JEE (Main)-2026 : Phase-1 (24-01-2026)-Evening

# Q1 of 58

The maximum value of n for which 40n divides 60! is equal to

Options :
A.

11


B.

12


C.

13


D.

14


Show Answer
Answer :

14


Solution :

40 = 23 × 5


Exponent of 2 in 60! = ⌊60/2⌋ + ⌊60/4⌋ + ⌊60/8⌋ + ⌊60/16⌋ + ⌊60/32⌋ + ⌊60/64⌋


= 30 + 15 + 7 + 3 + 1


= 56


Exponent of 5 in 60! = ⌊60/5⌋ + ⌊60/25⌋


= 12 + 2


= 14


Since 40 = 23 × 5


Maximum n = min(56/3 , 14)


= 14

Report

Questions