even odd problem?

Question

Five natural numbers D1,D2,D3,D4 and D5 ar such that their sum is odd. If the product of D3,D2 and D1 is odd and (D2D3+D3D4+D4D5) is not even , which of the following is true ?


(1)D4 is even , only if D5 is even

(2)D4 is always even

(3)D4 is always odd

(4)D4 is odd ,only if D5 is odd.


After some fight , i chose 1,3,5,7,11 which satisfies the conditions.

but while trying to choose the answer , i got stumbled because question has "if","always","only if" ....so not able to choose the answer.

How to solve this kind of problem ?

Answer 1

You can't approach this program by picking just one group of numbers for D1 to D5. If you happen to pick an odd D4, for example, that doesn't necessarily mean that D4 MUST BE odd.

Instead, you have to use the statements that they give, plus the properties of addition and multiplication of even and odd numbers, to figure out which numbers must be even or odd.

(1) "The product of D3,D2 and D1 is odd."

This statement gives away a lot.

You know D1, D2, and D3 must ALL be odd if their product is odd. If any of them were even, the even number would have a factor of 2, and their combined product would necessarily have a factor of 2 as well (i.e., be even as well).

(2) "[The five numbers'] sum is odd."

The sum of D1 + D2 + D3 is also necessarily odd, since (per (1) above) all three are odd. The sum of any three odd numbers is itself odd.

Since the sum of those three AND D4 and D5 is odd as well, you know that the sum D4+D5 is even.

You know this because the difference between any two odd numbers -- in this case the odd numbers D1+D2+D3 and D1+D2+D3+D4+D5 -- must be even.

If you know that the sum of D4 and D5 is even, then you know that either they must both be even, or they must both be odd. (If one were even and one odd, the sum would be odd.)

(3) "(D2D3+D3D4+D4D5) is not even."

You already know D2D3 is odd (because, per (1) above, you know both are odd, their product is odd).

If the sum D2D3+D3D4+D4D5 is odd, you know that D3D4+D4D5 must be even (because, same as we noted above, the difference between any two odd numbers -- in this case D2D3 and D2D3+D3D4+D4D5 -- must be even).

But that doesn't tell us anything new, because if D4 and D5 are both even, D3D4+D4D5 is even (since D4 is even and multiplied by both terms in the sum, both terms are even -- and the sum of two even numbers is even).

And if D4 and D5 are both odd, that sum is still even. This is because all five numbers are then odd. Then both D3D4 and D4D5 must be odd, and the sum of those two odd numbers is even.

======================
So... we've established that D1, D2, and D3 MUST all be odd, while D4 and D5 are EITHER both odd, or else both even.

Now let's look at the four statements that you have to evaluate:

(1) Is true. D4 and D5 must BOTH be even or both odd. So D4 can only be even if D5 is also even.

(2) Is false. D4 can be odd.

(3) Is false, D4 can be even.

(4) Is true. D4 and D5 must BOTH be even or both odd. D4 can only be odd if D5 is also odd.

=================
1,1,1,1,1 satisfies all conditions (sum of all five is odd [5], product of first three is odd [1], D2D3 + D3D4 + D4D5 is odd [3]).

1,1,1,2,2 satisfies all conditions as well (sum of all five is odd [7], product of the first three is odd [1], D2D3 + D3D4 + D4D5 is odd [7]).

Since D4 can be either 1 or 2 (as shown above), statements (2) and (3) aren't true.