If x = 2¹⁰ × 5⁶ then how many zeros will be there at the end of x?

Question

If x = 2¹⁰ × 5⁶ then how many zeros will be there at the end of x?

1. A. 6 (Correct answer)
2. B. 5
3. C. 4
4. D. 3

Correct Answer

Option A — A

Detailed Solution & Explanation

The correct answer is Option A.

Key Points

• The number of trailing zeros equals the number of times 10 divides the number, and 10 = 2 × 5. So count the pairs of 2 and 5.
• Here x = 2^10 × 5^6. The number of 5s is 6 and the number of 2s is 10. Each trailing zero needs one 2 and one 5, so the count is limited by the smaller power: min(10, 6) = 6.
• So x ends in 6 zeros.

Exam Tip

• trailing zeros = min(power of 2, power of 5).
• The 5s are usually the limiting factor, which is why factorial trailing-zero problems only count 5s.

Additional Information

• Trailing zeros come from factors of 10 = 2 × 5, so the count is the smaller of the powers of 2 and 5 — here min(10, 6) = 6.
• The same idea drives the classic factorial question: zeros at the end of n! are found by adding ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ …, because 5s are always scarcer than 2s.
• For 100! that gives 20 + 4 = 24 trailing zeros.