CSAT Solved Papers/ 2025/Q67

2025 CSAT — Q67

Quant Number theory 2.5 marks Medium

What is the remainder when $9^3 + 9^4 + 9^5 + 9^6 + \ldots + 9^{100}$ is divided by $6$ ?

A 0 Answer
B 1
C 2
D 3

Worked rationale

Reduce the base mod $6$ first: $9 \equiv 3 \pmod 6$ . Then every power behaves the same:

9^k \equiv 3^k \pmod 6, \qquad 3^1 = 3,\ 3^2 = 9 \equiv 3,\ 3^3 \equiv 3 \cdot 3 \equiv 3, \dots

so $9^k \equiv 3 \pmod 6$ for every $k \ge 1$ (multiplying $3$ by $3$ stays $\equiv 3$ ).

Count the terms: from $k = 3$ to $k = 100$ that is $100 - 3 + 1 = 98$ terms, each $\equiv 3$ :

\text{sum} \equiv 98 \times 3 = 294 \equiv 0 \pmod 6 \quad (294 = 49 \times 6).

Answer: (a) 0.

Why the other options miss

B

an arithmetic slip: mis-counts the terms (e.g. $97$ terms $\times 3 = 291 \equiv 3$ , then a further slip), or mishandles $294 \bmod 6$ .
C

an arithmetic slip: takes the per-term residue as $2$ (confusing $9 \bmod 6$ with $9 \bmod 7$ or similar) and propagates the error.
D

counted one too few: counts an odd number of terms (so the total $\equiv 3$ ), the classic fence-post error from $100 - 3 = 97$ instead of $98$ .

Specialist insight

Two moves win modular-sum items. First, collapse the base: $9 \equiv 3$ and $3^k \equiv 3 \pmod 6$ for all $k$ — a constant residue, not a cycle, because $3$ is a fixed point of multiplication-by- $3$ mod $6$ . Second, count terms with fence-post care: $k=3$ to $k=100$ inclusive is $98$ terms, and the parity of that count decides everything ( $98$ threes sum to a multiple of $6$ ; $97$ would leave $3$ ). The (d) distractor is engineered for exactly the off-by-one term count, so write $100 - 3 + 1$ deliberately. Reducing each term to a constant turns a $98$ -term sum into a single multiplication.

The trap, in one line

$9^k \equiv 3 \pmod 6$ for all $k$ ; the answer hinges on the term count $100-3+1 = 98$ (even) $\Rightarrow 98\times3 = 294 \equiv 0$ — an off-by-one gives the planted $3$ .

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