Difference between revisions of "2016 AMC 8 Problems/Problem 5"
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+ | == Problem == | ||
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The number <math>N</math> is a two-digit number. | The number <math>N</math> is a two-digit number. | ||
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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> | ||
− | == | + | ==Solutions== |
− | + | ===Solution 1=== | |
− | |||
− | ==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> in order for it to have the desired remainder<math>\pmod {10}.</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> in order for it to have the desired remainder<math>\pmod {10}.</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 ~ More efficient for proofs== | + | ===Solution 2 ~ More efficient for proofs=== |
This two digit number must take the form of <math>10x+y,</math> where <math>x</math> and <math>y</math> are integers <math>0</math> to <math>9.</math> However, if x is an integer, we must have <math>y=3.</math> So, the number's new form is <math>10x+3.</math> This needs to have a remainder of <math>1</math> when divided by <math>9.</math> Because of the <math>9</math> divisibility rule, we have <cmath>10x+3 \equiv 1 \pmod 9.</cmath> | This two digit number must take the form of <math>10x+y,</math> where <math>x</math> and <math>y</math> are integers <math>0</math> to <math>9.</math> However, if x is an integer, we must have <math>y=3.</math> So, the number's new form is <math>10x+3.</math> This needs to have a remainder of <math>1</math> when divided by <math>9.</math> Because of the <math>9</math> divisibility rule, we have <cmath>10x+3 \equiv 1 \pmod 9.</cmath> | ||
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==Video Solution== | ==Video Solution== | ||
− | + | https://youtu.be/7an5wU9Q5hk?t=574 | |
− | https:// | ||
{{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 03:52, 16 January 2021
Contents
Problem
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 ?
Solutions
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 in order for it to have the desired remainder 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 ~ More efficient for proofs
This two digit number must take the form of where and are integers to However, if x is an integer, we must have So, the number's new form is This needs to have a remainder of when divided by Because of the divisibility rule, we have We subtract the three, getting which simplifies to However, so and
Let the quotient of in our modular equation be and let our desired number be so and We substitute these values into and get so As a result,
- Alternatively, we could have also used a system of modular equations to immediately receive
To prove generalization vigorously, we can let be the remainder when is divided by Setting up a modular equation, we have Simplifying, If then we don't have a 2 digit number! Thus, and
Video Solution
https://youtu.be/7an5wU9Q5hk?t=574
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.