If k is an integer greater than 1 such that k^3 - k is a multiple of 240, what is the leas… — GRE General Test

If $k$ is an integer greater than $1$ such that $k^3 - k$ is a multiple of $240$ , what is the least possible value of $k$ ?

Answer: 9

The least possible value of

k

is 9.

Factoring

k^3 - k

gives

(k-1)k(k+1)

, representing the product of three consecutive integers. The prime factorization of

240

2^4 \times 3 \times 5

. To be divisible by

240

, the product

(k-1)k(k+1)

must contain at least four factors of 2, one factor of 3, and one factor of 5. Testing integers

k > 1

: for

k=9

, the product is

(8)(9)(10) = 720

, which is

3 \times 240

. Checking all integers

1 < k < 9

confirms that no smaller integer satisfies the condition.

Factor the algebraic expression

k^3 - k

k^3 - k = k(k^2 - 1) = (k-1)k(k+1)

This expresses the polynomial as a product of three consecutive integers.

Determine the prime factorization of 240.

240 = 2^4 \times 3 \times 5 = 16 \times 3 \times 5

For

(k-1)k(k+1)

to be a multiple of 240, the product must be divisible by 16, 3, and 5 simultaneously.

Test candidate values of

k > 1

to find the smallest valid integer.

For

k=9

(k-1)k(k+1) = (8)(9)(10) = 720

, which equals

3 \times 240

Testing smaller values:

k=5

yields

120

k=6

yields

210

k=7

yields

336

, and

k=8

yields

504

. None of these are divisible by 240.

Divisibility rules and prime factor distribution across consecutive integers

Estimated Time:2m 0s

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