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:

$$n^{\prime }=100h+10u+t$$

According to the problem:

$$n^{\prime }=\frac{17}{20}n$$

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

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

Let:

$$t=2k,0\le k\le 4$$

Then:

$$n=100h+10t=100h+20k$$ $$n^{\prime }=100h+t=100h+2k$$

Now compute the difference:

$$n-n^{\prime }=18k$$

Also, from the given:

$$n-n^{\prime }=n-\frac{17}{20}n=\frac{3}{20}n$$

Equating both expressions:

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

Since $k\ne 0$, we get:

$$n=120k$$

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

$$120,240,360,480$$

Their sum is:

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