Find the trailing zeroes in the product of first 100 multiple of 15.

Trailing zeroes are determined by factors of 10 = 2 × 5. The first 100 multiples of 15 are 15, 30, 45, ..., 1500. Factoring: 15k = 3 × 5 × k for k = 1 to 100. We need to count total factors of 2 and 5 in the product (15 × 30 × 45 × ... × 1500). Since 15 = 3 × 5, the product is 15^100 × 100! = (3 × 5)^100 × 100!. Counting factors of 5: from 15^100 we get 100 factors of 5, plus from 100! we get ⌊100/5⌋ + ⌊100/25⌋ + ⌊100/125⌋ = 20 + 4 + 0 = 24 factors, totaling 124. Counting factors of 2 from 100! gives ⌊100/2⌋ + ⌊100/4⌋ + ⌊100/8⌋ + ⌊100/16⌋ + ⌊100/32⌋ + ⌊100/64⌋ = 50 + 25 + 12 + 6 + 3 + 1 = 97. The minimum is 124 factors of 5, so 124 trailing zeroes.

Step-by-step Derivation:
Step 1: Express the product. First 100 multiples of 15 = 15·1, 15·2, ..., 15·100 = 15^100 × (1 × 2 × 3 × ... × 100) = 15^100 × 100!

Step 2: Factor into primes. 15^100 = (3 × 5)^100 = 3^100 × 5^100

Step 3: Count factors of 5 in the product.

• From 15^100: 100 factors of 5
• From 100!: ⌊100/5⌋ + ⌊100/25⌋ + ⌊100/125⌋ = 20 + 4 + 0 = 24 factors of 5
• Total factors of 5: 100 + 24 = 124

Step 4: Count factors of 2 in the product.

• From 15^100: 0 factors of 2 (15 is odd)
• From 100!: ⌊100/2⌋ + ⌊100/4⌋ + ⌊100/8⌋ + ⌊100/16⌋ + ⌊100/32⌋ + ⌊100/64⌋ = 50 + 25 + 12 + 6 + 3 + 1 = 97 factors of 2
• Total factors of 2: 0 + 97 = 97

Step 5: Trailing zeroes = min(factors of 2, factors of 5) = min(97, 124) = 124