Trailing Zeroes in 100!

24 zeros.

Trailing zeros come from factors of 10, and $10=2\times 5$10=2×5. We need both a factor of 2 and a factor of 5 for each trailing zero. Among the numbers 1 through 100, there are 50 even numbers but only 20 multiples of 5. Even accounting for higher powers (like 4, 8, 16... contributing extra 2s, and 25, 125... contributing extra 5s), factors of 2 vastly outnumber factors of 5. So the count of trailing zeros equals the count of factors of 5.

Every 5th number contributes at least one factor of 5: 5, 10, 15, 20, ..., 100. There are $100÷5=20$100÷5=20 such numbers.

However, some numbers contribute more than one factor of 5. Every 25th number (25, 50, 75, 100) contributes an additional factor of 5 since $25=5^2$25=52. There are $100÷25=4$100÷25=4 such numbers.

Higher powers of 5 like $125=5^3$125=53 would contribute a third factor of 5, but $125>100$125>100, so there are no such numbers in $100!$100!.

The total count is $20+4=24$20+4=24 factors of 5, which means 24 trailing zeros.