Bunuel wrote:
Official Solution:
If \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}\), what is the product of the tens and the units digits of \(\frac{x}{(8!)^3}-39\)?
A. 0
B. 6
C. 7
D. 12
E. 14
Apply \(a^2-b^2=(a-b)(a+b)\): \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}=\frac{((8!)^{5}-(8!)^3)((8!)^{5}+(8!)^3)}{(8!)^{5}-(8!)^3}=(8!)^{5}+(8!)^3\).
Next, \(\frac{x}{(8!)^3}-39=\frac{(8!)^{5}+(8!)^3}{(8!)^3}=\frac{(8!)^{5}}{(8!)^3}+\frac{(8!)^{3}}{(8!)^3}-39=(8!)^2+1-39=(8!)^2-38\).
Now, since \(8!\) has 2 and 5 as its multiples, then it will have 0 as the units digit, so \((8!)^2\) will have two zeros in the end, which means that \((8!)^2-38\) will have \(00-38=62\) as the last digits: \(6*2=12\).
Answer: D
why have we used 2 and 5 in the statement.I solved this question till this step,and did not know what to do further:
Now, since \(8!\) has 2 and 5 as its multiples, then it will have 0 as the units digit, so \((8!)^2\) will have two zeros in the end.
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Thanks,
Ankit
Target Score:730+
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