Three positive integers

Question

Find the smallest possible value of $ab+c$, where $a,b,c$ are positive integers with $a+bc=2016$.

(No computers! The puzzle has a nice direct solution.)

Answer 1

Proving that Zerris's answer is correct:

$$ab+c\ge2\sqrt{abc}=2\sqrt{a(2016-a)}=2\sqrt{1008^2-(1008-a)^2}$$

Because $2\sqrt{1008^2-1006^2}>96$, if $ab+c<96$, $\left|1008-a\right|$ must be greater than $1006$. The only possible values are $a=1$ and $a=2015$. Obviously $a=2015$ won't work, so $a=1$ and $bc=2015$. $2015=5\cdot13\cdot31$ only has four pairs of factors, and none of them sums to less than $96$.

Answer 2

The smallest possible value is

AB+C = 96

The smallest value will occur when

A = 1 and B = C

Unfortunately, 2016-1=2015 isn't a square, so do some factoring on paper to get the closest values of B and C as 31 * 65. A = 1, B = 31, C = 65 gives A+BC = 2016 and AB+C = 96. The theoretical minimum at A = 1, B = C would put B and C at around 45, for a total of just under 90.

Answer 3

If $a=1$, then $b=31$ or $65$, $c=65$ or $31$ and $ab+c=96$.

If $96>a>1$, $ab+c$ is at the minimum and $a$ is "known", then $bc$ is also known and...

$(ab+c)^2$ (min) $= a^2b^2 + c^2 + 2abc$ (known), $(ab+c)^2 - 4abc$ (known) $= (ab-c)^2$. so $|ab-c|$ must also be as low as possible. Even if $ab=c$, then $b=[sqrt((2016-a)/a)]$, making $ab+c=2*[sqrt((2016-a)*a)]$. The product inside is at least 4028, easily preventing us from reaching a better result than 96.