IPMAT Indore 2024 · Quantitative Ability · Question 27

Step 1 — the digit-count formula. For a positive integer N,

\text{number of digits}=⌊{\log}_{10}N⌋+1

Step 2 — apply to both numbers.

{D}_{1}=⌊2024{\log}_{10}2⌋+1,{D}_{2}=⌊2024{\log}_{10}5⌋+1

Step 3 — use the key identity.

2024{\log}_{10}2+2024{\log}_{10}5=2024\left({\log}_{10}2+{\log}_{10}5 \right)=2024{\log}_{10}10=2024

So the two quantities inside the floors are non-integers summing to exactly 2024 (neither is an integer because {\log}_{10}2 is irrational). For such a pair,

⌊\alpha ⌋+⌊2024-\alpha ⌋=2024-1=2023

Step 4 — total.

{D}_{1}+{D}_{2}=2023+1+1=2025

Sanity check with a small case. {2}^{3}=8 (1 digit), {5}^{3}=125 (3 digits); total 4=3+1 ✔

Answer: Option 2 — 2025

Shortcut. Memorise: d\left({2}^{n}\right)+d\left({5}^{n}\right)=n+1.

The reason is one line — {2}^{n}\times{5}^{n}={10}^{n}, and multiplying two numbers whose product is a power of ten always gives digit counts adding to n+1. So the answer is 2024+1=2025, with no logarithm tables and no computation at all. Option (d) 2000 and (a) 1987 exist purely to reward guessing-by-size.