UPSC CDS I 2025 Mathematics Paper

Given:
Expression: ‌

100!

100n

is an integer
Formula used:
We need to find the largest n such that 100n divides 100 !
100=22×52⇒100n=22n×52n
So, we must find maximum n such that both 22n and 52n divide 100 !
Calculations:
Powers of 2 in 100!
‌⇒[100∕2]+[100∕4]+[100∕8]+[100∕16]+[100∕32]+[100∕64]
‌⇒50+25+12+6+3+1=97
Powers of 5 in 100!
‌⇒[100∕5]+[100∕25]+[100∕125]
‌⇒20+4+0=24
Now: 22n must ≤297⇒2n≤97⇒n≤48.5
And: 52n must ≤524⇒2n≤24⇒n≤12
So, the limiting factor is powers of 5⇒n≤12
The largest value of n is 12 .