# 2005 PMWC Problems/Problem I5

## Problem

Consider the following conditions on the positive integer (natural number) $a$:

1. $3a + 5 > 40$

2. $49a \ge 301$

3. $20a \le 999$

4. $101a + 53 \ge 2332$

5. $15a-7 \ge 144$

If only three of these conditions are true, what is the value of $a$?

## Solution

We can solve each of these conditions for $a$, and also truncate the right-hand side of the inequality since we know $a$ is an integer.

1. $3a+5>40 \implies 3a>35 \implies a>\frac{35}3 \implies \underline{a>11}$

2. $49a\ge301 \implies a\ge\frac{43}7 \implies \underline{a>6}$

3. $20a\le999 \implies a\le\frac{999}{20} \implies \underline{a<50}$

4. $101a+53\ge2332 \implies 101a\ge2279 \implies a\ge\frac{2279}{101} \implies \underline{a>22}$

5. $15a-7\ge144 \implies 15a\ge151 \implies a\ge\frac{151}{15} \implies \underline{a>10}$

Exactly two of these statements are false.

Note that statement 4 implies statement 1, which implies statement 5, which implies statement 2. This means that 2 and 5 must both be true, or else 1, 4, and 5 are all false, which is in opposition to the fact that only two are false.

Note further that if 3 is false, $a\ge50$, which implies that all four of the other statements are true. This also disagrees with the stipulation that exactly three statements are true, and two are false. Therefore statement 3 is true.

This leaves 1 and 4 to be the only statements that are allowed to be false. If 1 and 4 are false, and the rest are true, we have $a\le11$ (the inverse of statement 1) and $a>10$ (statement 5). The only natural number that satisfies this is $\boxed{a=11}$.