harpritsn wrote:P and q are integers. If p is divided by 2, the remainder is 1, and if q is divided by 6, the remainder is 1, which of the following must be true?
1. p.q+1=even
2. p.q/2= is an integer
3. p.q is multiple of 12??

obviously both of them odd (q/6 the remainder is odd too), hence 1) is always true, odd*odd+1=even
next, 2) odd*odd/2 cannot be an integer (exclude)
3) p*q/12=integer implies p=2a+1 and q=6b+1 <> (2a+1)(6b+1)/12=integer. Let's see (12ab+2a+6b+1)/12=integer? Or (2a+6b+1)/12=integer? Or (a+3b)/6 +1/12=integer?

pick 1

harpritsn wrote:P and q are integers. If p is divided by 2, the remainder is 1, and if q is divided by 6, the remainder is 1, which of the following must be true?
1. p.q+1=even
2. p.q/2= is an integer
3. p.q is multiple of 12??

p is divided by 2, the remainder is 1 => p is an odd number and greater than 2
q is divided by 6, the remainder is 1 => q is a prime number. (Prime numbers are of the form 6k+1 or 6k-1)

Check the conditions:
1. p.q+1=even
This will be true because odd*odd =odd and odd+1 = even
Ex: 3*7 = 21 and 21+1 = 22 (Even)
2. p.q/2= is an integer
This is not true because odd*odd =odd
3. p.q is multiple of 12??
This is also not true
Ex: 3*7 = 21 and 21 is not a multiple od 12.

Answer 1

harpritsn wrote:P and q are integers. If p is divided by 2, the remainder is 1, and if q is divided by 6, the remainder is 1, which of the following must be true?
1. p.q+1=even
2. p.q/2= is an integer
3. p.q is multiple of 12??

We can pick numbers. Let p=3 and q=7. NOTE: they both are odd.

1. 3*7 + 1 = 22. TRUE (odd*odd=odd and odd+1=even)
2. Odd/2 is never an integer. FALSE
3. Odd is never a multiple of an even number. FALSE.

Is it advisable to always go for substitution in such type of problems to get a better view of the anwsers??

harpritsn wrote:Is it advisable to always go for substitution in such type of problems to get a better view of the anwsers??

For the questions involving number properties (odd/even, remainder etc), substitution is USUALLY the efficient method.