Review question

# Given this equation, what's the value of $a+b+c+d$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6219

## Solution

Four positive integers $a$, $b$, $c$ and $d$ are such that $abcd+abc+bcd+cda+dab+ab+bc+cd+da+ac+bd+a+b+c+d=2009.$

What is the value of $a+b+c+d$?

If we add 1 to both sides, the left hand side factorises. We get $(a+1)(b+1)(c+1)(d+1)=2010.$

Expressing $2010$ as a product of primes gives $2010 = 2 \times 3 \times 5 \times 67$.

As $2010$ has only four prime factors, these must be $a+1$, $b+1$, $c+1$, and $d+1$ in some order (since $a$, $b$, $c$ and $d$ are positive and cannot be zero).

This means that $a$, $b$, $c$, and $d$ must be $1$, $2$, $4$, and $66$ in some order.

Thus $a+b+c+d=1+2+4+66=73.$