Oxford Entrance Exam Problem 6

Probability Level 3

The functions $S$ and $T$ are defined for real numbers by

$S(x)=x+1$ and $T(x)=-x$ .

The function $S$ is applied $s$ times and the function $T$ is applied $t$ times, in some order, to produce the function

$F(x)=8-x$ .

It is possible to deduce that

s=8

and

t=1

. none of the other options.

s

is even and

t

is odd.

s

and

t

are powers of

2

s

is odd and

t

is even.

2 solutions

Vishwak Srinivasan
Jun 6, 2015

$F(x) = s(S(x)) + t(T(x)) \Rightarrow F(x) = s(x+1)+ t(-x) \Rightarrow F(x) = (s-t)x + s$

$8 - x = (s-t)x + s$

Comparing coefficients,

$s = 8 \text{ \& } t = 9$

Leonardo Watch
Nov 15, 2017

By quickly applying S 8 times and T once in the order $S^8(T(x))$ , you can deduce that s = 8 and t = 1 is a possible answer. However, this does not mean that it is the only way to create that answer. This rules out both the fourth (as 8 is even and 1 is odd) and the first answer.

You can prove there are additional answers because $S^8(T^3(x))$ also gives the solution $8-x$ . Since there are two answers, the second option cannot be right, and since $t$ is not a power of two, the fifth option cannot be right either. This leaves the third option as the correct answer.

In truth, there are multiple solutions that can be found via the formula $S^8(T^{2n+1}(x))$ .

Oxford Entrance Exam Problem 6

2 solutions

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