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If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink]
 10 Dec 2010, 07:20
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21% (02:18) correct
78% (01:33) wrong based on 17 sessions
If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest value of n? A. 1 B. 2 C. 3 D. 4 E. 5
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did it a hard way, my approach wont help anyone, lets waits for quant geeks
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aalriy wrote: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest value of n?
A. 1
B. 2
C. 3
D. 4
E. 5 So we should determine the # of trailing zeros 10! - 2*(5!)^2 has (trailing zeros are a sequence of 0s in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow). Note that every 5 and 2 in prime factorization will give one more trailing zero. 10!=5!*6*7*8*9*10=5!*(2*3)*7*(4*2)*9*(5*2)=5!*(2*3*4*5)*(7*2*9*2)*=252*(5!)^2; 252*(5!)^2-2*(5!)^2=250*(5!)^2=250*120*120=5^2*12*12*1,000 --> 5^2 with two 2-s from 12 will provide with 2 more zeros, so 10! - 2*(5!)^2 has total of 5 trailing zeros. So, max value of n for which above is divisible by 10^n is 5. Answer: E. P.S. Final value is 252*(5!)^2-2*(5!)^2=250*(5!)^2=250*120*120=1,000*25*12*12=1,000*300*12=12*300,000=3,600,000. Hope it's clear.
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Bunuel, I don't understand: the number of trailing zeros for 10! is 2 according to the formula you explained there: gmat-club-m12-100599.htmlThe number of trailing zeros for (5!)^2 is 2 (one for 5!, again, according to the same formula and common sense). So the difference of two numbers should give 2 trailing zeros: xxx00 - yyy00 -------- zzz00 Could you please explain again? Thank you.
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nonameee wrote: Bunuel, I don't understand: the number of trailing zeros for 10! is 2 according to the formula you explained there: gmat-club-m12-100599.htmlThe number of trailing zeros for (5!)^2 is 2 (one for 5!, again, according to the same formula and common sense). So the difference of two numbers should give 2 trailing zeros: xxx00
-
yyy00
--------
zzz00
Could you please explain again? Thank you. The red part is not correct. Consider this: 123,000 -23,000 ------- 100,000 By the way: 10!-2*(5!)^2=3,628,800-28,800=3,600,000 - five trailing zeros.
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Yes, you are right. I will read your explanation above. Thank you.
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink]
21 Dec 2012, 01:07
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aalriy wrote: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest value of n?
A. 1
B. 2
C. 3
D. 4
E. 5 =10*9*8*7*6*5! - 2*5!*5*4*3*2*1=5!*10*3*8(125)=5*(2*2)*3*2*(5*2)*24*5*5*5=5*2*5*2*5*2*5*2*24*5*3=10^4*24*5*3=10^4*5*2*12*3=10^5*36Answer: E
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink]
23 May 2013, 21:56
Can anyone make it a little easy ??
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink]
23 May 2013, 22:15
10! = 2^8*3^4*5^2*7= 2 zeros (10*5*2)
after rearranging 10! = 100*2^6*3^4*7
2*5!^2 = 2^7*3^2*5^2=2 zeroes(5*2*5*2)
after rearranging 100*2^5*3^2
now 10!-2*5!^2 = 100(2^6*3^4*7 - 2^5*3^2)
= 100*2^5*3^2(2*3^2*7 - 1)
=100*2^5*3^2*125
=100 *2^5*3^2 *5^3
= 10^5 *2^2*3^2
OA:E
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest
[#permalink]
23 May 2013, 22:15
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