2017 AMC 10B Problems/Problem 14
Contents
Problem
An integer is selected at random in the range . What is the probability that the remainder when is divided by is ?
Solution 1
Notice that we can rewrite as . By Fermat's Little Theorem, we know that if . Therefore for all we have . Since , and is divisible by , of the possible are divisible by . Therefore, with probability or .
Solution 2
Note that the patterns for the units digits repeat, so in a sense we only need to find the patterns for the digits . The pattern for is , no matter what power, so doesn't work. Likewise, the pattern for is always . Doing the same for the rest of the digits, we find that the units digits of , ,, ,, , and all have the remainder of when divided by , so .
Explanation:
Since the remainder is 1, the units digit is either 1 or 6. Since we only care about the units digits, we don't need to think about anything beyond units digits.
We count in modulus ten (fancy word of saying ignore everything except the units digit)
For 1:
1, 1, 1, 1... anything with a units digit one will always have powers that have a units digit one.
For 2:
2, 4, 8, 6, 2, 4, 8, 6... we see that it repeats with a cycle of 4. Since we are looking for the 16th power, the units digit will be 6, and this works.
For 3:
3, 9, 7, 1, 3, 9, 7, 1... similarly to 2, a cycle of 4 and the 16th power has a units digit of 1. This works.
For 4:
4, 6, 4, 6... a cycle of 4, 16th power has a units digit of 6. This works.
For 5:
5, 5, 5, 5... the units digit will always be 5, and this doesn't work.
For 6:
6, 6, 6, 6... the units digit will always be 6, and this works.
For 7:
7, 9, 3, 1, 7, 9, 3, 1... a cycle of 4, 16th power has a units digit of 1. This works.
For 8:
8, 4, 2, 6, 8, 4, 2, 6... a cycle of 4, 16th power has a units digit of 6. This works.
For 9:
9, 1, 9, 1... a cycle of 2, 16th power has a units digit of 1. This works.
Last but not least, 0:
0, 0, 0... the units digit will always be 0, and this does not work.
To sum up: There are ten choices for the units digit. Of these ten choices, 1, 2, 3, 4, 6, 7, 8, 9 all work, and the probability is therefore which is answer choice .
~Explanation by JH. L
Solution 3 (Casework)
We can use modular arithmetic for each residue of
If , then
If , then
If , then
If , then
If , then
In out of the cases, the result was , and since each case occurs equally as , the answer is
Video Solution 1 by OmegaLearn
https://youtu.be/zfChnbMGLVQ?t=2410
~ pi_is_3.14
Video Solution 2
~savannahsolver
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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All AMC 10 Problems and Solutions |
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