Difference between revisions of "2016 AMC 8 Problems/Problem 5"
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+ | == Problem == | ||
+ | |||
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> | ||
− | ==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> in order for it to have the desired remainder<math>\pmod {10}.</math> We now look for this one: | ||
<math>9(1)=9\\ | <math>9(1)=9\\ | ||
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9(8)=72</math> | 9(8)=72</math> | ||
− | The number <math>72+1=73</math> satisfies both conditions. We subtract the biggest multiple of 11 less than 73 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== | ||
+ | |||
+ | 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> | ||
+ | We subtract the three, getting <cmath>10x \equiv -2 \pmod 9.</cmath> | ||
+ | which simplifies to <cmath>10x \equiv 7 \pmod 9.</cmath> | ||
+ | However, <math>9x \equiv 0 \pmod 9,</math> so <cmath>10x - 9x \equiv 7 - 0 \pmod 9</cmath> and <cmath>x \equiv 7 \pmod 9.</cmath> | ||
+ | |||
+ | Let the quotient of <math>9</math> in our modular equation be <math>c,</math> and let our desired number be <math>z,</math> so <math>x=9c+7</math> and <math>z = 10x+3.</math> We substitute these values into <math>z = 10x+3,</math> and get <cmath> z = 10(9c+7) + 3</cmath> so <cmath> z = 90c+73.</cmath> As a result, <math>z \equiv 73 \pmod {90}.</math> | ||
+ | |||
+ | *Alternatively, we could have also used a system of modular equations to immediately receive <math>z \equiv 73 \pmod {90}.</math> | ||
+ | |||
+ | To prove generalization vigorously, we can let <math>a</math> be the remainder when <math>z</math> is divided by <math>11.</math> Setting up a modular equation, we have <cmath>90c + 73 \equiv a \pmod {11}.</cmath> Simplifying, <cmath>90c+7 \equiv a \pmod {11}</cmath> If <math>c = 1,</math> then we don't have a 2 digit number! Thus, <math>c=0</math> and <math>a=\boxed { \textbf{(E) }7}</math> | ||
+ | |||
+ | ==Solution 3== | ||
+ | We know that the number has to be one more than a multiple of 9, because of the remainder of one, and the number has to be 3 more than a multiple of 10, which means that it has to end in a <math>3</math>. Now, if we just list the first few multiples of 9 adding one to the number we get: <math>10, 19, 28, 37, 46, 55, 64, 73, 82, 91</math>. As we can see from these numbers, the only one that has a three in the denominator is <math>73</math>, thus we divide <math>73</math> by <math>11</math>, getting <math>6</math> <math>R7</math>, hence, <math>\boxed{\textbf{(E) }7}</math>. | ||
+ | -fn106068 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/7an5wU9Q5hk?t=574 | ||
+ | |||
{{AMC8 box|year=2016|num-b=4|num-a=6}} | {{AMC8 box|year=2016|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:32, 1 April 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 ?
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
Solution 3
We know that the number has to be one more than a multiple of 9, because of the remainder of one, and the number has to be 3 more than a multiple of 10, which means that it has to end in a . Now, if we just list the first few multiples of 9 adding one to the number we get: . As we can see from these numbers, the only one that has a three in the denominator is , thus we divide by , getting , hence, . -fn106068
Video Solution
https://youtu.be/7an5wU9Q5hk?t=574
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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All AJHSME/AMC 8 Problems and Solutions |
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