Pattern Recognition is the Best

Algebra Level 4

Let $f(x)$ be the least positive integer that can be added to $x$ to obtain a perfect square . For example, $f(1) = 3 , f(2) = 2, f(3) = 1 , f(4) = 5 , f(5) = 4$ and so on.

To what value of $x$ , starting from $x = 1$ , is the $100^\text{th}$ recurrence of 2017 as a function value?

The answer is 1225647.

1 solution

Christian Daang
Jan 30, 2017

The pattern can be recognized as $f(n^2) = (n+1)^2 - n^2 = 2n + 1$ .

hence, the $1^{st}$ showing up of 2017 as a function value will occur at 2n + 1 = 2017 $\implies n = 1008$

and therefore, its $100^{th}$ recurrence will show up after n = 1008 + 99 = 1107 .

So,

$f(1107^2) = 2(1107) + 1 = 2215 \\ f(1107^2 + 1) = 2214 \\ \vdots \\ f(1107^2 + h) = 2017$

$\implies 1107^2 + 2215 = 1107^2 + h + 2017 \\ h = 198 \\ \text{therefore, the 100th recurrence of 2017 as a function value will occur at} \ x = 1107^2 + 198 \ \text{or simply} \ \boxed{x = 1225647} .$

The last few steps can be simplified as follows:

$\text {Since the next (closest) square number after } 1107^2 \text { is } 1108^2 \text { , therefore: }$

$x = 1108^2 - 2017 = \boxed {1225647}$

Zee Ell - 4 years, 4 months ago

Pattern Recognition is the Best

The answer is 1225647.

1 solution

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