Question

Pls solve this question.
The solution is urgently needed.
#Well-Explained answer
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Answer 1

Hey mate,

As you know firstly we should take up the values ,

2^200-2^192 ×31+2^n

=2^192(2^8-31+2(n-192))

=2^192(256-31+2^k) {let k = n-192}

=2^192(225+2^k)

=>2^192 is a perfect square. Choose k such that 225+2^k will also become a perfect square. By trial k = from 1 to 5 no one satisfy this requirment. But k= 6 can satisfy it.

=>(225+2^6=289 is perfect)

k= 6

So, 6=n-192

n=192+6

n=198

So, it can be perfect square.

Hope it will help you.

✨It's. M.S.V.

Answer 2

Answer:

198

Step-by-step explanation:

$2^{200}-2^{192}.31+2^n\\\\\implies 2^{192}(2^8-31+2^{n-192})$

$2^{192}\\\implies (2^{96})^2)\\\\\textsf{Hence it is a perfect square}$

So this means :

$2^8-31+2^{n-192}\\\\\implies 256-31+2^{n-192}\\\\\implies 225+2^{n-192}$

n can be 198 so that 225 + 64 = 289 is a perfect square .

So n = 198 is a solution .