CSAT Solved Papers/ 2024/Q15

2024 CSAT — Q15

Quant Number theory 2.5 marks Medium

$222^{333} + 333^{222}$ is divisible by which of the following numbers?

A 2 and 3 but not 37
B 3 and 37 but not 2 Answer
C 2 and 37 but not 3
D 2, 3 and 37

Worked rationale

Do not compute the monster — test divisibility prime by prime. Factor the bases:

222 = 2 \times 3 \times 37, \qquad 333 = 3^2 \times 37.

A sum is divisible by $p$ iff both terms are (or their residues cancel). Check each candidate prime:

By $3$ : $222 \equiv 0$ and $333 \equiv 0 \pmod 3$ , so both powers $\equiv 0$ ; sum $\equiv 0$ . Divisible by 3. ✓
By $37$ : $222 = 6\cdot37 \equiv 0$ and $333 = 9\cdot37 \equiv 0 \pmod{37}$ , so both powers $\equiv 0$ ; sum $\equiv 0$ . Divisible by 37. ✓
By $2$ (parity): $222^{333}$ is even (even base), $333^{222}$ is odd (odd base). even $+$ odd $=$ odd. NOT divisible by 2. ✗

Divisible by $3$ and $37$ , not by $2$ .

Answer: (b) 3 and 37 but not 2.

Why the other options miss

A

two slips at once — a parity mistake and a missed factor: gets parity backwards (thinks even+something is even) and misses that $333$ also carries the factor $37$ . Two errors that happen to land on a plausible option.
C

wrong formula: spots $37$ in both bases but botches the parity (claims the sum is even) and forgets $333 = 3^2\cdot37$ is divisible by $3$ .
D

missed a case: the most seductive trap — sees that

Specialist insight

The entire template is “factor the bases, then a sum is divisible by $p$ only if every term is.” The examiner’s trap is always parity: one base is even, the other odd, so the sum is odd — kill divisibility by $2$ in two seconds and you’ve eliminated half the options before touching $3$ or $37$ . Generalise it: even base + odd base $\Rightarrow$ odd sum, never divisible by $2$ , regardless of the (positive) exponents. That one observation alone forces the answer to be (b), because (a), (c), (d) all claim divisibility by $2$ . This is the “read the parity first” discipline — it turns a scary $222^{333}$ into a five-second decision.

The trap, in one line

even $^{}$ + odd $^{}$ = odd, so the sum is never divisible by $2$ — the parity check alone eliminates three options.

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