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Positive integers less than 10,000
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Updated on: 17 Mar 2019, 01:25
Updated on: 17 Mar 2019, 01:25
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Question Stats:
21% (01:50) correct
78% (01:58) wrong based on 139 sessions
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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
Re: Positive integers less than 10,000
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26 Apr 2020, 12:24
26 Apr 2020, 12:24
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Text Explanation
A great place to start is figuring out the prime factorization of 210:
210 = 21 * 10 = (7 * 3) * (5 * 2) = 2 * 3 * 5 * 7
We can now look at this prime factorization to help us determine which factors of 210 can be used as the digits of our positive integers.
It's also important to stop for a moment and think about what "digits" are. Digits are the integers from 0 through 9. We obviously know 2, 3, 5, and 7 are possible digits from our prime factorization above. But we also need to keep in mind the digit 1. Any number multiplied by 1 remains that number, so that might be a sneaky case to watch out for! Likewise, is there any way we could combine some of our prime factors to produce another digit? For example, 2 * 3 = 6, which is a digit. However, any other combination of prime factors will yield a two-digit number, which won't work. (For example, 2 * 5 = 10, which is too big to be a single digit.) Finally, we also know that 0 is NOT a possible digit because anything multiplied by 0 is 0 and therefore could not equal 210.
There are thus 3 different cases we need to consider:
Case i: We use the prime factorization as is to create unique 4-digit numbers consisting of a 2, 3, 5, and 7. (For example, 2357, 7523, and 3527.) To determine all the possibilities, we can just use the Fundamental Counting Principle. We'll have 4 options for our first digit, then 3 for the second, 2 for the third, and 1 for the last. In other words:
4 * 3 * 2 * 1 = 24 possibilities
Case ii: A factor of 210 that's also a digit is the number 6. This comes from the fact that we can multiply our prime factors of 2 and 3 to make 6. Thus, there are unique 3-digit numbers consisting of a 5, 6, and 7 whose product is 210. Once again using the FCP, we see that there are:
3 * 2 * 1 = 6 possibilities
Case iii: Lastly, we need to recognize that 1 is also a factor of every number and that any number multiplied by 1 remains that number. Thus, there are also unique 4-digit numbers consisting of a 1, 5, 6, and 7 whose product is 210. Using the FCP again, we see that there are:
4 * 3 * 2 * 1 = 24 possibilities
Combining all of the possibilities from each case, we see that the total possible number of positive integers whose digits make a product of 210 is:
24 + 6 + 24 = 54 possibilities
Note that we do NOT need to consider a fourth case where the digit 1 is included among the digits 2, 3, 5, and 7 to form unique 5-digit numbers. This case is ruled out due to the restriction of the problem that we are creating 4-digit numbers less than 10,000.
A great place to start is figuring out the prime factorization of 210:
210 = 21 * 10 = (7 * 3) * (5 * 2) = 2 * 3 * 5 * 7
We can now look at this prime factorization to help us determine which factors of 210 can be used as the digits of our positive integers.
It's also important to stop for a moment and think about what "digits" are. Digits are the integers from 0 through 9. We obviously know 2, 3, 5, and 7 are possible digits from our prime factorization above. But we also need to keep in mind the digit 1. Any number multiplied by 1 remains that number, so that might be a sneaky case to watch out for! Likewise, is there any way we could combine some of our prime factors to produce another digit? For example, 2 * 3 = 6, which is a digit. However, any other combination of prime factors will yield a two-digit number, which won't work. (For example, 2 * 5 = 10, which is too big to be a single digit.) Finally, we also know that 0 is NOT a possible digit because anything multiplied by 0 is 0 and therefore could not equal 210.
There are thus 3 different cases we need to consider:
Case i: We use the prime factorization as is to create unique 4-digit numbers consisting of a 2, 3, 5, and 7. (For example, 2357, 7523, and 3527.) To determine all the possibilities, we can just use the Fundamental Counting Principle. We'll have 4 options for our first digit, then 3 for the second, 2 for the third, and 1 for the last. In other words:
4 * 3 * 2 * 1 = 24 possibilities
Case ii: A factor of 210 that's also a digit is the number 6. This comes from the fact that we can multiply our prime factors of 2 and 3 to make 6. Thus, there are unique 3-digit numbers consisting of a 5, 6, and 7 whose product is 210. Once again using the FCP, we see that there are:
3 * 2 * 1 = 6 possibilities
Case iii: Lastly, we need to recognize that 1 is also a factor of every number and that any number multiplied by 1 remains that number. Thus, there are also unique 4-digit numbers consisting of a 1, 5, 6, and 7 whose product is 210. Using the FCP again, we see that there are:
4 * 3 * 2 * 1 = 24 possibilities
Combining all of the possibilities from each case, we see that the total possible number of positive integers whose digits make a product of 210 is:
24 + 6 + 24 = 54 possibilities
Note that we do NOT need to consider a fourth case where the digit 1 is included among the digits 2, 3, 5, and 7 to form unique 5-digit numbers. This case is ruled out due to the restriction of the problem that we are creating 4-digit numbers less than 10,000.
General Discussion
Re: Positive integers less than 10,000
[#permalink]
17 Mar 2019, 01:27
17 Mar 2019, 01:27
3
Expert Reply
The tag tanle1 b as the name Blank (i) suggests is ONLY for the verbal questions text completion. NOT for quant
210=1*2*3*5*7=1*6*5*7. (Only 2*3 makes the single digit 6).
So, four digit numbers with combinations of the digits {1,6,5,7} and {2,3,5,7} and three digit numbers with combinations of digits {6,5,7} will have the product of their digits equal to 210.
{1,6,5,7} # of combinations 4!=24
{2,3,5,7} # of combinations 4!=24
{6,5,7} # of combinations 3!=6
24+24+6=54.
Answer: D.
Regards
210=1*2*3*5*7=1*6*5*7. (Only 2*3 makes the single digit 6).
So, four digit numbers with combinations of the digits {1,6,5,7} and {2,3,5,7} and three digit numbers with combinations of digits {6,5,7} will have the product of their digits equal to 210.
{1,6,5,7} # of combinations 4!=24
{2,3,5,7} # of combinations 4!=24
{6,5,7} # of combinations 3!=6
24+24+6=54.
Answer: D.
Regards
