Difference between revisions of "2017 AMC 10B Problems/Problem 14"
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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>. | ||
− | ==Solution 3 ( | + | ==Solution 3 (Casework)== |
− | We can use modular arithmetic for each | + | We can use modular arithmetic for each residue of <math>n \pmod 5</math> |
− | If <math>n \equiv 0 | + | If <math>n \equiv 0 \pmod 5</math>, then <math>n^{16} \equiv 0^{16} \equiv 0 \pmod 5</math> |
− | If <math>n \equiv 1 | + | If <math>n \equiv 1 \pmod 5</math>, then <math>n^{16} \equiv 1^{16} \equiv 1 \pmod 5</math> |
− | If <math>n \equiv 2 | + | If <math>n \equiv 2 \pmod 5</math>, then <math>n^{16} \equiv (n^2)^8 \equiv (2^2)^8 \equiv 4^8 \equiv (-1)^8 \equiv 1 \pmod 5</math> |
− | If <math>n \equiv 3 | + | If <math>n \equiv 3 \pmod 5</math>, then <math>n^{16} \equiv (n^2)^8 \equiv (3^2)^8 \equiv 9^8 \equiv (-1)^8 \equiv 1 \pmod 5</math> |
− | If <math>n \equiv 4 | + | If <math>n \equiv 4 \pmod 5</math>, then <math>n^{16} \equiv 4^{16} \equiv (-1)^{16} \equiv 1 \pmod 5</math> |
− | In <math>4</math> out of the <math>5</math> cases, the result was <math>1 | + | In <math>4</math> out of the <math>5</math> cases, the result was <math>1 \pmod 5</math>, and since each case occurs equally as <math>2020 \equiv 0 \pmod 5</math>, 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 11:25, 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 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
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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