Which one of the following is equal to 1³⁷ + 2³⁷ + 3³⁷ + ⋯ + 88³⁷ in ℤ/89ℤ?

D. 89 is prime and (ℤ/89) is cyclic of order 88. For a generator g, Σ{k} k³⁷ = Σj g^{37j} is a geometric sum equal to 0 because 88 ∤ 37 (so g³⁷ ≠ 1).

Power sums mod p — CSIR NET December 2023 Part B, solved

Power sums mod p

Which one of the following is equal to $1^{37} + 2^{37} + 3^{37} + \cdots + 88^{37}$ in $\mathbb{Z}/89\mathbb{Z}$ ?

A.88
B.−88
C.−2
D.0✓ Correct

The idea, in plain words

Adding up the same power of every non-zero remainder is a sum with a hidden symmetry: the terms can be rearranged into a repeating cycle that cancels itself out. It fails to cancel only when the power lines up exactly with the length of the cycle, which is the case to check.

Solution

89 is prime and $(\mathbb{Z}/89)$ * is cyclic of order 88.

For a generator $g, \sum_{k}$ $k^{37} = \sum_j$ $g^{37j}$ is a geometric sum equal to 0 because 88 ∤ 37 (so $g^{37} \ne 1)$ .

Why each option is right or wrong

Checked against 4 symbolic computations

A.WrongBoundary and endpoint
False here

but it is the right answer to a neighbouring question — which is exactly why it is offered.

The power sum $\sum{}k^m mod p$ equals −1 ≡ p − 1 precisely when (p − 1) | m, and 0 otherwise.

SymPy computes both cases directly: $\sum{}k^{88} mod 89 = 88$

while $\sum{}k^{37} mod 89 = 0$ .

Since 88 ∤ 37 (verified: 37 mod 88 = 37), we are in the second case.
B.Wrong
False.

−88 ≡ 1 (mod 89), and the sum is 0, not 1.
C.Wrong
False.

The sum is 0; −2 arises from no step of the correct argument.
D.Correct
True — SymPy evaluates the sum directly and reduces it: $\sum_{k=1}^{88}$ $k^{37} \equiv 0 (mod 89)$ .

The argument: 89 is prime (verified)

so $(\mathbb{Z}/89)$ * is cyclic of order 88 with some generator g, and the sum becomes $\sum_{j=0}^{87}$ $(g^{37})^j$ — a geometric series.

Because 88 ∤ $37, g^{37} \ne 1$

so the series sums to $(g^{37}\cdot^{88} - 1)/(g^{37} - 1) = 0$ .

The trigger: a full power sum over $\mathbb{Z}/p$ should immediately raise the question 'does (p−1) divide the exponent?'

1 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Last revised 12 Sept 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.

Power sums mod p

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