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16 Sep 2014, 01:26
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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
A. 0
B. 6
C. 7
D. 12
E. 14
16 Sep 2014, 01:26
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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
We can simplify the expression for \(x\) by using the identity \(a^2-b^2=(a-b)(a+b)\), which gives \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}=\frac{((8!)^{5}-(8!)^3)((8!)^{5}+(8!)^3)}{(8!)^{5}-(8!)^3}=(8!)^{5}+(8!)^3\).
Substituting this value into the expression for \(\frac{x}{(8!)^3}-39\), we get \(\frac{x}{(8!)^3}-39=\frac{(8!)^{5}+(8!)^3}{(8!)^3} - 39=\frac{(8!)^{5}}{(8!)^3}+\frac{(8!)^{3}}{(8!)^3}-39=(8!)^2+1-39=(8!)^2-38\).
Since \(8!\) is divisible by both 2 and 5, its last digit is 0, and hence \((8!)^2\) will have at least two trailing zeros. Thus, \((8!)^2-38\) ends with the digits \(...00 - 38 = ...62\). Therefore, the product of the tens and units digits of \(\frac{x}{(8!)^3}-39\) is \(6*2=12\).
Answer: D
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
We can simplify the expression for \(x\) by using the identity \(a^2-b^2=(a-b)(a+b)\), which gives \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}=\frac{((8!)^{5}-(8!)^3)((8!)^{5}+(8!)^3)}{(8!)^{5}-(8!)^3}=(8!)^{5}+(8!)^3\).
Substituting this value into the expression for \(\frac{x}{(8!)^3}-39\), we get \(\frac{x}{(8!)^3}-39=\frac{(8!)^{5}+(8!)^3}{(8!)^3} - 39=\frac{(8!)^{5}}{(8!)^3}+\frac{(8!)^{3}}{(8!)^3}-39=(8!)^2+1-39=(8!)^2-38\).
Since \(8!\) is divisible by both 2 and 5, its last digit is 0, and hence \((8!)^2\) will have at least two trailing zeros. Thus, \((8!)^2-38\) ends with the digits \(...00 - 38 = ...62\). Therefore, the product of the tens and units digits of \(\frac{x}{(8!)^3}-39\) is \(6*2=12\).
Answer: D
General Discussion
