Arrange the following in ascending order: (A) Remainder of 4916 when divided…

2023

Arrange the following in ascending order:

(A) Remainder of 4916 when divided by 17

(B) Remainder of 2446 when divided by 9

(C) Remainder of 15517 when divided by 17

(D) Last digit of the number 745

Choose the correct answer from the options given below:

Answer: C. (A), (C), (B), (D)ConceptFor modular powers, first reduce the base modulo the divisor, then reduce the exponent using a valid power cycle or Fermat’s little theorem. For a last…

  1. A.

    (A), (B), (C), (D)

  2. B.

    (A), (B), (D), (C)

  3. C.

    (A), (C), (B), (D)

  4. D.

    (D), (C), (B), (A)

Attempted by 35 students.

Show answer & explanation

Correct answer: C

Concept

For modular powers, first reduce the base modulo the divisor, then reduce the exponent using a valid power cycle or Fermat’s little theorem. For a last digit, work modulo 10 and use the repeating last-digit cycle.

After evaluating each expression independently, compare the four numerical results to arrange them from least to greatest.

Application

  1. For (A), 49 ≡ −2 (mod 17). Since 17 is prime, Fermat’s little theorem gives 216 ≡ 1 (mod 17). Therefore, 4916 leaves remainder 1.

  2. For (B), powers of 2 modulo 9 repeat every 6 powers. Since 446 = 6 × 74 + 2, we get 2446 ≡ 22 ≡ 4 (mod 9). Therefore, the remainder is 4.

  3. For (C), 155 ≡ 2 (mod 17). Using 216 ≡ 1 (mod 17), we obtain 217 = 2 × 216 ≡ 2 (mod 17). Therefore, 15517 leaves remainder 2.

  4. For (D), the last digits of powers of 7 cycle as 7, 9, 3, 1. Since 45 = 4 × 11 + 1, 745 has last digit 7.

  5. Thus the values are (A) = 1, (B) = 4, (C) = 2, and (D) = 7. Sorting these gives 1 < 2 < 4 < 7, so the ascending order is (A), (C), (B), (D).

Cross-check

The reduced values 1, 2, 4, and 7 are already strictly increasing in the sequence (A), (C), (B), (D), confirming the ordering without any tie.

Result: (A), (C), (B), (D).

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