Squares everywhere

Number Theory Level 4

If $\overline {abcde}$ and $\overline {edcba}$ are distinct perfect squares, let $A= \sqrt{\overline{abcde}}$ and $B = \sqrt {\overline{edcba}}$ , with $A<B$ . If $A = \overline{pqr}$ , and $B = \overline{xyz}$ , determine which among the choices has the same value as

$pz + qy + rx$

(p + q+ r)(x+y+z)

x+y+z

(pq + qr)(xy + yz)

p^2 + q^2 + r^2

pqrxyz - pqr - xyz

px + qy + rz

1 solution

E Koh
Dec 23, 2020

abide = 10609, edcba = 90601. A = sqrt(10609) = 103, B = sqrt(90601) = 301. A = pqr = 103, B = xyz = 301. pz + qy + rx = (1)(1) + (0)(0) + (3)(3) = 1^2 + 0^2 + 3^2 => p^2 + q^2 + r^2.

Squares everywhere

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