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if q and r both are odd, addition and subtraction between q and r will always yield positive values

Product of q and r will always be odd
Squares of odd numbers are also odd

Hence D
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Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

If q and r are both odd numbers, which of the following must also be odd?

A. q – r
B. (q + r)^2
C. q(q + r)
D. (qr)^2
E. q/r




Since q and r are odd numbers we can put q and r as q=2*Q+1 and r=2*R+1, for some integers Q, R.
A. q-r=(2*Q+1)-(2*R+1)=2*(Q-R) ---> must be an even number.
B. (q + r)^2 =(2Q+2R+2)^2= 4*(Q+R+1)^2 ---> must be an even number.
C. q(q+r) = (2Q+1)(2Q+2R+2)= 2*(2Q+1)(Q+R+1) ---> must be an even number.
D. (qr)^2 = ((2Q+1)(2R+1))^2 = (4QR +2Q +2R +1)^2
by putting 2QR+Q+R = T we have (2T+1)^2= 4T^2 + 4T +1=2(2T^2 +2T)+1 ---> must be an odd number.
E. q/r cannot be an integer

The answer is, therefore, (D).
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Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

If q and r are both odd numbers, which of the following must also be odd?

A. q – r
B. (q + r)^2
C. q(q + r)
D. (qr)^2
E. q/r




Since q and r are odd numbers we can put q and r as q=2*Q+1 and r=2*R+1, for some integers Q, R.
A. q-r=(2*Q+1)-(2*R+1)=2*(Q-R) ---> must be an even number.
B. (q + r)^2 =(2Q+2R+2)^2= 4*(Q+R+1)^2 ---> must be an even number.
C. q(q+r) = (2Q+1)(2Q+2R+2)= 2*(2Q+1)(Q+R+1) ---> must be an even number.
D. (qr)^2 = ((2Q+1)(2R+1))^2 = (4QR +2Q +2R +1)^2
by putting 2QR+Q+R = T we have (2T+1)^2= 4T^2 + 4T +1=2(2T^2 +2T)+1 ---> must be an odd number.
E. q/r cannot be an integer

The answer is, therefore, (D).


Y cant q/r be odd?
let us assume q = 27 and r = 3
then q/r = 9 which is odd
Any specific relation between q and r is not specified.

Please explain why E is wrong option
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Hi nipunjain14,

You are correct, Q/R COULD be an integer. The simplest examples would include any time Q and R are the SAME odd integer or any time that R = 1. That type of thoroughness-of-thinking should serve you well as you continue to study.

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nipunjain14
MathRevolution
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

If q and r are both odd numbers, which of the following must also be odd?

A. q – r
B. (q + r)^2
C. q(q + r)
D. (qr)^2
E. q/r




Since q and r are odd numbers we can put q and r as q=2*Q+1 and r=2*R+1, for some integers Q, R.
A. q-r=(2*Q+1)-(2*R+1)=2*(Q-R) ---> must be an even number.
B. (q + r)^2 =(2Q+2R+2)^2= 4*(Q+R+1)^2 ---> must be an even number.
C. q(q+r) = (2Q+1)(2Q+2R+2)= 2*(2Q+1)(Q+R+1) ---> must be an even number.
D. (qr)^2 = ((2Q+1)(2R+1))^2 = (4QR +2Q +2R +1)^2
by putting 2QR+Q+R = T we have (2T+1)^2= 4T^2 + 4T +1=2(2T^2 +2T)+1 ---> must be an odd number.
E. q/r cannot be an integer

The answer is, therefore, (D).


Y cant q/r be odd?
let us assume q = 27 and r = 3
then q/r = 9 which is odd
Any specific relation between q and r is not specified.

Please explain why E is wrong option


Hi,
why E is wrong?
It is because the question asks you "which of the following must also be odd?"...
"must" word makes the choice wrong..
you can find ways of\(\frac{q}{r}\)being odd and ways of\(\frac{q}{r}\)being non integer, so it does not fall into must category..
E also would have been an answer, had the Q asked "which of the following CAN also be odd?"..
I Hope the Reasoning is clear
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MathRevolution
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

If q and r are both odd numbers, which of the following must also be odd?

A. q – r
B. (q + r)^2
C. q(q + r)
D. (qr)^2
E. q/r




Since q and r are odd numbers we can put q and r as q=2*Q+1 and r=2*R+1, for some integers Q, R.
A. q-r=(2*Q+1)-(2*R+1)=2*(Q-R) ---> must be an even number.
B. (q + r)^2 =(2Q+2R+2)^2= 4*(Q+R+1)^2 ---> must be an even number.
C. q(q+r) = (2Q+1)(2Q+2R+2)= 2*(2Q+1)(Q+R+1) ---> must be an even number.
D. (qr)^2 = ((2Q+1)(2R+1))^2 = (4QR +2Q +2R +1)^2
by putting 2QR+Q+R = T we have (2T+1)^2= 4T^2 + 4T +1=2(2T^2 +2T)+1 ---> must be an odd number.
E. q/r cannot be an integer

The answer is, therefore, (D).


Y cant q/r be odd?
let us assume q = 27 and r = 3
then q/r = 9 which is odd
Any specific relation between q and r is not specified.

Please explain why E is wrong option


Hi,
why E is wrong?
It is because the question asks you "which of the following must also be odd?"...
"must" word makes the choice wrong..
you can find ways of\(\frac{q}{r}\)being odd and ways of\(\frac{q}{r}\)being non integer, so it does not fall into must category..
E also would have been an answer, had the Q asked "which of the following CAN also be odd?"..
I Hope the Reasoning is clear

Thanks I understand now. We need to keepin mind all aspects when facing "Must be" type questions
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Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

Sorry for my lateness and hi, chetan2u!
Thank you for your kind explanation!

Well....I would like to add some more because of my apology.
"q/r is an integer if and only if q is a multiple of r" (that is, it can be paraphrased as"q/r is not an integer if and only if q is not a multiple of r).

q/r is an integer --> q is a multiple of r
pf) Since q/r is an integer we have q/r = n, for some integer n.
Then we have q= nr --> q is a multiple of r.

q is a multiple of r ---> q/r is an integer
pf) Similarly since q is a multiple of r we have some integer n which satisfies q= nr. ---> q/r = (nr)/r = n which is an integer...

So, since 27 is a multiple of 3, 27/3=9 is an odd integer.
But as chetan2u pointed out it cannot be an odd integer(not even an integer) if we choose q and r as not multiple-factor relation.
---> for example, q= 27, r=5 ---> q/r = 27/5 is not even an integer.

We cannot say, therefore, that q/r MUST be an odd integer.

Cheers!
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