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16 Sep 2014, 01:53
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Difficulty:
45%
(medium)
Question Stats:
63% (01:58) correct
38%
(02:15)
wrong
based on 104
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The product of the digits of the four-digit number \(h\) is 36. No two digits of \(h\) are identical. How many different numbers are possible values of \(h\)?
A. 6
B. 12
C. 24
D. 36
E. 48
A. 6
B. 12
C. 24
D. 36
E. 48
16 Sep 2014, 01:53
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Official Solution:
The product of the digits of the four-digit number \(h\) is 36. No two digits of \(h\) are identical. How many different numbers are possible values of \(h\)?
A. 6
B. 12
C. 24
D. 36
E. 48
We start with the prime factorization of \(h\), which is \(2^23^2\). In other words, \(h\) is the product of two 2's and two 3's. All the factors of 36, except for 1, can be constructed as products of some or all of these 2's and 3's: 2, 3, 4, 6, 9, 12, 18, and 36.
At this stage, we can see that the possible values of any digit of h only include 1, 2, 3, 4, 6, and 9. Any other digit would contain the wrong primes (5, 7) or too many 2's (8), or it would turn the product to zero (0).
We now test subgroups of four digits, imposing the conditions that the product of all four digits is 36 and that no two digits are the same. Start with one digit at a time; do not worry about the order of the digits yet.
9: If one of the digits is 9, then the product of the other three digits must be 4. The only possible sets of positive digits that have 3 members and multiply together to 4 are {2, 2, 1} and {4, 1, 1}, but these sets fail the condition that no two digits can be identical. Thus, 9 cannot be one of the digits of \(h\).
6: If one of the digits is 6, then the product of the other three digits must be 6. The only possible set of positive digits that have 3 non-identical members and multiply together to 6 is {3, 2, 1}, again ignoring order. Thus, one possible set of the digits of \(h\) is {6, 3, 2, 1}.
4: If one of the digits is 4, then the product of the other three digits must be 9. The only possible sets of positive digits that have 3 members and multiply together to 9 are {3, 3, 1} and {9, 1, 1}, but these sets fail the condition that no two digits can be identical. Thus, 4 cannot be one of the digits of \(h\).
If we examine the remaining digits 3, 2, and 1, we see that they can only be part of the unordered set {6, 3, 2, 1}, if we are to satisfy all the given conditions.
Thus, the possible values of \(h\) result from the rearrangement of these four digits. Since all the digits are distinct, the number of different rearrangements is simply \(4!\), or \((4)(3)(2)(1)\), which equals 24.
Answer: C
The product of the digits of the four-digit number \(h\) is 36. No two digits of \(h\) are identical. How many different numbers are possible values of \(h\)?
A. 6
B. 12
C. 24
D. 36
E. 48
We start with the prime factorization of \(h\), which is \(2^23^2\). In other words, \(h\) is the product of two 2's and two 3's. All the factors of 36, except for 1, can be constructed as products of some or all of these 2's and 3's: 2, 3, 4, 6, 9, 12, 18, and 36.
At this stage, we can see that the possible values of any digit of h only include 1, 2, 3, 4, 6, and 9. Any other digit would contain the wrong primes (5, 7) or too many 2's (8), or it would turn the product to zero (0).
We now test subgroups of four digits, imposing the conditions that the product of all four digits is 36 and that no two digits are the same. Start with one digit at a time; do not worry about the order of the digits yet.
9: If one of the digits is 9, then the product of the other three digits must be 4. The only possible sets of positive digits that have 3 members and multiply together to 4 are {2, 2, 1} and {4, 1, 1}, but these sets fail the condition that no two digits can be identical. Thus, 9 cannot be one of the digits of \(h\).
6: If one of the digits is 6, then the product of the other three digits must be 6. The only possible set of positive digits that have 3 non-identical members and multiply together to 6 is {3, 2, 1}, again ignoring order. Thus, one possible set of the digits of \(h\) is {6, 3, 2, 1}.
4: If one of the digits is 4, then the product of the other three digits must be 9. The only possible sets of positive digits that have 3 members and multiply together to 9 are {3, 3, 1} and {9, 1, 1}, but these sets fail the condition that no two digits can be identical. Thus, 4 cannot be one of the digits of \(h\).
If we examine the remaining digits 3, 2, and 1, we see that they can only be part of the unordered set {6, 3, 2, 1}, if we are to satisfy all the given conditions.
Thus, the possible values of \(h\) result from the rearrangement of these four digits. Since all the digits are distinct, the number of different rearrangements is simply \(4!\), or \((4)(3)(2)(1)\), which equals 24.
Answer: C
27 Jan 2015, 05:18
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Hi,
What I don't understand here is in this sentence "At this stage, we can see that the possible values of any digit of h only include 1, 2, 3, 4, 6, and 9. Any other digit would contain the wrong primes (5, 7) or too many 2's (8), or it would turn the product to zero (0)", why do we exclude 12 and 18? They have either 3,2,2 or 2,3,3 as their prime factors. Is it because they are already too large to be multiplied by any of the remaining factors and still yield 36?
What I don't understand here is in this sentence "At this stage, we can see that the possible values of any digit of h only include 1, 2, 3, 4, 6, and 9. Any other digit would contain the wrong primes (5, 7) or too many 2's (8), or it would turn the product to zero (0)", why do we exclude 12 and 18? They have either 3,2,2 or 2,3,3 as their prime factors. Is it because they are already too large to be multiplied by any of the remaining factors and still yield 36?
