How many positive integers less than 10,000 are such that the product

less than 10,000 means it has to be less than 5 digits.

1) 4 digits
-------------
210 = 2*3*5*7 ... total 24 ways
210 = 1*6*5*7 ... total 24 ways

2) 3 digits
----------
6 * 5 * 7 .. total 6 ways.

hence total 54 ways...

210=21*10=7*3*2*5

7325 can be arranged in 4! ways
3*2=6
765 can be arranged in 3! ways.
And
7651 can be arranged in 4! ways

Total=2*4!+3!=2*24+6=48+6=54

Ans: "D"

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
\(4!=24\)

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

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 is the answer when 2, 3, 5 and 7 are multiplied. 210 can also be arrive using 5,6 and 7 and 1, 5, 6 and 7.

So sum of arrangements of 2357, 567 and 1567. This translates to 4! +3! + 4!, this equals to 24 + 6 + 24 = 54, D is the answer.

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?

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!

4!+4!+3! =54

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

Merging topics.

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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

I love your sharing,i learn a lot from your post.There is a lot of very useful knowledge in your post.

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.

Posted from my mobile device

210 = (2)(3)(5)(7)

We need to consider 3 cases:

Case 1: 4-digit numbers using 2, 3, 5, 7
There are 4 digits, so this can be accomplished in 4! (24) ways

Aside: Notice that (2)(3) = 6

Case 2: 4-digit numbers using 1, 6, 5, 7
There are 4 digits, so this can be accomplished in 4! (24) ways

Case 3: 3-digit numbers using 6, 5, 7
There are 3 digits, so this can be accomplished in 3! (6) ways

Add up all 3 cases to get 24 + 24 + 6 = 54

So, the answer is D

Cheers,
Brent

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.

Posted from my mobile device

Hi All,

To answer this question, you have to be very careful to be thorough (2 of the answers are "partial work" answers, meaning that could very easily get to one of those answers and think that you're correct, but you're not):

We're dealing with numbers less than 10,000 and we need the product of the digits to equal 210, so let's deal with THAT first:

210 = (2)(3)(5)(7)

With THOSE 4 digits (2,3,5,7), we could end up with 4x3x2x1 = 24 different numbers

But the question DID NOT state that the digits had to be prime, so that's NOT the only way to get to 210 using 4 digits:

We could use (1,5,6,7), and end up with 4x3x2x1 = 24 additional numbers

AND we have to consider numbers that are NOT 4-digits; we could use (5,6,7), which gives us 3x2x1 = 6 additional numbers

24 + 24 + 6 = 54

Final Answer:

GMAT assassins aren't born, they're made,
Rich

If you use 2 * 3 * 5 * 7 * 1 as 210's factorization , then you have to choose 4 different numbers from a pool of 5 numbers and create integers (between 0000 and 9999<10000) meaning 5P2 = 5*4*3*2 = 120 integers.

BUT

One such case is 7152. But 7*1*5*2 = 70 not 210. so you can't use that method.

You can only use 1 when 6 is used instead of 2*3. Because 6 * 5 * 7 * 1 = 210

If you use 2 * 3 * 5 * 7 * 1 as 210's factorization , then you have to choose 4 different numbers from a pool of 5 numbers and create integers (between 0000 and 9999<10000) meaning 5P2 = 5*4*3*2 = 120 integers.

BUT

One such case is 7152. But 7*1*5*2 = 70 not 210. so you can't use that method.

You can only use 1 when 6 is used instead of 2*3. Because 6 * 5 * 7 * 1 = 210

210 = 1*2*3*5*7 = 1*5*6*7

3-digit numbers consisting of 5,6,7 = 3! = 6
4-digit numbers consisting of 2,3,5,7 = 4! = 24
4-digit numbers consisting of 1,6,5,7 = 4! = 24

Total numbers = 54

IMO D