Where Is The Second Equation?

Number Theory Level 2

If $a,b>1$ are integers satisfying $11(a-1) = 13(b-1)$ , find the minimum value of $a+b$ .

The answer is 26.

1 solution

A Former Brilliant Member
May 12, 2016

Relevant wiki: Greatest Common Divisor

$\gcd(11,13)=1$ . $13$ divides L.H.S and $11$ divides R.H.S

$a,b$ must be of the form $13n+1$ and $11m+1$ respectively where $n,m$ are integers.

As $a,b>1$ the minimum value of $a$ and $b$ occurs when $n,m=1$ or $a=14$ and $b=12$ .

Hence minimum value of $a+b=14+12=\boxed{26}$ .

Moderator note:

Great work! I would have phrased it this way:

Let $c = 11(a-1) = 13(b-1)$ , then $c$ must be a multiple of 11 and 13. Since we want to minimize $a,b> 1$ , then this suggests that should minimize $c$ as well. the minimum value of $c$ must be a multiple of 11 and 13. So a possible value of $c$ is $11\times13=143$ . Hence, $143 = 11(a-1) = 13(b-1) \Rightarrow a = \dfrac{143}{11} + 1 = 14 , b = \dfrac{143}{13} + 1 = 12$ . The result follows.

Where Is The Second Equation?

The answer is 26.

1 solution

Moderator note:

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