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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

210=1*2*3*5*7=1*6*5*7. (Only 2*3 makes the single digit 6).

So, four digit numbers with combinations of the digits {1,6,5,7} and {2,3,5,7} and three digit numbers with combinations of digits {6,5,7} will have the product of their digits equal to 210.

{1,6,5,7} # of combinations 4!=24 {2,3,5,7} # of combinations 4!=24 {6,5,7} # of combinations 3!=6

Re: How many positive integers less than 10,000 are such that the product [#permalink]

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30 Jun 2011, 02:34

1. read question carefully--it says no \(< 10,000\) that means \(10,000 < 4 digit > 999\)and 3 digit \(< 1000\)2. so now 210 has factors \(7,5,3,2,1\)

case 1 : four digit is possible with 7,5,3,2 because multiplication of digit\(=210\) \(4!= 24\) case 2 : we take 3X2=6 and then we can include 1 for four digit no, so no are 7,5,6,1 \(4!=24\)

Quote:

see other muliplication or cases cannot be included because multiplication goes to 2 digit no . ex \(7 X 3 =21\). which is not possible

case 3 : 3 digit no, we can only take 7,6,5 so \(3!=6\) adding all the case \(1,2,3= 54\)that is our answer D. see gmat will not go complicate these kind of question further so all the best
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how many positive integers less than 9999 are such that the product of their digits is 210.

A.24 B.30 C.48 D.56 E.72

Posted from my mobile device

The prime factorization of 210 is 2*3*5*7. So one way to make the right kind of number is to use those four digits, in any of the 4! = 24 orders you can put them in.

Notice though that we can also get 210 as a product by multiplying 5, 6 and 7. So we can make some 3-digit numbers with the right product: 3! = 6 of them to be exact.

But we can also get the right product using the digit 1 along with the digits 5, 6, and 7. Again we can arrange those digits in 4! = 24 orders.

So adding up the possible ways to make the right kinds of number, there are 24+24+6 = 54 ways. I think there might be a typo in your answer choices?
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Re: How many positive integers less than 10,000 are such that the product [#permalink]

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29 Sep 2013, 08:26

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Re: How many positive integers less than 10,000 are such that the product [#permalink]

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06 Sep 2014, 21:59

cvsmech 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

at first & got no answer for this question by missing the possibility for 3 digit number

prime factor of 210- 2 3 5 7 ... we could make 4! combination using this.. 6(2*3) , 5, 7 , 1 again 4! ... since number could be also three digits 6, 5 ,7 in 3!

how many positive integer less than 10,000 such that product of their digits is 210? ans- 54 how come??

hi anik1989, 210= 2*3*5*7.... now there cannot be any two digits numbers satisfying the condition, as only one set of the numbers (2,3) will give you a single digit... 1) three digits number will consist of 6,5,7.. ways to arrange these three digits = 3!=6.. 2) four digits number will consist of 2,3,5,7 or 1,6,5,7... each will have 4!=24.. TOTAL 24*2=48... TOTAL 1+2=6+48=54... Hope it helped
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Re: How many positive integers less than 10,000 are such that the product [#permalink]

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12 Mar 2016, 03:49

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Re: How many positive integers less than 10,000 are such that the product [#permalink]

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20 Jul 2016, 05:14

I have a question regarding this, I know it will be so basic, but I don't understand why 2x3x5x7x1 is not an option and 6x5x7x1 it is. I mean, why we count 1 in certain options and not in others? Thanks.

Re: How many positive integers less than 10,000 are such that the product [#permalink]

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20 Jul 2016, 05:49

I'm sorry but I have to insist. I know that steps, but I don't know the reason why the '1' is not included in the first prime factorisation 2x3x5x7, and we include it in 6x5x7x1, making that different from 6x5x7 (the '1' gives us two different factorisation). I don't know if I explain my request well.

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11 Aug 2017, 03:41

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