An algebra problem by Rindell Mabunga

Algebra Level pending

$(3^2 - 2)(9^2 - 8)(513^2 - 512)(134217729^2 - 134217728) = 2^a - b$

Given that $a$ and $b$ are integers and $b$ is minimized, what is the value of $a + b$ ?

2 solutions

Rindell Mabunga
Mar 17, 2014

Note that $(a + 1)^2 - a = a^2 + a + 1$

and

$(a - 1)(a^2 + a + 1) = a^3 - 1$

So,

$(3^2 - 2)(9^2 - 8)(513^2 - 512)(134217729^2 - 134217728)$

= $(2^2 + 2 +1)(2^6 + 2^3 + 1)(2^{18} + 2^9 + 1)(2^{54} + 2^{27} + 1)$

= $(2 - 1)(2^2 + 2 +1)(2^6 + 2^3 + 1)(2^{18} + 2^9 + 1)(2^{54} + 2^{27} + 1)$

(Note: Since $1 + 1 = 2$ , then $2 - 1 = 1$ )

$(2 - 1)(2^2 + 2 +1)(2^6 + 2^3 + 1)(2^{18} + 2^{9} + 1)(2^{54} + 2^{27} + 1)$

= $(2^3 - 1)(2^6 + 2^3 + 1)(2^{18} + 2^9 + 1)(2^{54} + 2^{27} + 1)$

= $(2^9 - 1)(2^{18} + 2^9 + 1)(2^{54} + 2^{27} + 1)$

= $(2^{27} - 1)(2^{54} + 2^{27} + 1)$

= $2^{81} - 1$

Therefore, $a = 81$ and $b = 1$

and $a + b = 81 + 1 = 82$

Mas Mus
Apr 8, 2014

(3^2-2)=2^2+2^1+2^0=(2^3-1)/(2-1)

(9^2-8)=8^2+8^1+8^0=(8^3-1)/(8-1)=(2^9-1)/(2^3-1)

(〖513〗^2-512)=〖512〗^2+〖512〗^1+〖512〗^0=(〖512〗^3-1)/(512-1)=(2^27-1)/(2^9-1)

(〖134217729〗^2-134217728)=〖134217728〗^2+〖134217728〗^1+〖134217728〗^0=(〖134217728〗^3-1)/(134217728-1)=(2^81-1)/(2^27-1)

So, we get

(3^2-2)(9^2-8)(〖513〗^2-512)(〖134217729〗^2-134217728)=(2^3-1)/(2-1)×(2^9-1)/(2^3-1)×(2^27-1)/(2^9-1)×(2^81-1)/(2^27-1)=2^81-1

There for a=81 and b=1. Hence, a+b=82