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
