Question

find the value of n, if
2^8 + 2^11 + 2^n be a perfect square

Answer 1

its n=12. it's a nice problem, not tricky.

Answer 2

how come to get it ???

Answer 3

group the first two terms
\[\large (2^8+2^{11})+2^n\] factor out 2^ 8 \[\large 2^8(1+2^3)+2^n\]

Answer 4

next ?

Answer 5

\[\large 2^8(9) + 2^n\]

Answer 6

factor the 2^8 from both terms \[\large 2^8(9+2^{n-8})\] rewrite 9 as 3^2 \[\large 2^8(3^2+2^{n-8)}\]

Answer 7

the expression inside the parentheses is the familiar \[\large3^2 + 4^2 = 5^2\] right triangle.
so we rewrite the problem as \[\huge 2^8(3^2+4^{\frac {n-8}{2}})\] so \[\large \frac{ n-8 }{ 2 }=2\] just solve n.

Answer 8

thank you very much, that's make sense for me