CSAT Solved Papers/ 2022/Q74

2022 CSAT — Q74

Quant Number theory 2.5 marks Medium

If $15 \times 14 \times 13 \times \cdots \times 3 \times 2 \times 1 = 3^m \times n$ , where $m$ and $n$ are positive integers, then what is the maximum value of $m$ ?

A 7
B 6 Answer
C 5
D 4

Worked rationale

The product is $15!$ . The maximum $m$ is the exponent of $3$ in $15!$ , given by Legendre’s formula:

m = \left\lfloor \frac{15}{3} \right\rfloor + \left\lfloor \frac{15}{9} \right\rfloor + \left\lfloor \frac{15}{27} \right\rfloor = 5 + 1 + 0 = 6.

(The $\lfloor 15/9\rfloor = 1$ term counts the extra factor of $3$ from $9$ ; $27 > 15$ so it stops there.)

Answer: (b) 6.

Why the other options miss

A

miscounted a repeated factor: over-counts, e.g. adds a phantom $\lfloor 15/27\rfloor$ contribution.
C

miscounted a repeated factor: counts only the multiples of $3$ ( $\lfloor 15/3\rfloor = 5$ ), forgetting that $9 = 3^2$ contributes a second factor.
D

an arithmetic slip: miscounts the multiples of $3$ up to $15$ .

Specialist insight

The exponent of a prime $p$ in $N!$ is $\sum_{k\ge 1}\lfloor N/p^k\rfloor$ . The single most-missed term is the $\lfloor 15/9\rfloor = 1$ : the number $9$ carries two threes, so it must be counted again. Multiples of $3$ give $5$ , the extra from $9$ gives $1$ , total $6$ . Always run the powers $p, p^2, p^3,\dots$ until the quotient hits $0$ .

The trap, in one line

$\nu_3(15!) = \lfloor 15/3\rfloor + \lfloor 15/9\rfloor = 5 + 1 = 6$ (don't forget $9 = 3^2$ ) $\Rightarrow$ (b).

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