AMC 10 Step by Step

2024 AMC 10A Problem 10

P1-10~3 minPrint

Consider the following operation. Given a positive integer nn , if nn is a multiple of 33 , then you replace nn by n3\frac{n}{3} . If nn is not a multiple of 33 , then you replace nn by n+10n+10 . For example, beginning with n=4n=4 , this procedure gives 4→14→24→8→18→6→2→12→⋯4\to14\to24\to8\to18\to6\to2\to12\to\cdots . Suppose you start with n=100n=100 . What value results if you perform this operation exactly 100100 times?

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Key insight

Work out the first ten or so steps. From step 6 the values repeat 20, 30, 10, so step 100 matches step 7, which is 30.

Solution

Doing 100100 steps by hand is too many, but the numbers might start repeating. So do the first few steps and watch.

  1. Start at 100100. It is not a multiple of 33, so add 1010: 110110. Keep going:
step1234567891011
value110120405060203010203010

(For example, 120120 is a multiple of 33, so it becomes 4040; 6060 becomes 2020; 3030 becomes 1010; 1010 is not a multiple of 33, so it becomes 2020.)
2. At step 99 we are back at 2020, the value from step 66. From then on the same three numbers repeat forever: 20,30,10,20,30,10,…20, 30, 10, 20, 30, 10, \ldots
3. Match each step with its value:
- steps 6,9,12,…6, 9, 12, \ldots (multiples of 33) give 2020,
- steps 7,10,13,…7, 10, 13, \ldots (one more than a multiple of 33) give 3030,
- steps 8,11,14,…8, 11, 14, \ldots (two more than a multiple of 33) give 1010.
4. 100=99+1100 = 99 + 1, and 99=3⋅3399 = 3 \cdot 33. So 100100 is one more than a multiple of 33, just like step 77. The value is 3030.

The answer is (C) 30\boxed{\textbf{(C)}\ 30}.

Why this works

A rule that always does the same thing to the same number must start repeating as soon as a number comes back. Work out steps until you see a repeat, note how long the repeating block is, and then use division to find where the target step lands in the block. Line it up using a step you have actually computed, so you do not end up one step off.

Concise solution

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

Apply the rule and record the result after each step:

step1234567891011
value110120405060203010203010

From step 66 on the values repeat 20,30,1020, 30, 10 with period 33: step 6,9,12,…6, 9, 12, \dots (multiples of 33) give 2020; steps 7,10,13,…7, 10, 13, \dots give 3030; steps 8,11,14,…8, 11, 14, \dots give 1010.

Since 100=3⋅33+1100 = 3\cdot 33 + 1 is one more than a multiple of 33 (like step 77), the value after 100100 operations is 3030.

The answer is (C) 30\boxed{\textbf{(C)}\ 30}.

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.