Difference between revisions of "2017 AMC 10B Problems/Problem 14"
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Note that the patterns for the units digits repeat, so in a sense we only need to find the patterns for the digits <math>0-9</math> . | Note that the patterns for the units digits repeat, so in a sense we only need to find the patterns for the digits <math>0-9</math> . | ||
The pattern for <math>0</math> is <math>0</math>, no matter what power, so <math>0</math> doesn't work. Likewise, the pattern for <math>5</math> is always <math>5</math>. Doing the same for the rest of the digits, we find that the units digits of <math>1^{16}</math>, <math>2^{16}</math> ,<math>3^{16}</math>, <math>4^{16}</math> ,<math>6^{16}</math>, <math>7^{16}</math> ,<math>8^{16}</math> and <math>9^{16}</math> all have the remainder of <math>1</math> when divided by <math>5</math>, so <math>\boxed{\textbf{(D) } \frac 45}</math>. | The pattern for <math>0</math> is <math>0</math>, no matter what power, so <math>0</math> doesn't work. Likewise, the pattern for <math>5</math> is always <math>5</math>. Doing the same for the rest of the digits, we find that the units digits of <math>1^{16}</math>, <math>2^{16}</math> ,<math>3^{16}</math>, <math>4^{16}</math> ,<math>6^{16}</math>, <math>7^{16}</math> ,<math>8^{16}</math> and <math>9^{16}</math> all have the remainder of <math>1</math> when divided by <math>5</math>, so <math>\boxed{\textbf{(D) } \frac 45}</math>. | ||
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+ | ==Solution 3 (casework)== | ||
+ | We can use modular arithmetic for each case of <math>n(\text{mod }5)</math> | ||
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+ | If <math>n \equiv 0(\text{mod }5)</math>, then <math>n^{16} \equiv 0^{16} \equiv 0 (\text{mod }5)</math> | ||
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+ | If <math>n \equiv 1(\text{mod }5)</math>, then <math>n^{16} \equiv 1^{16} \equiv 1 (\text{mod }5)</math> | ||
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+ | If <math>n \equiv 2(\text{mod }5)</math>, then <math>n^{16} \equiv (n^2)^8 \equiv 4^8 \equiv (-1)^8 \equiv 1 (\text{mod }5)</math> | ||
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+ | If <math>n \equiv 3(\text{mod }5)</math>, then <math>n^{16} \equiv (n^4)^4 \equiv 81^4 \equiv (1)^4 \equiv 1 (\text{mod }5)</math> | ||
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+ | If <math>n \equiv 4(\text{mod }5)</math>, then <math>n^{16} \equiv (-1)^{16} \equiv 1 (\text{mod }5)</math> | ||
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+ | In <math>4</math> out of the <math>5</math> cases, the result was <math>1(\text{mod }5)</math>, and since each case occurs equally, the answer is <math>\boxed{\textbf{(D) }\frac{4}{5}}</math> | ||
{{AMC10 box|year=2017|ab=B|num-b=13|num-a=15}} | {{AMC10 box|year=2017|ab=B|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:19, 26 December 2017
Problem
An integer is selected at random in the range . What is the probability that the remainder when is divided by is ?
Solution 1
By Fermat's Little Theorem, when N is relatively prime to 5. However, this happens with probability .
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 .
Solution 3 (casework)
We can use modular arithmetic for each case 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, the answer is
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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