What is the remainder when the positive integer n is divided by 2?

When a no. Is divided by 2. It either leave remainder 0 or 1

Statement 1: When n is divided by 13, the remainder is 3
Therefore value of n = 3,16,29,42.............
When odd/2 leaves 1 as remainder
When even/2 leaves 0 as remainder
(Not sufficient)

Statement 2: n + 2 is a multiple of 7
N+2=7x
Where x is a positive natural no. = 1,2,3.........
X = 1
07 = N+2 => N = 5
X = 2
14 = N+2 => N = 12
X = 3
21 = N+2 => N = 19
Values of N = 5,12,19,26.......
When odd/2 leaves 1 as remainder
When even/2 leaves 0 as remainder
(Not sufficient).

Statement 1&2 together: Set values of N = 3,5,12,16,19,26,29,42,26......
When odd/2 leaves 1 as remainder
When even/2 leaves 0 as remainder
(Not sufficient)

Answer is E

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I DO HOPE IT IS E
1. N/13 THEN R=3. IT COULD BE 3 SO 3/2 HAS A REMAINDER OR 16 WHICH HAS NO REMINDER AT ALL. NOT SUFFICIENT
2. N+2 IS 7X SO N CAN BE 5 OR 12 . 5 LEAVES A RAMAINDER BUT 12 NOT. NOT SUFFICIENT

BOTH:
n=13K+3
n=7W-2
2n=13K+7w+1
n=6.5K+3.5w+1 (since N is an integer- K and W could be only even numbers so that N will be an integer-- ------ 0.5*3 etc. is a fraction. 0.5*4 or *6 or * 8 is a whole number)

now if K and W are i.e, 2 N is odd ( 6.5*2+ 3.5*2+1), so when N is divided by 2, N will have a remainder.

but if K is 2 and W is 4, i.e, N is even (6.5*2+ 3.5*4+1), so he won't have a reminder.

81 and 146, i.e applies to both statements

How to think for a=12
I checked by putting values for a and got for a=5 -> n=68
but to reach a=12 n-159 would be lengthy and time consuming.
How to do I think that it'll further give a value with a different remainder?

(1) When n is divided by 13, the remainder is 3

n is of the form:

\(n=13q+3\)-----> if we divide 13q+3 by 2, we will get multiple values for remainders.

if \(q=;1 n=13+3=16\) in that case remainder will be 0.
if \(q=2;n=26+3=29\) in this case remainder will be 1.

clearly Not sufficient.

(2) n + 2 is a multiple of 7.

this means, \(7q=n+2\), where q is any integer: 1,2,3,4...

\(n=7q-2\)

\(7q-2 / 2\) gives multiple remainders as well.

if q=1; \( n=7*2-2= 12\). in this case, when divided by 2, remainder will be 0
if q=3; \(n=7*3-2=19\).in this case, when divided by 2, remainder will be 1

Not sufficient.

1+2 together.

\(n=13q+3\) and n+2 is a multiple of 7

therefore, \(n+2= 13q+3+2\)----> \(n+2=13q+5\) is also a multiple of 7. again, q can be any integer=1,2,3,4...

you don't need to "solve" further it's clear we're kinda in a circle now. since q can be any integer this will not give one single result.

Hence, E.

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