Problem:
How many zeros are at the end of the product
25 Γ 25 Γ 25 Γ 25 Γ 25 Γ 25 Γ 25 Γ 8 Γ 8 Γ 8 ? 25 \times 25 \times 25 \times 25 \times 25 \times 25 \times 25 \times 8 \times 8 \times 8?
2 5 Γ 2 5 Γ 2 5 Γ 2 5 Γ 2 5 Γ 2 5 Γ 2 5 Γ 8 Γ 8 Γ 8 ?
Answer Choices:
A. 3 33
B. 6 66
C. 9 99
D. 10 101 0
E. 12 121 2
Solution:
Since 2 Γ 5 = 10 2 \times 5=102 Γ 5 = 1 0 , each zero at the end of the product comes from a product of 2 22 and 5 55 in the prime factorization of the number. Since 25 = 5 Γ 5 25=5 \times 52 5 = 5 Γ 5 and 8 = 2 Γ 2 Γ 2 8=2 \times 2 \times 28 = 2 Γ 2 Γ 2 , it follows that there are fourteen factors of 5 55 and 9 99 factors of 2 22 . This yields 9 99 pairs of 2 Γ 5 2 \times 52 Γ 5 and results in 9 99 zeros at the end of the product.
OR \textbf{OR}
OR
Multiplying the given numbers using a calculator gives an answer equivalent to 3.125 Γ 1 0 12 3.125 \times 10^{12}3 . 1 2 5 Γ 1 0 1 2 which equals 3 , 125 , 000 , 000 , 000 3,125,000,000,0003 , 1 2 5 , 0 0 0 , 0 0 0 , 0 0 0 . This results in 9 99 zeros at the end of the product.
OR \textbf{OR}
OR
Factoring each number yields
2 Γ 2 Γ 2 β 8 Γ 2 Γ 2 Γ 2 β 8 Γ 2 Γ 2 Γ 2 β 8 Γ 5 Γ 5 β 25 Γ 5 Γ 5 β 25 Γ 5 Γ 5 β 25 Γ 5 Γ 5 β 25 Γ 5 Γ 5 β 25 Γ 5 Γ 5 β 25 Γ 5 Γ 5 β 25 . \begin{gathered}
\overbrace{2 \times 2 \times 2}^{8} \times \overbrace{2 \times 2 \times 2}^{8} \times \overbrace{2 \times 2 \times 2}^{8} \\
\times\ \underbrace{5 \times 5}_{25} \times \underbrace{5 \times 5}_{25} \times \underbrace{5 \times 5}_{25} \times \underbrace{5 \times 5}_{25} \times \underbrace{5 \times 5}_{25} \times \underbrace{5 \times 5}_{25} \times \underbrace{5 \times 5}_{25}.
\end{gathered}
2 Γ 2 Γ 2 β 8 β Γ 2 Γ 2 Γ 2 β 8 β Γ 2 Γ 2 Γ 2 β 8 β Γ 2 5 5 Γ 5 β β Γ 2 5 5 Γ 5 β β Γ 2 5 5 Γ 5 β β Γ 2 5 5 Γ 5 β β Γ 2 5 5 Γ 5 β β Γ 2 5 5 Γ 5 β β Γ 2 5 5 Γ 5 β β . β
Pairing each factor of 2 22 with a factor of 5 55 yields ( 2 Γ 5 ) 9 Γ 5 5 = 1 0 9 Γ ( (2 \times 5)^{9} \times 5^{5}=10^{9} \times(( 2 Γ 5 ) 9 Γ 5 5 = 1 0 9 Γ ( an odd number). Thus the product ends in nine zeros.
OR \textbf{OR}
OR
Since 25 Γ 25 Γ 8 = 25 Γ 200 = 5000 25 \times 25 \times 8=25 \times 200=50002 5 Γ 2 5 Γ 8 = 2 5 Γ 2 0 0 = 5 0 0 0 , regrouping yields
( 25 Γ 25 Γ 8 ) Γ ( 25 Γ 25 Γ 8 ) Γ ( 25 Γ 25 Γ 8 ) Γ 25 = 5000 Γ 5000 Γ 5000 Γ 25 = 125 , 000 , 000 , 000 Γ 25 = 3 , 125 , 000 , 000 , 000 \begin{aligned}
&(25 \times 25 \times 8) \times (25 \times 25 \times 8) \times (25 \times 25 \times 8) \times 25 \\
&= 5000 \times 5000 \times 5000 \times 25 \\
&= 125,000,000,000 \times 25 \\
&= 3,125,000,000,000
\end{aligned}
β ( 2 5 Γ 2 5 Γ 8 ) Γ ( 2 5 Γ 2 5 Γ 8 ) Γ ( 2 5 Γ 2 5 Γ 8 ) Γ 2 5 = 5 0 0 0 Γ 5 0 0 0 Γ 5 0 0 0 Γ 2 5 = 1 2 5 , 0 0 0 , 0 0 0 , 0 0 0 Γ 2 5 = 3 , 1 2 5 , 0 0 0 , 0 0 0 , 0 0 0 β
Answer: C \boxed{C}C β .
The problems on this page are the property of the MAA's American Mathematics Competitions