IPMAT Indore 2022 · Quantitative Ability · Question 38

Step 1 — count all factors. For N={3}^{7}{2}^{8}{17}^{3}, a divisor is {3}^{p}{2}^{q}{17}^{r} with

0\le p\le 7, 0\le q\le 8, 0\le r\le 3

\text{total}=\left( 7+1 \right)\left( 8+1 \right)\left( 3+1 \right)=8\times 9\times 4=288

Step 2 — count the perfect-square factors. A divisor is a perfect square exactly when every exponent is even:

p\in \{0,2,4,6\} — 4 choices

q\in \{0,2,4,6,8\} — 5 choices

r\in \{0,2\} — 2 choices

\text{squares}=4\times 5\times 2=40

Step 3 — divide.

P=\frac{40}{288}=\frac{5}{36}

Answer: Option 1 — \frac{5}{36}

Shortcut. Two parallel counts, then one fraction.

\text{all factors}=∏\left({a}_{i}+1 \right), \text{square factors}=∏\left( \left\lfloor{a}_{i}/2 \right\rfloor+1 \right)

Here 8\cdot 9\cdot 4=288 against 4\cdot 5\cdot 2=40, giving \frac{40}{288}=\frac{5}{36}.

The even-exponent count is where errors creep in: for an exponent of 7 the even choices are 0,2,4,6 — that is ⌊7/2⌋+1=4, not 3. Counting 0 as an even exponent is essential, since {3}^{0}=1 is a perfectly good square factor.

A quick check on the answer’s size: roughly one factor in seven should be a square here, and 5/36\approx 0.14 sits right.