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Let abcd be a general four-digit number. How many odd four-digits nu

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Magoosh GMAT Instructor
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Let abcd be a general four-digit number. How many odd four-digits nu  [#permalink]

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New post 27 Feb 2017, 17:13
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Let abcd be a general four-digit number. How many odd four-digits numbers abcd exist such that the four digits are all distinct, no digit is zero, and the product of a and b is the two-digit number cd?

(A) 4
(B) 6
(C) 12
(D) 24
(E) 36


This 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 Questions

Mike :-)

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Let abcd be a general four-digit number. How many odd four-digits nu  [#permalink]

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New post 27 Feb 2017, 22:46
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mikemcgarry wrote:
Let abcd be a general four-digit number. How many odd four-digits numbers abcd exist such that the four digits are all distinct, no digit is zero, and the product of a and b is the two-digit number cd?

(A) 4
(B) 6
(C) 12
(D) 24
(E) 36


This 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 Questions

Mike :-)


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 four-digit number. How many odd four-digits nu  [#permalink]

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New post 09 Apr 2017, 11:37
Hi Mike,

Is 3*5=15 not a possible case?
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Let abcd be a general four-digit number. How many odd four-digits nu  [#permalink]

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New post 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 four-digit number. How many odd four-digits nu  [#permalink]

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New post 18 Apr 2017, 05:39
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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 4-digit numbers = 6
Hence, Correct choice B
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Let abcd be a general four-digit number. How many odd four-digits nu  [#permalink]

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New post 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 four-digit number. How many odd four-digits nu  [#permalink]

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New post 17 May 2017, 10:41
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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 two-digit number cd?

Thus, "cd" does NOT denote a product of two single digits, but instead a single two-digit 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 two-digit number "cd" has to be odd. This, in turn, means that a & b must be odd, because their product is an odd two-digit 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 four-digit number. How many odd four-digits nu  [#permalink]

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New post 17 May 2017, 10:48
mikemcgarry,
now I got it! thanks a lot:)
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Re: Let abcd be a general four-digit number. How many odd four-digits nu  [#permalink]

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New post 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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Re: Let abcd be a general four-digit number. How many odd four-digits nu &nbs [#permalink] 07 Feb 2018, 02:49
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