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 = (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

j_shreyans wrote:Hi Brent ,

Thanks for the explanation one thing more i need to clear.

(5)(7) = 35

so 1,6,35

this above case can also be formed?

Please advise and correct me if i am wrong.

Thanks,

Shreyans

Hi Shreyans,

The question says that the product of the digits is 210
35 isn't a digit, so we need not consider that case.

Cheers,
Brent

j_shreyans wrote: ↑

Fri May 15, 2015 8:47 am

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

Solution:

Note that the prime factorization of 210 = 2 x 105 = 2 x 3 x 5 x 7.

Since no two-digit numbers multiply to 210, the integer must be either 3 digits or 4 digits (if it’s less than 10,000).

If it’s a 3-digit integer, then the 3 digits must be {5, 6, 7} (notice that 5 x 6 x 7 = 210). Since there are 3! = 6 ways to permute the 3 digits, there are six 3-digit integers such that the product of their digits is 210.

If it’s a 4-digit integer, then the 4 digits must be {2, 3, 5, 7} (notice that 2 x 3 x 5 x 7 = 210) or {1, 5, 6, 7} (notice that 1 x 5 x 6 x 7 = 210). Since there are 4! = 24 ways to permute the 4 digits, there are 24 four-digit integers in each group such that the product of their digits is 210. In total, there are 24 + 24 = 48 four-digit integers such that the product of their digits is 210.

Therefore, there are 6 + 48 = 54 such integers.

Answer: D