Fun with digits

Number Theory Level 2

Let $A$ be a $2n$ -digit number whose digits are all $1$ 's and let $B$ be a $n$ -digit number whose digits are all $2$ 's.

Then which of the following options is true :

AB - 2

is a prime number

A-B

is a perfect square

A^2 - B^2

is a perfect cube

AB + 1

is a perfect cube

1 solution

Danish Ahmed
Jan 10, 2015

$A = \dfrac{10^{2n} - 1}{9}$ and $B = \dfrac{2(10^n - 1)}{9}$

$\begin{aligned}A - B &= \dfrac{10^{2n} - 1}{9} - \dfrac{2(10^n - 1)}{9}\\ &= \dfrac{10^{2n}-2*10^n+1}{9}\\ &= \dfrac{(10^n - 1)^2}{3^2}\\ &= (\dfrac{3(10^n - 1)}{9})^2\\ \end{aligned}$

Since $\dfrac{10^n - 1}{9}$ is an integer so $A - B$ is a perfect square.

$A - B = (333....33)^2$ thus the square root of A-B consist of $n$ 3's.

Fun with digits

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