Question

Find the number of zeros at end of 10 × 15 × 20 × 25 × … × 195.
1. 35
2. 38
3. 39
4. 46

Answer 1

Correct Answer - Option 1 : 35

Given:

10 × 15 × 20 × 25 × … × 195.

Concept Used:

Number of zeros = Number of pairs of 2 × 5

n! = n(n – 1)(n – 2)…1

Maximum power of 5 in n! = n/5 + n/5² + n/5³ +... (Consider integer values only)
Maximum power of 2 in n! = n/2 + n/2² + n/2³ +... (Consider integer values only)

Formula Used:

n = (an – a)/d + 1 where a is first term, an is nth term and d is common difference

Calculation:

Number of terms in the AP series 10 × 15 × … × 195 = (195 –10)/5 + 1 = 38 

The given expression 10 × 15 × 20 × 25 × … × 195 = 5³⁸(2 × 3 × 4 × 5 × … × 39)

⇒ 5³⁸(1 × 2 × 3 × 4 × 5 × … × 39)

⇒ 5³⁸ × 39!

Power of 5 in 39! = 39/5 + 39/5² = 7 + 1 = 8

Power of 2 in 39! = 39/2 + 39/2² + 39/2³ + 39/2⁴ + 39/2⁵ = 19 + 9 + 4 + 2 + 1 = 35

In the given expression 5³⁸ × 39! maximum power of 5 is 46 and maximum power of 2 is 35. So, there are 35 pairs of 2 × 5.

∴ The number of zeros is 35.