AMC 10 Step by Step

2000 AMC 10 Problem 23

P21-25~7 minPrint

When the mean, median, and mode of the list

10,2,5,2,4,2,x10,2,5,2,4,2,x

are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of xx ?

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Key insight

The mode is always 2 and the mean is (25 + x)/7. The median is 2, x or 4, so check each case for three evenly spaced values.

Solution

1. The mode. The number 22 appears three times, and adding one xx cannot give any other number three appearances. So the mode is always 22.
2. The mean. The numbers other than xx add to 10+2+5+2+4+2=2510 + 2 + 5 + 2 + 4 + 2 = 25, so the mean is 25+x7\frac{25 + x}{7}.
3. The median. Without xx, the sorted list is 2,2,2,4,5,102, 2, 2, 4, 5, 10. With xx there are 77 numbers, and the median is the 44th one. So:
- if x≤2x \le 2, the median is 22;
- if xx is between 22 and 44, the median is xx;
- if x≥4x \ge 4, the median is 44.
4. "Arithmetic progression" means the three values, in order, are evenly spaced, like 2,3,42, 3, 4. "Non-constant" means they are not all equal. Now check each case for the median.

Case 1: median 22. Then the mode and the median are both 22. Evenly spaced numbers with two equal would all be equal, which is not allowed. No solutions.

Case 2: median 44 (x≥4x \ge 4). We have 22, 44 and the mean. The mean can go in three places:
- after them: 2,4,62, 4, 6, so the mean is 66, giving 25+x=4225 + x = 42 and x=17x = 17;
- between them: 2,3,42, 3, 4, so the mean is 33, giving x=−4x = -4;
- before them: 0,2,40, 2, 4, so the mean is 00, giving x=−25x = -25.

Only x=17x = 17 fits x≥4x \ge 4. Check: the mean is 427=6\frac{42}{7} = 6, the median is 44, the mode is 22. That is 2,4,62, 4, 6. It works.

Case 3: median xx (between 22 and 44). We have 22, xx and the mean, with 22 smaller than xx.
- Order 2,x,mean2, x, \text{mean}: the middle one is the average of the other two, so 2x=2+25+x72x = 2 + \frac{25 + x}{7}. Multiply by 77: 14x=14+25+x14x = 14 + 25 + x, so 13x=3913x = 39 and x=3x = 3. The mean is 287=4\frac{28}{7} = 4, giving 2,3,42, 3, 4. It works.
- Order 2,mean,x2, \text{mean}, x: 2⋅25+x7=2+x2 \cdot \frac{25 + x}{7} = 2 + x gives 50+2x=14+7x50 + 2x = 14 + 7x, so x=365=7.2x = \frac{36}{5} = 7.2. That is not between 22 and 44.
- Order mean,2,x\text{mean}, 2, x: 4=25+x7+x4 = \frac{25 + x}{7} + x gives 28=25+8x28 = 25 + 8x, so x=38x = \frac38. Not between 22 and 44.

  1. The possible values are x=17x = 17 and x=3x = 3. Their sum is 2020.

The answer is (E) 20\boxed{\textbf{(E)}\ 20}.

Why this works

Two of the three statistics are easy: the mode never changes and the mean is a simple formula. Only the median depends on where xx lands, so split into cases by the median. Each case gives a simple equation, and every answer must be checked against the range of xx that defined its case. The word "non-constant" rules out the whole first case.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The value 22 occurs three times and nothing else can catch up, so the mode is 22 regardless of xx. The mean is 25+x7\frac{25 + x}{7}. Sorted without xx the list is 2,2,2,4,5,102, 2, 2, 4, 5, 10; the median is the fourth of seven values, so it is 22 if x≤2x \le 2, it is xx if 2≤x≤42 \le x \le 4, and it is 44 if x≥4x \ge 4.

Median 22 (x≤2x \le 2). Mode and median both equal 22; a progression with a repeated term is constant. Rejected.

Median 44 (x≥4x \ge 4). The three numbers are 22, 44 and the mean. For them to form a progression, the mean must be 66 (giving 2,4,62, 4, 6), 33 (giving 2,3,42, 3, 4) or 00 (giving 0,2,40, 2, 4). Solving 25+x7=6,3,0\frac{25 + x}{7} = 6, 3, 0 gives x=17,−4,−25x = 17, -4, -25. Only 1717 satisfies x≥4x \ge 4.

Median xx (2<x<42 < x < 4). The numbers are 22, xx and the mean, with 2<x2 < x.
- Order 2,x,mean2, x, \text{mean}: 2x=2+25+x72x = 2 + \frac{25 + x}{7}, so 14x=39+x14x = 39 + x and x=3x = 3. Then the mean is 44 and 2,3,42, 3, 4 works.
- Order 2,mean,x2, \text{mean}, x: 2(25+x)7=2+x\frac{2(25 + x)}{7} = 2 + x gives x=365x = \frac{36}{5}, outside the range.
- Order mean,2,x\text{mean}, 2, x: 25+x7=4−x\frac{25 + x}{7} = 4 - x gives x=38x = \frac38, outside the range.

The possible values are 1717 and 33, with sum 2020.

The answer is (E) 20\boxed{\textbf{(E)}\ 20}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site. Also appeared as 2000 12 #14.