 222²²² + 888⁸⁸⁸
has units digit 4 + 6 = 10
Hence, the units digit is 0

To find the units digit, we only need the units digit of each base.
So we consider 3⁴⁶ × 6⁷³ × 5⁸²
Step 1: Find the units digit of 3⁴⁶
The units digits of powers of 3 follow the cycle 3, 9, 7, 1
The cycle length is 4.
Since 46 ÷ 4 leaves remainder 2
the units digit of 3⁴⁶ is the 2nd digit in the cycle 9
Step 2: Find the units digit of 6⁷³
Every positive power of 6 ends in 6.
∴ 6⁷³ → 6
Step 3: Find the units digit of 5⁸²
Every positive power of 5 ends in 5.
∴ 5⁸² → 5
Now multiply the units digits 9 × 6 × 5 = 270
Therefore, the units digit of the original product is 0

Let x = Nq₁ + 4,376 and y = Nq₂ + 2,986
Adding:
x + y = N(q₁ + q₂) + 4,376 + 2,986
∴ x + y = N(q₁ + q₂) + 7,362
But when x + y is divided by N, the remainder is 2,361.
So the excess 7,362 − 2,361 = 5,001 must be exactly divisible by N.
∴ N is a factor of 5,001.
Factorising 5,001 = 3 × 1,667
Since 4,376 is a remainder, the divisor must be greater than 4,376.
The only factor of 5,001 greater than 4,376 is 5,001
N = 5,001
Verification
The sum of the two remainders is:
4,376 + 2,986 = 7,362
And
7,362 = 5,001 + 2,361
So when the sum is divided by 5,001, the remainder is indeed 2,361.

We only need to consider the remainder of each term when divided by 4.
For numbers divisible by 4 4⁴, 8⁸, 12¹², …, 100¹⁰⁰ each term leaves remainder 0.
For numbers that leave remainder 2 when divided by 4
2², 6⁶, 10¹⁰, …
Since their powers are even, each leaves remainder 0 modulo 4.
For numbers that leave remainder 1 when divided by 4 are 1, 5, 9, 13, … any positive power leaves remainder 1.
For numbers that leave remainder 3 when divided by 4 are 3, 7, 11, 15, …
Since the exponents here are odd, each leaves remainder 3.
There are 25 numbers in each of the four residue classes from 1 to 100.
Therefore, the total remainder is
25 × 1 + 25 × 3
= 25 + 75
= 100
Since 100 ÷ 4 leaves remainder 0,
∴ Remainder = 0.

We have:
n⁵ − n = n(n⁴ − 1)
= n(n² − 1)(n² + 1)
= n(n − 1)(n + 1)(n² + 1)
Now, (n − 1), n and (n + 1) are three consecutive integers. Therefore, their product is divisible by 6.
Also, among any two consecutive integers, one is even. Hence n⁵ − n is divisible by 2.
To show divisibility by 5, consider the possible remainders of n when divided by 5:
If n leaves remainder 0, then n⁵ − n leaves remainder 0.
If n leaves remainder 1, then 1⁵ − 1 = 0.
If n leaves remainder 2, then 2⁵ − 2 = 30, divisible by 5.
If n leaves remainder 3, then 3⁵ − 3 = 240, divisible by 5.
If n leaves remainder 4, then 4⁵ − 4 = 1020, divisible by 5.
Thus, n⁵ − n is divisible by 5.
Therefore, n⁵ − n is divisible by 2 × 3 × 5 = 30
∴ Remainder = 0

We can reduce each number by dividing it by 13:
1330 = 13 × 102 + 4
So, 1330 ≡ 4 (mod 13)
1356 = 13 × 104 + 4
So, 1356 ≡ 4 (mod 13)
1363 = 13 × 104 + 11
So, 1363 ≡ 11 (mod 13)
1368 = 13 × 105 + 3
So, 1368 ≡ 3 (mod 13)
1397 = 13 × 107 + 6
So, 1397 ≡ 6 (mod 13)
Therefore, the required remainder is the remainder of
4 × 4 × 11 × 3 × 6
Simplify
4 × 4 = 16 ≡ 3 (mod 13)
So
3 × 11 × 3 × 6
= 9 × 11 × 6
= 99 × 6
Since 99 ≡ 8 (mod 13)
8 × 6 = 48
and
48 = 13 × 3 + 9
∴ Remainder = 9

1. Allowing the remainder to be equal to or greater than the divisor: A remainder is always non-negative and less than the divisor.
2. Using an incorrect form of the Division Algorithm: Always write the dividend as Dividend = Divisor × Quotient + Remainder.
3. Changing the divisor without rewriting the remainder: When the divisor changes, express the original remainder in terms of the new divisor before determining the new remainder.
4. Assuming two numbers with the same remainder are equal: Such numbers differ by a multiple of the divisor, not necessarily by zero.
5. Adding the divisor instead of subtracting the remainder: To make a number exactly divisible, add Divisor − Remainder, not the divisor itself.
6. Subtracting more than the remainder: The greatest number that can be subtracted to make a number exactly divisible is the remainder itself.
7. Ignoring the repeating cycle while solving large powers: Always determine the cycle length before evaluating the exponent.
8. Using the exponent directly instead of its position in the cycle: Divide the exponent by the cycle length and use the remainder to identify the correct term in the cycle.
9. Forgetting to reduce the base before finding cyclicity: If the base is larger than the divisor, first replace it with its remainder upon division by the divisor.
10. Squaring or multiplying the entire expression unnecessarily: In expressions such as (Divisor × Quotient + Remainder)², only the remainder affects the final remainder. The terms containing the divisor are always exactly divisible by the divisor.

Question 1: When a certain number is divided by 693, it leaves a remainder of 58. What remainder will the same number leave when divided by 21?

Question 2: A number leaves a remainder of 37 when divided by 64. What remainder will be obtained when four times the number is divided by 64?

Question 3: When two numbers are divided by the same divisor, they leave the same remainder. If the numbers are 1527 and 1971, find the greatest possible divisor.

Question 4: Find the smallest number that should be added to 5678 so that it becomes exactly divisible by 81.

Question 5: Find the greatest number that should be subtracted from 8745 so that the result is exactly divisible by 56.

Question 6: Find the remainder when 19²⁰²⁵ is divided by 8.

Question 7: Find the remainder when 27¹²³ is divided by 11.

Question 8: A number leaves a remainder of 24 when divided by 41. What remainder will be obtained when the square of the number is divided by 41?

Question 9: When a number is divided by 72, it leaves a remainder of 53. What remainder will be obtained when seven times the number is divided by 72?

Question 10: When a number is divided by 84, it leaves a remainder of 29. What remainder will be obtained when the cube of the number is divided by 84?

Divisibility Rules

Percentages

Simplification

Geometery