2025 Mathcounts state sprint #22 problem and solution : level 1 +

 

2025 Mathcounts state sprint

#22: Let n be a positive integer less than or equal to 1000. If the last two digits of n are reversed, the resulting integer is exactly 85 percent of n. What is the sum of the possible values of n?

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Let n be a positive integer less than or equal to 1000. If the last two digits of n are reversed, the resulting integer is exactly 85 percent of n. What is the sum of the possible values of n?

Let the original number be:

$$n = 100h + 10t + u$$

The number formed by swapping the tens and units digits is:

$$n' = 100h + 10u + t$$

According to the problem:

$$n' = \frac{17}{20}n$$

So $n$ has to be divisible by 20 (make sure you know why). This implies:

$$u = 0, \quad t \text{ is even}$$

Let:

$$t = 2k, \quad 0 \leq k \leq 4$$

Then:

$$n = 100h + 10t = 100h + 20k$$ $$n' = 100h + t = 100h + 2k$$

Now compute the difference:

$$n - n' = 18k$$

Also, from the given:

$$n - n' = n - \frac{17}{20}n = \frac{3}{20}n$$

Equating both expressions:

$$18k = \frac{3}{20}n \Rightarrow n = 120k$$

Since $k \neq 0$, we get:

$$n = 120k$$

Valid values for $k \in \{1, 2, 3, 4\}$, so the numbers are:

$$120, \quad 240, \quad 360, \quad 480$$

Their sum is:

$$120 + 240 + 360 + 480 = 120(1 + 2 + 3 + 4) = 120 \times 10 = \boxed{1200}$$