Tuesday, May 27, 2025

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?


Try this question first. Then scroll down for solution. 















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:

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

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