Difference between revisions of "2021 Fall AMC 12B Problems/Problem 25"
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− | Upon adding one to <math>n</math>, consider each individual remainder. Either it will increase by 1, or it will "wrap around', going from <math>5\rightarrow 0</math> mod 6 and in general, <math>(n-1) \rightarrow 0</math> mod n. We will use '<math>+1</math>' to refer to remainders that increase by 1, and 'wrap-around's to refer to remainders that go to 0. | + | Upon adding one to <math>n</math>, consider each individual remainder. Either it will increase by 1, or it will "wrap around', going from <math>5\rightarrow 0</math> (mod 6) and in general, <math>(n-1) \rightarrow 0</math> (mod n). We will use '<math>+1</math>' to refer to remainders that increase by 1, and 'wrap-around's to refer to remainders that go to 0. |
Revision as of 19:32, 4 November 2024
Contents
[hide]Problem
For a positive integer, let be the sum of the remainders when is divided by , , , , , , , , and . For example, . How many two-digit positive integers satisfy
Solution 1
Note that we can add to to get , but must subtract for all . Hence, we see that there are four ways to do that because . Note that only is a plausible option, since indicates is divisible by , indicates that is divisible by , indicates is divisible by , and itself indicates divisibility by , too. So, and is not divisible by any positive integers from to , inclusive, except and . We check and get that only and give possible solutions so our answer is .
- kevinmathz
Solution 2
Denote by the remainder of divided by . Define .
Hence,
Hence, this problem asks us to find all , such that .
: .
We have .
Therefore, there is no in this case.
: and .
The condition implies . This further implies . Hence, .
To get , we have .
However, we have .
Therefore, there is no in this case.
: for and .
The condition implies with . Hence, and .
To get , we have .
However, we have .
Therefore, there is no in this case.
: for and .
To get , we have .
Hence, we must have and for .
Therefore, .
: for and .
The condition implies with . Hence, and .
To get , we have .
However, we have .
Therefore, there is no in this case.
: for and .
To get , we have .
This can be achieved if , , .
However, implies . This implies . Hence, . We get a contradiction.
Therefore, there is no in this case.
: for and .
The condition implies with . Hence, .
To get , we have . This implies .
Because and , we have . Hence, . However, in this case, we assume . We get a contradiction.
Therefore, there is no in this case.
: for and .
To get , we have . This is infeasible.
Therefore, there is no in this case.
: for .
To get , we have . This is infeasible.
Therefore, there is no in this case.
Putting all cases together, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 3
To get from to , would add by for each remainder . However, given that some of these remainders can "round down" to given the nature of mods, we must calculate the possible values of such that the remainders in "rounds down" by a total of , effectively canceling out the adding by initially.
To do so, we will analyze the "rounding down" for each of :
: subtract by
: subtract by
: subtract by , but this also implies mod , so subtract by .
: subtract by
: subtract by , but this also implies mod and , so subtract by : too much
: subtract by
: subtract by , but this also implies mod and , so subtract by : too much
: subtract by , but this also implies mod , so subtract by : too much
: subtract by : too much
Notice that . By testing these sums, we can easily show that the only time when the total subtraction is is when AND . By CRT, :
As in solution 1, then, only and give possible solutions, so our answer is .
~xHypotenuse
Solution 4
Upon adding one to , consider each individual remainder. Either it will increase by 1, or it will "wrap around', going from (mod 6) and in general, (mod n). We will use '' to refer to remainders that increase by 1, and 'wrap-around's to refer to remainders that go to 0.
Clearly, s isn't possible, since then .
If there are s and wrap-around, the wrap-around must be equal to , which is the case for . However, if is mod , it clearly must also be mod , meaning there must more be one wrap-around, and this case won't work.
If there are s and wrap-arounds, these two wrap-arounds must add to . For the possible modulo, we could have , , and . Clearly, won't work since if it is mod , then it must also be mod , meaning won't be the only wrap-arounds. Similarly, doesn't work since mod implies that mod will also be a wrap-around. That leaves . The number must be mod and mod , or in other words, mod and mod , meaning n will be mod . Testing all such two digits numbers that are equivalent to mod , we see that and are the only two that work.
If there are s and wrap-arounds, the only possible combination of modulo is . Thus, must be mod . However, this means that mod will also be a wrap around, so this case won't work.
Notice that there can be no more cases, as for s, no matter what mods wrap around, the s will not be able to balance them out, as it's magnitude is too small. Therefore, there are only numbers.
Video Solution
~MathProblemSolvingSkills.com
Video Solution by Mathematical Dexterity
https://www.youtube.com/watch?v=Fy8wU4VAzkQ
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