Solved Examples

Example 1: Find the remainder when 2^100 is divided by 7. - 2³ = 8 ≡ 1 (mod 7) - 100 = 3×33 + 1, so 2^100 = (2³)^33 × 2¹ ≡ 1^33 × 2 ≡ 2 (mod 7) Answer: Remainder = 2

Example 2: Find the remainder when 17^23 is divided by 16. - 17 ≡ 1 (mod 16), so 17^23 ≡ 1^23 ≡ 1 (mod 16) Answer: Remainder = 1

Example 3 (Negative remainder trick): Find the remainder when 6^83 is divided by 7. - 6 ≡ −1 (mod 7), so 6^83 ≡ (−1)^83 = −1 ≡ 7 − 1 = 6 (mod 7) Answer: Remainder = 6

Example 4 (Factorial remainder): Find the remainder when (1! + 2! + 3! + … + 100!) is divided by 5. - For n ≥ 5, n! contains the factor 5, so n! ≡ 0 (mod 5) for all n ≥ 5. - Only 1! + 2! + 3! + 4! matters: 1 + 2 + 6 + 24 = 33 - 33 mod 5 = 3 Answer: Remainder = 3

Example 5: A number, when divided by 39, leaves remainder 20. What remainder will be obtained when the same number is divided by 13? - Number = 39k + 20. Since 39 = 13×3, 39k is exactly divisible by 13. - Remainder from the number ÷ 13 = remainder from 20 ÷ 13 = 7

Combined remainder problems (finding a number satisfying two remainder conditions): A common SSC/RRB pattern gives two (or more) separate divisor–remainder pairs and asks for the smallest number satisfying both. The systematic approach: list out numbers satisfying the first condition (add the divisor repeatedly to the first remainder) until one of them also satisfies the second condition.

Example 6: Find the smallest number which leaves a remainder of 2 when divided by 3, and a remainder of 3 when divided by 5. - Numbers leaving remainder 2 when divided by 3: 2, 5, 8, 11, 14, 17, … - Check each against “remainder 3 when divided by 5”: 2÷5 rem2 (no); 5÷5 rem0 (no); 8÷5 rem3 (yes!) Answer: 8 (General solution: all numbers of the form 15k + 8, since LCM(3,5)=15)

A special common-difference shortcut: Whenever (divisor − remainder) works out to be the same value for every condition, the smallest such number is simply LCM(all divisors) − that common difference. Example: Find the smallest number which leaves remainder 1 when divided by 2, remainder 2 when divided by 3, remainder 3 when divided by 4, and remainder 4 when divided by 5. - Check divisor−remainder in each case: 2−1=1, 3−2=1, 4−3=1, 5−4=1 — the same value (1) every time, so the shortcut applies. - LCM(2,3,4,5) = 60. Answer = 60 − 1 = 59. - Verify: 59÷2 = 29 rem1 ✓; 59÷3 = 19 rem2 ✓; 59÷4=14 rem3 ✓; 59÷5=11 rem4 ✓. This is distinct from the “same remainder in every case” pattern used in Section 16, Q. B11, where the answer is instead LCM(divisors) + remainder — always check which of the two patterns (constant remainder, or constant divisor−remainder) the question actually follows before picking the formula.

Example 7 (Remainder on division by 10 = unit digit): Find the remainder when 3^97 is divided by 10. - The remainder on division by 10 is always exactly the unit digit of the number — this is a very common “disguised cyclicity” question. - Unit digit of base 3 → cycle {3,9,7,1}, length 4. 97 ÷ 4 = 24 remainder 1 → 1st term = 3. Answer: Remainder = 3

Example 8 (Combined remainder + cyclicity of remainders): Find the remainder when (2^100 + 3^100) is divided by 7. - From Example 1 above, 2^100 mod 7 = 2. - For 3^100 mod 7: build the cycle of powers of 3 mod 7 (period must divide 6, since 7 is prime): 3¹=3, 3²=9≡2, 3³≡6, 3⁴≡18≡4, 3⁵≡12≡5, 3⁶≡15≡1 (mod 7) — cycle {3,2,6,4,5,1}, length 6. - 100 ÷ 6 = 16 remainder 4 → 4th term of the cycle = 4. So 3^100 ≡ 4 (mod 7). - Total remainder = (2 + 4) mod 7 = 6 mod 7 = 6 Answer: Remainder = 6