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M26-28

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M26-28 [#permalink]

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New post 16 Sep 2014, 00:26
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Question Stats:

32% (00:02) correct 68% (01:50) wrong based on 19 sessions

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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
[Reveal] Spoiler: OA

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Re M26-28 [#permalink]

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New post 16 Sep 2014, 00:26
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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


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
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Re: M26-28 [#permalink]

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New post 22 May 2016, 10:23
Would anyone perhaps have any similar questions for practice? Many many thanks!
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Re M26-28 [#permalink]

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New post 01 Oct 2017, 21:47
How do we arrive at 00-32=62 in the above given solution

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Re: M26-28 [#permalink]

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New post 01 Oct 2017, 21:51
In the above given solution how do we arrive at 00-38=62 ? Please help.

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New post 01 Oct 2017, 21:57

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Re: M26-28 [#permalink]

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New post 01 Oct 2017, 22:02
Got it. Thanks a lot.

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Re: M26-28   [#permalink] 01 Oct 2017, 22:02
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