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Bunuel
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M26-28 10 Jun 2026, 12:40
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dhruva09
Hi Bunuel. I had a doubt with my solution. I tried solving this by factoring-
(8!)^6((8!)^4-1)/(8!)^3((8!)^2-1).

I approximated 1 to be negligible and proceede, only to end up with (8!)^2 as a solution x/(8!)^3.
Can you please confirm, what did I go wrong with.
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Your factoring is fine, but the mistake is approximating 1 as negligible. This is an exact digit question, so no term can be ignored.
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M26-28 11 Jun 2026, 07:22
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I don’t quite agree with the solution. Value of 8! is 40320 ( only one trailing zero.) Hence 40320 - 38 = 40282.
So the product of last two digits is 8 X 2 = 16.
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M26-28 11 Jun 2026, 13:29
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VenkataSai
I don’t quite agree with the solution. Value of 8! is 40320 ( only one trailing zero.) Hence 40320 - 38 = 40282.
So the product of last two digits is 8 X 2 = 16.
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You are subtracting 38 from 8!, but the expression is (8!)^2 - 38, not 8! - 38. Pleaser review carefully. Hope it helps.
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M26-28 26 Jul 2026, 02:09
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Bunuel

I took (8!)^10 as a very big number compared to (8!)^6 thus ignored (8!)^6 and (8!)^3

Have seen such question in the past where it did not ask for approximate value but still did this. I might not be able to find it but you definitely would be aware

How do we differentiate those from this

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M26-28 26 Jul 2026, 04:02
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Bunuel

I took (8!)^10 as a very big number compared to (8!)^6 thus ignored (8!)^6 and (8!)^3

Have seen such question in the past where it did not ask for approximate value but still did this. I might not be able to find it but you definitely would be aware

How do we differentiate those from this

Bunuel
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Please check this post: https://gmatclub.com/forum/m26-184467.html#p3769521
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KarishmaB
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M26-28 26 Jul 2026, 23:17
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Bunuel
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
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\(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}\)

So \(\frac{x}{(8!)^3} = \frac{1}{(8!)^3} =\frac{(8!)^{10}-(8!)^6}{(8!)^{8}-(8!)^6} = \frac{(8!)^{10}-(8!)^6}{(8!)^{8}-(8!)^6} = \frac{(8!)^{4}-1}{(8!)^{2}-1}\)

Using difference of squares, we know \((8!)^{4}-1 = ((8!)^{2}+1)((8!)^{2}-1)\)

Hence, \(\frac{x}{(8!)^3} = ((8!)^{2}+1)\)

We know that 8! will end with a 0 since there is a 5 and a 2 in 8!. So \((8!)^2\) will end with two 0s. When we add 1 to it, we will get something like XXX..X01

\(\frac{x}{(8!)^3}-39 = XXX..X01 - 39 = XXX...Y62\)

So product of units and tens digits is 2*6 = 12. Answer (D)

Approximation is a great tool but we cannot use it here. Why? Because the question is asking about the units and tens digits.
We know that x = 10001 can easily be approximated to x=10000 in most cases but if the question asks us the value of the units digit of x, we cannot answer 0. Hence we need to find the exact values of units and tens digits here.
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