Hello everyone! In today’s post we will be solving 2021 AIME I #4. The way this problem is written may make the question may make it look simple. However, keep in mind that this is a math competition problem so don’t let it deceive you!

For those of you who want to first try this question out yourselves, feel free to do so! However, please stop here as below the stop sign we will be discussing the solution to this problem.

Alright based upon what we know define the following variables:

• a be the amount of coins in the first pile
• b be the amount of coins in the second pile
• c be the amount of coins in the third pile

The total of these variables becomes 66 meaning a + b + c = 66. We are also given that the third pile has the most coins, second pile has second most, and the first pile has the least number of coins. Or a < b < c.

Using this new insight, let the difference between the third and second pile be m and the difference between the third and first pile be n. We know the n > m because the first pile is the smallest and the third pile is the largest while the second pile is in between the two piles in terms of items. From this idea we can rewrite our equation to the following:

a+b+c = a + (a + m) + (a+n) = 3a + m + n = 66

As we want a to be an integer, we want 66 – (m + n) to be divisible by 3. So, now we can do casework upon m as m is smaller to figure possible values for n.

• m = 1: This would mean 3a = 65 – n. We notice that for 65 – n to be divisible by 3 we need n to give a remainder of 2 when divided by 3 through modular arithmetic and greater than m as we previously defined n > m. This gives the following numbers to work: 2, 5, 8, 11, 14, … 62. We stop at 62 because 65 would make there exist an empty pile which we want to avoid. This means 21 numbers work.
• m = 2: This would mean 3a = 64 – n. Using a similar strategy we need n to give a remainder of 1 when divided by 3. This gives the following numbers that work: 4, 7, 10, 13 … 61. This means 20 numbers work.
• m = 3: This would mean 3a = 63 – n. This would mean that n must be divisible by 3. This gives the following numbers to work: 6, 9, 12, …, 60. This means 19 numbers work
• m = 4: This would mean 3a = 62 – n. Again n must give remainder 2 when divided by 3. This gives the same numbers that worked for m = 1 except for n = 2 because it is less than 4 and n = 62 as that’d result in an empty pile so 21-2 = 19.
• m = 5: This would mean 3a = 61 – n. n gives remainder 1 when divided by 3. Again we can use similar logic to what we did with m = 4 and remove n = 61 and n = 4 which would leave 18 numbers to work
• m = 6: This would mean 3a = 60 – n. Again n must be divisible by 3. This takes away 60 and 6 from the sequence giving 17 numbers to work:
• m = 7: This would mean 3a = 59 – n. Again n must give remainder 2 when divided by 3. This gives 17 numbers.

From trying small cases for m we can figure out a pattern when adding the cases which leads to the following:

(21+20+19) + (19+18+17)+(17+16+15)+(15+14+13)+(13+12+11)+(11+10+9)+(9+8+7)+(7+6+5)+(5+4+3)+(3+2+1) + 1.

From the equation above it seems that most of the sum makes sense because we see the pattern still exists of the number of ways decreasing for 3 terms staying constant for one and decreasing for 3 terms again. However, what seems off in this pattern is this +1 portion. Why should we do this?

The +1 portion also follows the pattern because like we saw in the pattern of the cases was decreasing for 3 terms and staying constant for one and then decreasing again. This + 1 accounts for the staying constant for one term and as as the terms less than one are zero the other two terms will not impact our sum as those will just not be possible.

By adding the equation we formed using special sum formulas such as sum of odds and sum of first n integers we get:

As we can see we have our answer of 331. The key insight of this problem rather than setting up the equation would be the idea of noticing the pattern of the casework and being consistent with it so the +1 will be accounted for.

Thank you all for reading and see you next post!