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# M26-28

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Math Expert
Joined: 02 Sep 2009
Posts: 50623

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16 Sep 2014, 00:26
2
25
00:00

Difficulty:

95% (hard)

Question Stats:

45% (01:59) correct 55% (01:58) wrong based on 191 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
Joined: 02 Sep 2009
Posts: 50623

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16 Sep 2014, 00:26
10
1
5
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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Manager
Joined: 17 May 2016
Posts: 67
Concentration: Finance, International Business
GMAT 1: 720 Q49 V39
GPA: 3.7
WE: Analyst (Investment Banking)

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22 May 2016, 10:23
Would anyone perhaps have any similar questions for practice? Many many thanks!
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Please kindly +Kudos if my posts or questions help you!

My debrief: Self-study: How to improve from 620(Q39,V36) to 720(Q49,V39) in 25 days!

Intern
Joined: 08 Apr 2017
Posts: 6

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01 Oct 2017, 21:47
How do we arrive at 00-32=62 in the above given solution
Intern
Joined: 08 Apr 2017
Posts: 6

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01 Oct 2017, 21:51
In the above given solution how do we arrive at 00-38=62 ? Please help.
Math Expert
Joined: 02 Sep 2009
Posts: 50623

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01 Oct 2017, 21:57
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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Joined: 08 Apr 2017
Posts: 6

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01 Oct 2017, 22:02
Got it. Thanks a lot.
Intern
Joined: 13 Oct 2017
Posts: 39

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28 Feb 2018, 02:26
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
Joined: 02 Sep 2009
Posts: 50623

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28 Feb 2018, 04:48
1
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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Intern
Joined: 13 Oct 2017
Posts: 39

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28 Feb 2018, 09:44
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
Joined: 05 Nov 2016
Posts: 88

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24 Aug 2018, 19:00
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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Kudos are always welcome ... as well your suggestions

Math Expert
Joined: 02 Sep 2009
Posts: 50623

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25 Aug 2018, 03:17
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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Manager
Joined: 29 May 2017
Posts: 103
Location: Pakistan
Concentration: Social Entrepreneurship, Sustainability

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26 Aug 2018, 07:00
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
Joined: 02 Sep 2009
Posts: 50623

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26 Aug 2018, 11:00
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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Re: M26-28 &nbs [#permalink] 26 Aug 2018, 11:00
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# M26-28

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