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# Prime Factors

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Intern
Joined: 03 May 2003
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Prime Factors [#permalink]  21 Aug 2003, 07:04

If m = (2^x)(5^y)(7^z) and both 350 and 280 are factors of m, what is the minimum value of xyz ?
(A) 700
(B) 70
(C) 24
(D) 6
(E) 5
Intern
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My attempt [#permalink]  21 Aug 2003, 07:12
Find prime factors of 350 & 280

350
35 10
7 5 2 5 ------> 2^1 * 5^2 * 7^1

280
28 10
4 7 2 5
2 2 7 2 5 ------> 2^3 * 5^1 * 7^1

so least common is (2^3)(5^2)(7^1) --> m = 1400
x=3
y=2
z=1

xyz = 6

?? However its not the answer given??
Manager
Joined: 10 Jun 2003
Posts: 210
Location: Maryland
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Re: My attempt [#permalink]  21 Aug 2003, 08:29
exy18 wrote:
Find prime factors of 350 & 280

350
35 10
7 5 2 5 ------> 2^1 * 5^2 * 7^1

280
28 10
4 7 2 5
2 2 7 2 5 ------> 2^3 * 5^1 * 7^1

so least common is (2^3)(5^2)(7^1) --> m = 1400
x=3
y=2
z=1

xyz = 6

?? However its not the answer given??

I arrived at the exact same answer that you did...xyz=6. I believe the answer key is wrong here. Does the book you got it from explain answers?
Manager
Joined: 02 Jul 2003
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B.=70

share 7x5x2=70

Could be wrong

Rich
Manager
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Location: Maryland
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rich28 wrote:
B.=70

share 7x5x2=70

Could be wrong

Rich

As a check, (2^3)(5^2)(7^1) --> m = 1400

1400/350 = 4
1400/280 = 5

So xyz=6 is definitely valid. Unless there is some smaller value, then this is correct. What does the book say is the correct answer?
Intern
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Official Ans [#permalink]  21 Aug 2003, 15:48
According to PR (hard math set #3) the official answer is 70.
The explination they give is not helpful at all.
I think its probably a typo.

Thanks for you help!
t-2 and counting
CEO
Joined: 15 Aug 2003
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Re: Official Ans [#permalink]  24 Aug 2003, 22:13
exy18 wrote:
According to PR (hard math set #3) the official answer is 70.
The explination they give is not helpful at all.
I think its probably a typo.

Thanks for you help!
t-2 and counting

What is the official explanation for the problem ?

Thanks
Praetorian
Re: Official Ans   [#permalink] 24 Aug 2003, 22:13
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