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Magoosh GMAT Instructor
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Let abcd be a general fourdigit number. How many odd fourdigits nu [#permalink]
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27 Feb 2017, 17:13
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Let abcd be a general fourdigit number. How many odd fourdigits numbers abcd exist such that the four digits are all distinct, no digit is zero, and the product of a and b is the twodigit number cd?
(A) 4 (B) 6 (C) 12 (D) 24 (E) 36This is one of the easier questions from a bank of 15 challenging questions. You can see the whole set, as well as the OE for this question, here: Challenging GMAT Math Practice QuestionsMike
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Let abcd be a general fourdigit number. How many odd fourdigits nu [#permalink]
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27 Feb 2017, 22:46
mikemcgarry wrote: Let abcd be a general fourdigit number. How many odd fourdigits numbers abcd exist such that the four digits are all distinct, no digit is zero, and the product of a and b is the twodigit number cd?
(A) 4 (B) 6 (C) 12 (D) 24 (E) 36This is one of the easier questions from a bank of 15 challenging questions. You can see the whole set, as well as the OE for this question, here: Challenging GMAT Math Practice QuestionsMike Odd integer : 1, 3, 5, 7 and 9 a b c d\(a*b=cd\) \(7*9=63\) \(9*7=63\) \(9*3=27\) \(3*9=27\) \(7*3=21\) \(3*7=21\) Answer : 6
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Re: Let abcd be a general fourdigit number. How many odd fourdigits nu [#permalink]
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09 Apr 2017, 11:37
Hi Mike,
Is 3*5=15 not a possible case?



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Let abcd be a general fourdigit number. How many odd fourdigits nu [#permalink]
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09 Apr 2017, 13:48
kavyagupta wrote: Hi Mike,
Is 3*5=15 not a possible case? That would be a repeated number (2 5s), which is not allowed. Kudos if you agree!



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Let abcd be a general fourdigit number. How many odd fourdigits nu [#permalink]
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18 Apr 2017, 05:39
Lets list out all the constraints first: d odd / a,b,c,d distinct and Non Zero (range 1 to 9) / a x b = [c] [d]
Based on given constraints, let's derive: a x b > 10 , distinct and odd. Therefore, both a and b are odd numbers.
Now lets draw out combinations basis above: a cannot be 1 as max b is 9. So product will be less than 10. a=3 ; b=5 or 7 or 9. Numbers formed 3515 (non distinct) / 3721 / 3927 = 2 numbers a=5 ; b=3 or 7 or 9. Numbers formed 5315 (non distict) / 5735 (non distict) / 5945 (non distinct). = 0 number a=7 ; b=3 or 5 or 9. Numbers formed 7321 / 7535 (non distict) / 3927. = 2 numbers a=9 ; b=3 or 5 or 7. Numbers formed 9327 / 9545 (non distict) / 9763. = 2 numbers
Total possible 4digit numbers = 6 Hence, Correct choice B



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Let abcd be a general fourdigit number. How many odd fourdigits nu [#permalink]
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17 May 2017, 09:56
Thanks mikemcgarry for the good question. Could you please point on mistake in my solution? I don't understand why a, b, c can't be even? If abcd is odd, it means that d is odd. But the product cd can be even, if c is even. It gives us combinations (the last digit in the sets is c; a,b have 2 options): d=1: 2,4,8; 2,3,6; d=3: 6,4,8; 1,6,2; 2,6,4; 2,9,6 d=5: d=7  no options; d=9: 3,6,2; So, we have 7 options and applying permutations we are getting 4 options for each of the numbers (2*2*1*1): 7*4=28. Please point on my mistake.



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Re: Let abcd be a general fourdigit number. How many odd fourdigits nu [#permalink]
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17 May 2017, 10:41
StaicyT wrote: Thanks mikemcgarry for the good question. Could you please point on mistake in my solution? I don't understand why a, b, c can't be even? If abcd is odd, it means that d is odd. But the product cd can be even, if c is even. It gives us combinations (the last digit in the sets is c; a,b have 2 options): d=1: 2,4,8; 2,3,6; d=3: 6,4,8; 1,6,2; 2,6,4; 2,9,6 d=5: d=7  no options; d=9: 3,6,2; So, we have 7 options and applying permutations we are getting 4 options for each of the numbers (2*2*1*1): 7*4=28. Please point on my mistake. Dear StaicyT, I'm happy to respond. I think what you may be missing is what exactly is denoted by " cd." Consider the very end of the prompt: . . . and the product of a and b is the twodigit number cd?Thus, " cd" does NOT denote a product of two single digits, but instead a single twodigit number. Thus, if c = 8 and d = 1, " cd" is NOT the product (8)(1) = 8, but the number 81. Obviously, we can't have two different single digits numbers that have a product of 81. If d is odd, then the twodigit number " cd" has to be odd. This, in turn, means that a & b must be odd, because their product is an odd twodigit number. It is possible for c to be even: for example, the number 3721 satisfies the conditions of the question. Does all this make sense? Mike
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Re: Let abcd be a general fourdigit number. How many odd fourdigits nu [#permalink]
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17 May 2017, 10:48
mikemcgarry, now I got it! thanks a lot:)



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Re: Let abcd be a general fourdigit number. How many odd fourdigits nu [#permalink]
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07 Feb 2018, 02:49
Since it is an odd number, the last digits can be 1,3,5,7,9. Now you have to choose cd in such a way that it is a composite number. Also the the two numbers a * b shouldn't be double digit. After screening you get 21, 63 and 27 for cd. Hence the numbers are 3721, 7321, 9763, 7963, 9327, 3927. The answer hence is 6. That's B
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