IPMAT Indore 2026 · Quantitative Ability · Question 17

Step 1 — write n in terms of the required gcd. If \gcd \left( n,{10}^{7}\right)={10}^{3}, then {10}^{3} divides n, so put

n={10}^{3}m

Step 2 — translate the gcd condition onto m. Since {10}^{7}={10}^{3}\times{10}^{4},

\gcd \left({10}^{3}m,{10}^{3}\cdot{10}^{4}\right)={10}^{3}\gcd \left( m,{10}^{4}\right)={10}^{3}\Rightarrow \gcd \left( m,{10}^{4}\right)=1

So m must share no factor of 2 or 5 with {10}^{4} — that is, m must be coprime to 10.

Step 3 — find the range of m. From n\le{10}^{7} we get m\le{10}^{4}.

Step 4 — count integers up to {10}^{4} coprime to 10. Remove multiples of 2 and of 5:

{10}^{4}\times \left( 1-1/2 \right)\left( 1-1/5 \right)={10}^{4}\times 1/2\times 4/5=4000

Answer: Option 4 — 4000

Shortcut. Every “gcd equals d” question becomes “write n=dm and make m coprime to the rest”. Here that leaves counting numbers up to {10}^{4} coprime to 10, which is just \frac{1}{2}\times \frac{4}{5} of them — 40\%, so 4000.