Problem #10

Probability Level 3

There have been 9 problems before this, and this is the 10th. To celebrate, let's have a fun problem before we carry on with normal, serious ones!

Let $a_{i}, i=1, 2, 3, 4, 5, 7, 8, 9, 10$ be the answer to Problem # $i$ in the set Easy Problems. (The link can be found below.)

There exists a unique triple of distinct positive integers from 1 to 10 $(m, n, p)$ such that $a _{m}+a_{n}=a_{p}$ . Find $m+n+p$ .

Hint: solve systematically!

This problem is part of the set Easy Problems

The answer is 24.

1 solution

Jared Low
Nov 30, 2014

ANSWERS A.K.A. SPOILERS:

$1) 210.25$

$2) 6$

$3) 6$

$4) 13$

$5) 600001$

$6)$ Inadmissable (MCQ)

$7) 3,$

$8) 1,$

$9) 2$ .

From these 9 previous given answers, we know that $m,n,p$ cannot hold the values $1, 6$ due to the nature of the solutions ( $a_1$ is the only non-integer solution). We also know that they cannot hold the value $5$ due to the high value of $a_5$ . They also cannot hold the value 10 because if so we would have the value $m+n+p \geq a_{10} = m+n+p$ , a contradiction.

Among the six remaining answers, we finally have:

$1+2=3$

$\Rightarrow a_8+a_9=a_7$

$\Rightarrow (m,n,p)=(8,9,7)$ , so we have $m+n+p=\boxed{24}$

Problem #10

The answer is 24.

1 solution

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