IPMAT Indore 2023 · Quantitative Ability · Question 22

Step 1 — factorise.

{10}^{2023}={2}^{2023}\times{5}^{2023}

Step 2 — count all divisors. A divisor has the form {2}^{a}{5}^{b} with 0\le a,b\le 2023:

\text{total}=2024\times 2024={2024}^{2}

Step 3 — count the favourable divisors. For {2}^{a}{5}^{b} to be a multiple of {10}^{2001}={2}^{2001}{5}^{2001}, we need both exponents at least 2001:

2001\le a\le 2023 \text{and}2001\le b\le 2023

Each range contains 2023-2001+1=23 values, so

\text{favourable}=23\times 23={23}^{2}

Step 4 — divide.

P=\frac{{23}^{2}}{{2024}^{2}}

Answer: Option 2 — \frac{{23}^{2}}{{2024}^{2}}

Shortcut. Both counts are squares, so the answer must be a square. {10}^{n} has two prime factors with identical exponent ranges, so the divisor count is {\left( n+1 \right)}^{2} and the favourable count is {k}^{2} — options (a) and (d), which are not squares, cannot be right.

Then just get the two numbers right:

2024, not 2023 — the exponent runs from 0 inclusive.

23, not 22 — the range 2001 to 2023 includes both ends.

P={\left( \frac{23}{2024}\right)}^{2}

Both off-by-one errors are represented in the options, which is the whole design of the question. Count inclusive ranges as \text{high}-\text{low}+1 every single time.