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

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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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!

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

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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mikemcgarry,
now I got it! thanks a lot:)

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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no , cz in qs we need distinct int we will get 5 as 2 times so its not true

@mikemcgarryGiven: Let abcd be a general four-digit number.
Asked: 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?
Since abcd is odd 4-digit number, unit digit d is odd = {1,3,5,7,9}
a = {1,3,5,7,9}
b = {1,3,5,7,9} 
Since product of a & b is odd number 2-digit number cd.

Since no digit is 0 and 4 digits are all distinct
a & b can not be 1 or 5, 
a = {3,7,9}
b = {3,7,9}
Since a & b are also distinct

All possible choices for a & b = 3*2 = 6

3*7 = 21; abcd = 3721 
3*9 = 27; abcd = 3927 
7*3 = 21; abcd = 7321
7*9 = 63; abcd = 7963
9*3 = 27; abcd = 9327
9*7 = 63; abcd = 9763

Total 4-digit numbers abcd = 6

IMO B