If $f(x + 2) = f(x) + f(x + 1)$ for all positive integers $x$, and $f(11) = 91$, $f(15) = 617$, then $f(10)$ equals

We have a recurrence relation: $f(x + 2) = f(x) + f(x + 1)$

This means each term equals the sum of the two previous terms - similar to how Fibonacci numbers work!

We know: $f(11) = 91$ and $f(15) = 617$

We need to find: $f(10)$

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Since we know $f(11)$ and $f(15)$, but need $f(10)$, we'll work forward from $f(11)$ to find the values in between, then work backward to find $f(10)$.

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Let's call $f(12) = a$ (we'll find this value)

Using our recurrence relation $f(x + 2) = f(x) + f(x + 1)$:

$f(13) = f(11) + f(12) = 91 + a$

$f(14) = f(12) + f(13) = a + (91 + a) = 91 + 2a$

$f(15) = f(13) + f(14) = (91 + a) + (91 + 2a) = 182 + 3a$

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We know $f(15) = 617$, so:

$182 + 3a = 617$

$3a = 617 - 182 = 435$

$a = 145$

Therefore: $f(12) = 145$

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We can rearrange our recurrence relation!

From $f(x + 2) = f(x) + f(x + 1)$, we can solve for $f(x)$:

$f(x) = f(x + 2) - f(x + 1)$

Using this rearranged formula:

$f(10) = f(12) - f(11) = 145 - 91 = 54$

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$f(10) = 54$

When dealing with recurrence relations, you can work both forward and backward by rearranging the equation. This flexibility often helps solve problems where you have information about terms that aren't consecutive!