Question

Find the number of trailing zeroes in 75! + 79!
1. 18
2. 19
3. 20
4. 17

Answer 1

Correct Answer - Option 1 : 18

Concept Used:

n! = n(n – 1)(n – 2)…1

Number of zeros = Maximum Power of 5 in factorial

In the case of factorial, the power of 5 is the limiting factor because 5 is less likely to occur than 2.

Maximum power of 5 in n! = n/5 + n/5² + n/5³ +... (Consider integer values only)
Calculation:

75! = 75 × 74 × ... × 1

Maximum power of 5 in 75! = 75/5 + 75/5² = 15 + 3 = 18

79! = 79 × 78 × … × 1

Maximum power of 5 in 79! = 79/5 + 79/5² = 15 + 3 = 18

Here we consider integer values only and ignore the remainder.

The minimum power of 5 in both terms is 18.

∴ The number of zeros in 75! + 79! is 18.