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
(A) 24
(B) 30
(C) 48
(D) 54
(E) 72
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Edited:GmatKiss 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
210 = 2*3*5*7.GmatKiss 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
Good catch - I completely missed the fact that 2x3 creates a new digit. My badGMATGuruNY wrote:210 = 2*3*5*7.GmatKiss 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
Case 1: 4-digit integers composed of the digits 2, 3, 5 and 7.
Number of ways to arrange these 4 digits = 4! = 24.
Case 2: 4-digit integers composed of the digits 1, 5, 6, and 7.
Number of ways to arrange these 4 digits = 4! = 24.
Case 3: 3-digit integers composed of the digits 5, 6 and 7.
Number of ways to arrange these 3 digits = 3! = 6.
Total ways = 24+24+6 = 54.
The correct answer is D.
A video explanation is provided in this linkBrent@GMATPrepNow wrote:Good catch - I completely missed the fact that 2x3 creates a new digit. My badGMATGuruNY wrote:210 = 2*3*5*7.GmatKiss 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
Case 1: 4-digit integers composed of the digits 2, 3, 5 and 7.
Number of ways to arrange these 4 digits = 4! = 24.
Case 2: 4-digit integers composed of the digits 1, 5, 6, and 7.
Number of ways to arrange these 4 digits = 4! = 24.
Case 3: 3-digit integers composed of the digits 5, 6 and 7.
Number of ways to arrange these 3 digits = 3! = 6.
Total ways = 24+24+6 = 54.
The correct answer is D.
Cheers,
Brent
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