How many trailing zeros are there in the product \(20 \times 25 \times 30 \times 35 \times 40\)?
Answer: B
To find the number of trailing zeros, we need to count the pairs of 2 and 5 in the prime factorization of the product.
\(20 = 2^2 \times 5^1\)
\(25 = 5^2\)
\(30 = 2 \times 3 \times 5^1\)
\(35 = 5 \times 7\)
\(40 = 2^3 \times 5^1\)
Total number of 2s = \(2+1+3 = 6\). Total number of 5s = \(1+2+1+1+1 = 6\). Wait, 35 is 5x7. Yes. 40 is 8x5. Yes. Total 5s is 6. Total 2s is 6. So there are 6 pairs. The answer should be 6. Let me re-count. 20 (one 5), 25 (two 5s), 30 (one 5), 35 (one 5), 40 (one 5). Total 5s = 1+2+1+1+1=6. Total 2s. 20 (two 2s), 30 (one 2), 40 (three 2s). Total 2s = 2+1+3=6. So number of zeros is 6. The answer key says 5. Why? Let me check again. 20x25 = 500. 30x35=1050. 40. So 500x1050x40. 5x105x4 x 10000. 20x105x10000. 2100 x 10000. 21 x 10^6. So 6 zeros. The answer key is wrong. I will correct it to C.