If n is a positive integer, is n divisible by 2?

Question

(1) 7n - 8 is divisible by 20
(2) 3n^2 + 2n + 5 is a prime number

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ArunSharma12 · Answer

Given: n > 0

statement 1: 7n - 8 = 20k
7n = 20k + 8
7n = 7(2k + 1) + 6k + 1
n = (2k+1) + (6k+1/7)
 6k+1/ 7 can be even or odd. 
6k+1/ 7 can only be odd
n will be even
sufficient

statement 2: 
3n^2 + 2n + 5 = prime
let, 3n^2 + 2n + 5 = 2
3n^2 + 2n + 3 = 0; but n will not be a positive integer
3n^2 + 2n + 5 = odd prime
n has to be even and hence divisible by 2
sufficient
Ans:  B D

firas92 · Answer

(1)  7n - 8  is divisible by  20

7n-8=20a  or  7n=20a+8=4(5a+2)

n=\frac{4(5a+2)}{7}

Since  n  is an integer and  4    is not multiple of  7 ,  (5a+2)  must be a multiple of  7  and so  n  must be a multiple of  4

So  n  is even and divisible by  2

1 is sufficient

(2)  3n^2 + 2n + 5  is a prime number

Minimum value of  (3n^2 + 2n + 5)  is  10  so  (3n^2 + 2n + 5)  must be a prime number greater than  10  and so  (3n^2 + 2n + 5)  must be odd

If  n  is odd,  (3n^2 + 2n + 5) = odd+even+odd = even  ---------- Not possible since we know that  (3n^2 + 2n + 5)  is odd

So  n  is not odd which means  n  is even and divisible by  2

2 is sufficient

Answer is  (D)

Archit3110 · Answer

target find if n is divisible by 2
#1
7n - 8 is divisible by 20
at n = 4,24 ; yes sufficeint
#2
3n^2 + 2n + 5 is a prime number
n=0 prime is 5 ; n = 4 prime is 61 ; yes
sufficient
OPTION D

NitishJain · Answer

IMO  D

To find n is Even/odd

(1) 7n - 8 is divisible by 20
for (7n-8) to be divisible by 20, it has to be even
=>7n-8 = Even
8 is even
Even - Even = even
Odd-Even = odd x

so 7n has to be even, for 7n to be even, n has to. be even (Even*Odd = Even; Odd*Odd= Odd)

SUFFICIENT

(2) 3n^2 + 2n + 5 is a prime number
Sine n is +ve, value for the given expression will be >5 and only 2 is even integer
==> 3n^2 + 2n + 5 = odd
2n will always be even & Even+Odd = Odd
so 3n^2 + 5 = odd. 
Now we know, Odd + Odd = Even & Odd+ Even = Odd
so 3n^2 has to be even
=> n has to be even (becoz: Even * odd = Even)

SUFFICIENT

HoneyLemon · Answer

D ,  IMO 
(1) 7n - 8 is divisible by 20

7n = 20 I + 8 => 7n = even .. so n must be even .. Sufficient .

(2) 3n^2 + 2n + 5 is a prime number

3n^2 + 2n must be even => 3n^2 must be even => n must be even -- Sufficient .

Hence D is the ans .

sambitspm · Answer

Kindly see the attachment

bumpbot · Answer

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If n is a positive integer, is n divisible by 2?

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