If k is a positive integer such that k is divisible by 6 and k + 1 is divisible by 5, what… — GRE General Test

If $k$ is a positive integer such that $k$ is divisible by $6$ and $k + 1$ is divisible by $5$ , what is the remainder when $k^2 + 5k$ is divided by $30$ ?

A
$0$
B
$4$
$6$ Answer
D
$14$
E
$24$

Answer

The remainder when

k^2 + 5k

is divided by

30

6

The correct answer is

6

. Since

k

is a multiple of

6

and leaves a remainder of

4

when divided by

5

, the general form of

k

modulo

30

24

. Evaluating

k^2 + 5k

modulo

30

yields

24^2 + 5(24) = 576 + 120 = 696

, which gives a remainder of

6

when divided by

30

Step-by-Step Solution

Determine the congruence class of

k

modulo 6 and modulo 5.

k \equiv 0 \pmod 6

and

k \equiv 4 \pmod 5

Given that

k

is divisible by 6, its remainder modulo 6 is 0. Since

k + 1

is divisible by 5,

k + 1 \equiv 0 \pmod 5

, which implies

k \equiv 4 \pmod 5

Find the smallest positive integer value of

k

modulo 30 satisfying both conditions.

k \equiv 24 \pmod{30}

The multiples of 6 are 0, 6, 12, 18, 24, 30, ... Among these, 24 gives a remainder of 4 when divided by 5. Since 5 and 6 are coprime,

k \equiv 24 \pmod{30}

Substitute

k \equiv 24 \pmod{30}

into the expression

k^2 + 5k

and compute the remainder modulo 30.

The remainder is

6

k^2 + 5k = k(k+5) \equiv 24(24+5) = 24(29) \pmod{30}

. Using modular arithmetic,

24 \equiv -6 \pmod{30}

and

29 \equiv -1 \pmod{30}

, so

(-6)(-1) = 6 \pmod{30}

Key Concept

Chinese Remainder Theorem and Modular Arithmetic Properties

Estimated Time:1m 30s

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