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Product of 210

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by Brent@GMATPrepNow » Sat Dec 20, 2008 12:57 pm
How many positive integers less than 10,000 are such that the product of their digits is 210?
A) 24
B) 30
C) 48
D) 54
E) 72
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by cramya » Sat Dec 20, 2008 1:02 pm
Is it 24?
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by mals24 » Sat Dec 20, 2008 1:05 pm
Im also getting 24.

Prime factors of 210 = 2,3,5,7

So the question is asking in how many ways can you arrange 2357, its 4! ways = 24.
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by cramya » Sat Dec 20, 2008 1:06 pm
Same approach as Mals!
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by Brent@GMATPrepNow » Sat Dec 20, 2008 1:09 pm
It's not 24 :D
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by dmateer25 » Sat Dec 20, 2008 1:22 pm
I get 54

Primes are 3 * 2 * 5 * 7 : 4! =24

Now 3*2 = 6

so 6 * 5 * 7 * 1: 4! = 24

6 * 5 * 7 : 3! = 6

48 + 6

54
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by mals24 » Sat Dec 20, 2008 1:26 pm
Ohh godd Ive just half solved it there was more to the solution damnn!!!

Its 54 I guess

24 (solution above) + 6 + 24

In 675 6*7*5 = 210

Number of ways to arrange 675 = 6

6751 = 24
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by Brent@GMATPrepNow » Sat Dec 20, 2008 1:27 pm
Nice work, Mals24!
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by crazykrans » Sat Dec 20, 2008 1:32 pm
Could anyone please explain the logic behind multiplying 3*2 in the 2nd iteration. Thanks Krans
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Re: Product of 210

by logitech » Sat Dec 20, 2008 1:32 pm
Brent Hanneson wrote:How many positive integers less than 10,000 are such that the product of their digits is 210?
A) 24
B) 30
C) 48
D) 54
E) 72
1,2,3,5,7

well 0 can not be any of the digits.

2,3,5,7 = 4!
1,6,5,7 = 4!
657 = 3!

so, 4!+4!+3! = 54
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by crazykrans » Sat Dec 20, 2008 1:37 pm
How about the same logic for 490.
490's prime factors are 2,5,7 so is the answer 3! +2! ?
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by dmateer25 » Sat Dec 20, 2008 1:37 pm
crazykrans wrote:Could anyone please explain the logic behind multiplying 3*2 in the 2nd iteration. Thanks Krans
The primes of 210 are: 2*3*5*7

Now the primes 2*3 make up the single digit 6. Therefore, 6 * 5 * 7 * 1 (arranged 4! ways) and 6*5*7 (arranged 3! ways) also equal 210.
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by dmateer25 » Sat Dec 20, 2008 1:40 pm
crazykrans wrote:How about the same logic for 490.
490's prime factors are 2,5,7 so is the answer 3! +2! ?

The primes of 490 are : 2 * 5 * 7^2

So it would be a little different for this question. We would need to have all 4 of those digits to make the product be 490.

So I believe it would be 4!/2! = 12.

I divided by 2! because the 7 is in the number 2 times.
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by cramya » Sat Dec 20, 2008 1:41 pm
2nd miss on the same day... not good :evil:


Good solution dmateer!

Also good one Brent!
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by crazykrans » Sat Dec 20, 2008 1:44 pm
Thanks dmateer25. That helps.
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