CMI2016-B-7aiii

Question

Consider the funciton $M$ defined as follows:

$M(n) = \begin{cases} n-10 & \text{ if } n > 100 \\ M(M(n+11)) & \text{ if } n \leq 100 \end{cases}$

Compute the following$: M(87)$

Answer 1

M(87) $= M(M(98))\\ =M(M(M(109)))\\ =M(M(99))\\ =M(M(M(110)))\\ =M(M(100))\\ =M(M(M(111)))\\ =M(M(101)) = M(91)$

from above we can see that when $90 \leq n \leq 100$ n is incremented by 1 so,

$M(91) = M(92) =M(93) =M(94) =M(95) =M(96) =M(97) =M(98) = M(99) =M(100) = M(101) = 91$ 

So, M(87) = 91