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Math Expert V
Joined: 02 Sep 2009
Posts: 58445

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28 00:00

Difficulty:   95% (hard)

Question Stats: 48% (01:52) correct 52% (01:50) wrong based on 232 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

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Math Expert V
Joined: 02 Sep 2009
Posts: 58445

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10
1
6
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: $$6*2=12$$.

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Would anyone perhaps have any similar questions for practice? Many many thanks!
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Intern  B
Joined: 09 Apr 2017
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How do we arrive at 00-32=62 in the above given solution
Intern  B
Joined: 09 Apr 2017
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In the above given solution how do we arrive at 00-38=62 ? Please help.
Math Expert V
Joined: 02 Sep 2009
Posts: 58445

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GeetikaR wrote:
How do we arrive at 00-32=62 in the above given solution

(8!)^2 = (1*2*3*4*5*6*7*8)^2 will have two zeros at the end. So, ...00 - 38 = ...62.
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Got it. Thanks a lot.
Intern  B
Joined: 13 Oct 2017
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Hi Bunuel,

I have a two part question:

1. How did you know to use a^2 - b^2 ...plus also how come the powers reduced by half when you used difference of squares i.e. from 8!^10 to *8!^5

2. I understand how the tens digit and the units digit will both be zero...and thus how 8!^2 - 38 = 00-38...I just don't know how you get 62...leading to 6*2=12.
Math Expert V
Joined: 02 Sep 2009
Posts: 58445

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ttaiwo wrote:
Hi Bunuel,

I have a two part question:

1. How did you know to use a^2 - b^2 ...plus also how come the powers reduced by half when you used difference of squares i.e. from 8!^10 to *8!^5

2. I understand how the tens digit and the units digit will both be zero...and thus how 8!^2 - 38 = 00-38...I just don't know how you get 62...leading to 6*2=12.

Consider easier examples:

1. $$x^{10} - y^{10} = (x^5)^2 - (y^5)^2 = (x^5 - y^5)(x^5 + y^5)$$

2. 100 - 38 = 62.
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Bunuel wrote:
ttaiwo wrote:
Hi Bunuel,

I have a two part question:

1. How did you know to use a^2 - b^2 ...plus also how come the powers reduced by half when you used difference of squares i.e. from 8!^10 to *8!^5

2. I understand how the tens digit and the units digit will both be zero...and thus how 8!^2 - 38 = 00-38...I just don't know how you get 62...leading to 6*2=12.

Consider easier examples:

1. $$x^{10} - y^{10} = (x^5)^2 - (y^5)^2 = (x^5 - y^5)(x^5 + y^5)$$

2. 100 - 38 = 62.

I should've seen 2....but I've genuinely haven't come across 1. before ...thanks Manager  S
Joined: 05 Nov 2016
Posts: 82

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Thanks @Bunnel for the great solution.

To elaborate further on the (8!)$$2$$−38 = 62. As mentioned earlier (8!)$$2$$has 00 in the end(i.e., units and tens are zeroes). We do have no concern for the third digit(so lets take it as x. where x lies from 0 - 9). So any combination gives the last two digits as 62. (200-38 = 162 or 500-38=462). Since we were asked only to find the product of last two digits its 6*2 = 12
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Math Expert V
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Thanks @Bunnel for the great solution.

To elaborate further on the (8!)$$2$$−38 = 62. As mentioned earlier (8!)$$2$$has 00 in the end(i.e., units and tens are zeroes). We do have no concern for the third digit(so lets take it as x. where x lies from 0 - 9). So any combination gives the last two digits as 62. (200-38 = 162 or 500-38=462). Since we were asked only to find the product of last two digits its 6*2 = 12

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Yes, that's correct.
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Bunuel wrote:
GeetikaR wrote:
How do we arrive at 00-32=62 in the above given solution

(8!)^2 = (1*2*3*4*5*6*7*8)^2 will have two zeros at the end. So, ...00 - 38 = ...62.

Hi...

can you point me to where i can learn about multiples giving trailing zeros?
Math Expert V
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Mansoor50 wrote:
Bunuel wrote:
GeetikaR wrote:
How do we arrive at 00-32=62 in the above given solution

(8!)^2 = (1*2*3*4*5*6*7*8)^2 will have two zeros at the end. So, ...00 - 38 = ...62.

Hi...

can you point me to where i can learn about multiples giving trailing zeros?

Everything about Factorials on the GMAT
Power of a Number in a Factorial Problems
Trailing Zeros Problems

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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Joined: 11 Jun 2017
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GMAT 1: 630 Q44 V33 GMAT 2: 680 Q47 V37 GPA: 3.2

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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: $$6*2=12$$.

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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The answer would be the same if 8! was replaced by any n!, given n>5.. Right? Re: M26-28   [#permalink] 10 Dec 2018, 23:28
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