Question

Q4. Find n, if 2²⁰⁰ – 2¹⁹² × 31 + 2ⁿ is a perfect square.​

Answer 1

Given :-

$\sf \bullet \;\; 2^{200} -2^{192} \times 31 + 2^{n}$

To Find :-

• Value of n if given expression is perfect square  

Solution :-  

$\sf : \; \implies 2^{200} -2^{192} \times 31 +2^{n}$

~By taking 2¹⁹² as common

$\sf : \; \implies 2^{192}( 2^{8} -31) + 2^{n}$

$\sf : \; \implies 2^{192}( 256-31) + 2^{n}$

$\sf : \; \implies 2^{192} \times 225 + 2^{n}$

~By taking  2¹⁹² as common again

$\sf : \; \implies 2^{192} ( 15^{2} + 2^{n-192})$

~We need a number closest to 15 whose square difference will be a perfect square.

# By hit and trial method ::  

→ 16² = 256  

→ 256-225  

→ 31  

It is not a perfect square  

→ 17² = 289

→ 289 -225  

→ 64  

It is a perfect square.  

H E N C E ,  

$\sf : \; \implies 2^{n-192} = 64$

$\sf : \; \implies 2^{n-192} = 2^{6}$

$\sf : \; \implies n-192 = 6$

$\sf : \; \implies n = 192 + 6$

$\boxed{\bf{ \star \;\; n = 198 }}$

Answer 2

ANSWER-;

______________

= 2²⁰⁰-2¹⁹² ( 2⁵-1) +2ⁿ

= 2²⁰⁰-2¹⁹⁷+ 2¹⁹² + 2ⁿ

= 2¹⁹⁷× 7 + 2¹⁹² +2ⁿ

= 2¹⁹² [ 1+32×7] +2ⁿ

= 2¹⁹² × 225 +2ⁿ

➜ 2¹⁹² is a perfect square By trial n= from 1 to 5 No one satisfy this But k= 6 can satisfy it .

= 2⁶ = 192

n =6

6= n- 192

n= 192+6

n= 198 Ans✓