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> | ||
| − | == | + | ==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 02:52, 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 ![\[10x+3 \equiv 1 \pmod 9.\]](http://latex.artofproblemsolving.com/8/1/b/81bd1dd4cc2d0a7a8fbdc0621527fbf09a247538.png)
We subtract the three, getting ![\[10x \equiv -2 \pmod 9.\]](http://latex.artofproblemsolving.com/c/c/a/cca71cc504fbbbbb7cf16ea9db3f692875897cf8.png)
which simplifies to ![\[10x \equiv 7 \pmod 9.\]](http://latex.artofproblemsolving.com/4/d/e/4deff7bb0d81272c1c6815181dc6926512698872.png)
However, 
so ![\[10x - 9x \equiv 7 - 0 \pmod 9\]](http://latex.artofproblemsolving.com/8/a/2/8a2138dd6dcf59c80125e8a3e29ab2b643cf8c4c.png)
and ![\[x \equiv 7 \pmod 9.\]](http://latex.artofproblemsolving.com/2/c/4/2c4300571cd8069b6f97f8141dc2e0cd6ab43afc.png)
Let the quotient of in our modular equation be 
and let our desired number be 
so 
and 
We substitute these values into 
and get ![\[z = 10(9c+7) + 3\]](http://latex.artofproblemsolving.com/f/c/7/fc7e72589a215f83b691eb52b7384a56d732c5da.png)
so ![\[z = 90c+73.\]](http://latex.artofproblemsolving.com/8/a/d/8adddfa401df143bfe433152276c3963db2d582f.png)
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 ![\[90c + 73 \equiv a \pmod {11}.\]](http://latex.artofproblemsolving.com/e/4/f/e4fe56412012a1f52beafb9ffaa7754f4eafa04e.png)
Simplifying, ![\[90c+7 \equiv a \pmod {11}\]](http://latex.artofproblemsolving.com/1/4/c/14c916c9ef4c36df1ecf74bfdbfe89e096f24f2e.png)
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 | ||
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