Difference between revisions of "2016 AMC 8 Problems/Problem 5"
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<math>\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7</math> | <math>\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7</math> | ||
− | ==Solution== | + | ==Solution 1== |
From the second bullet point, we know that the second digit must be <math>3</math>. Because there is a remainder of <math>1</math> when it is divided by <math>9</math>, the multiple of <math>9</math> must end in a <math>2</math>. We now look for this one: | From the second bullet point, we know that the second digit must be <math>3</math>. Because there is a remainder of <math>1</math> when it is divided by <math>9</math>, the multiple of <math>9</math> must end in a <math>2</math>. We now look for this one: | ||
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The number <math>72+1=73</math> satisfies both conditions. We subtract the biggest multiple of <math>11</math> less than <math>73</math> to get the remainder. Thus, <math>73-11(6)=73-66=\boxed{\textbf{(E) }7}</math>. | The number <math>72+1=73</math> satisfies both conditions. We subtract the biggest multiple of <math>11</math> less than <math>73</math> to get the remainder. Thus, <math>73-11(6)=73-66=\boxed{\textbf{(E) }7}</math>. | ||
+ | ==Solution 2== | ||
+ | |||
+ | We can use modular arithmetic to solve this. | ||
+ | Firstly, we can form the equations: | ||
+ | |||
+ | <math>1\equiv N \pmod{9}\\ | ||
+ | 3\equiv N \pmod{10}.</math> | ||
+ | |||
+ | Therefore, <math>N = 9x + 1</math> and <math>N = 10y + 3</math>. | ||
+ | Since the divisibility rule for <math>10</math> is that the last digit has to be <math>0</math>, we can say that a number that has a remainder of <math>3</math> when divided by <math>10</math> ends in <math>3</math>. | ||
+ | |||
+ | As the number is <math>1</math> more than a multiple of <math>9</math>, the multiple of <math>9</math> ends in <math>2</math>. The numbers that are greater than <math>9</math> and end in <math>2</math> are: | ||
+ | <math>12, 22, 32, 42, 52, 62, 72, 82, 92</math> and so on. We can see that <math>72</math> is the smallest positive multiple of <math>9</math> that ends in a <math>2</math>, so <math>N</math> must equal <math>72 + 1 = 73</math>. | ||
+ | |||
+ | Now, we need to find the remainder when <math>73</math> is divided by <math>11</math>. The largest multiple of <math>11</math> that is less than <math>73</math> is <math>66</math>, so <math>73 - 66 = 7</math> is the remainder when <math>N</math> is divided by <math>11</math>. | ||
+ | |||
+ | Our answer is <math>\boxed{\textbf{(E) }7}</math>. | ||
{{AMC8 box|year=2016|num-b=4|num-a=6}} | {{AMC8 box|year=2016|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:37, 21 December 2016
The number is a two-digit number.
• When is divided by , the remainder is .
• When is divided by , the remainder is .
What is the remainder when is divided by ?
Solution 1
From the second bullet point, we know that the second digit must be . Because there is a remainder of when it is divided by , the multiple of must end in a . We now look for this one:
The number satisfies both conditions. We subtract the biggest multiple of less than to get the remainder. Thus, .
Solution 2
We can use modular arithmetic to solve this. Firstly, we can form the equations:
Therefore, and . Since the divisibility rule for is that the last digit has to be , we can say that a number that has a remainder of when divided by ends in .
As the number is more than a multiple of , the multiple of ends in . The numbers that are greater than and end in are: and so on. We can see that is the smallest positive multiple of that ends in a , so must equal .
Now, we need to find the remainder when is divided by . The largest multiple of that is less than is , so is the remainder when is divided by .
Our answer is .
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.