M31-23

Question

How many positive integers less than 10,000 have a product of their digits equal to 30?

A. 12 
B. 24
C. 36
D. 38 
E. 50

Bunuel · Accepted Answer

Official Solution:

How many positive integers less than 10,000 have a product of their digits equal to 30?

A. 12 
B. 24
C. 36
D. 38 
E. 50

The number 30 can be expressed as the product of single digit numbers, excluding 1's, in only two ways:  2*3*5  or  6*5 .
 
Thus, we need to count the number of positive integers less than 10,000 that consist of the digits {2, 3, 5} or {5, 6}, accompanied with any necessary number of 1's.
 
 For 2-digit numbers: 
 
Composed of digits {5, 6}, there are 2 combinations: 56 and 65.
 
 For 3-digit numbers: 
 
Composed of digits {1, 5, 6}, there are 3! (or 6) combinations: 156, 165, 516, 561, 615, and 651.
 
Composed of digits {2, 3, 5}, there are also 3! (or 6) combinations.
 
 For 4-digit numbers: 
 
Composed of digits {1, 1, 5, 6}, there are 4!/2! (or 12) combinations.
 
Composed of digits {1, 2, 3, 5}, there are 4! (or 24) combinations.
 
Adding these up, the total is  2 + 6 + 6 + 12 + 24 = 50 .

Answer: E

Senthil7 · Answer

I think this is a  high-quality  question and I agree with explanation.

SemperLiberi · Answer

Just to clarify:

For {1, 1, 5, 6} - the number of combinations = 4!/2! = 12. you choose 2! because two of the four places would produce the same number?

Bunuel · Answer

Yes.

THEORY:

Permutations of  n  things of which  P_1  are alike of one kind,  P_2  are alike of second kind,  P_3  are alike of third kind ...  P_r  are alike of  r_{th}  kind such that:  P_1+P_2+P_3+..+P_r=n  is:

\frac{n!}{P_1!*P_2!*P_3!*...*P_r!} .

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is  \frac{6!}{2!2!} , as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be  \frac{9!}{4!3!2!} .

Shiprasingh1100 · Answer

I think this is a  high-quality  question and I agree with explanation. This is a very tricky question and a true rep of GMAT type Qs. Thanks, Bunuel.

saranasser · Answer

in (1, 1 , 5, 6) why it is a permutation while the order does not matter ???

Bunuel · Answer

We are interested in numbers less than 10,000 such that the product of their digits is 30.

One combination which gives the product of 30 is (1, 1 , 5, 6). But with this combination gives different numbers, isn't it? 1156 is different from 6511 and each of them has the product of their digits equal to 30. Thus we need all numbers which we can get from each group.

Hope it helps.

danklin · Answer

I think this is a  high-quality  question and I agree with explanation.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

bumpbot · Answer

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M31-23

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