Topic: Post logical mathematical reasoning problems here.

[AdrenalineXV]

Question 1:

With the digits 7, 3, 5, 6; 24 four-digit numbers can be formed. From these 24 numbers, the quantity of odd numbers is the same as a:

So, I know that 4! / (4-4)! = 24. There are 3 odd numbers and one even number. What is next?

Regards.

[[MAF]Agus]

[[MAF]mooman]

I'm so confused.

[AdrenalineXV]

I'm so confused.

mooman, Agus:

It is the same as above: If 24 numbers (four-digit each one) can be formed with: 7, 3, 5 ,6. Find the quantity of odd numbers from these.

Regards.

[[MAF]mooman]

Anything multiplied by 10, 100 or 1000 is even and adding an even number to an odd/even number results in an odd/even number. Therefore only the last digit matters and every number will be odd except for those ending in 6, of which there are 3! = 6.

[AdrenalineXV]

Anything multiplied by 10, 100 or 1000 is even and adding an even number to an odd/even number results in an odd/even number. Therefore only the last digit matters and every number will be odd except for those ending in 6, of which there are 3! = 6.

Ok mooman, I have got your point. Now: 7, 3, 5, 6. There are 3 odd numbers, that means that just 3 of four numbers can be used to get a four-digit odd number; so: 3 out of four numbers.

3 possibilities will be multiplied for the rest of available digits: 3*3*2*1 = odd numbers.

1 possibility will be multiplied in order to get the total of even numbers: 1*3*2*1 = even numbers.

Thanks.