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. | ||
• When <math>N</math> is divided by <math>9</math>, the remainder is <math>1</math>. | • When <math>N</math> is divided by <math>9</math>, the remainder is <math>1</math>. | ||
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• When <math>N</math> is divided by <math>10</math>, the remainder is <math>3</math>. | • When <math>N</math> is divided by <math>10</math>, the remainder is <math>3</math>. | ||
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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>, for a number divisible by <math>10</math> ends in zero. Since there is a remainder of <math>1</math> when <math>N</math> is divided by <math>9</math>, the multiple of <math>9</math> must end in a <math>2</math> for it to have the desired remainder<math>\pmod {10}.</math> We now look for this one: | ||
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| + | <math>9(1)=9\\ | ||
| + | 9(2)=18\\ | ||
| + | 9(3)=27\\ | ||
| + | 9(4)=36\\ | ||
| + | 9(5)=45\\ | ||
| + | 9(6)=54\\ | ||
| + | 9(7)=63\\ | ||
| + | 9(8)=72</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>. | ||
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| + | ~CHECKMATE2021 | ||
| + | |||
| + | ==Solution 2== | ||
| + | We know that the number has to be one more than a multiple of <math>9</math>, because of the remainder of one, and the number has to be <math>3</math> more than a multiple of <math>10</math>, which means that it has to end in a <math>3</math>. Now, if we just list the first few multiples of <math>9</math> 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 units place 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 | ||
| + | |||
| + | We could also remember that, for a two-digit number to be divisible by <math>9</math>, the sum of its digits has to be equal to <math>9</math>. Since the number is one more than a multiple of <math>9</math>, the multiple we are looking for has a ones digit of <math>2</math>, and therefore a tens digit of <math>9-2 = 7</math>, and then we could proceed as above. -vaisri | ||
| + | |||
| + | ==Video Solution== | ||
| + | |||
| + | https://youtu.be/d-bCEDoZEjg?si=VFLhpgyJ_vHhE7h3 | ||
| + | |||
| + | A solution so simple a 12-year-old made it! | ||
| + | |||
| + | ~Elijahman~ | ||
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/7an5wU9Q5hk?t=574 | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/aKWQl7kEMy0 | ||
| + | |||
| + | ~savannahsolver | ||
| + | ==See Also== | ||
{{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 15:52, 1 July 2024
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 , for a number divisible by ends in zero. Since there is a remainder of when is divided by , the multiple of must end in a 
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, 
.
~CHECKMATE2021
Solution 2
We know that the number has to be one more than a multiple of , because of the remainder of one, and the number has to be more than a multiple of , which means that it has to end in a . Now, if we just list the first few multiples of adding one to the number we get: 
. As we can see from these numbers, the only one that has a three in the units place is , thus we divide by , getting 
, hence, 
.
-fn106068
We could also remember that, for a two-digit number to be divisible by , the sum of its digits has to be equal to . Since the number is one more than a multiple of , the multiple we are looking for has a ones digit of , and therefore a tens digit of 
, and then we could proceed as above. -vaisri
Video Solution
https://youtu.be/d-bCEDoZEjg?si=VFLhpgyJ_vHhE7h3
A solution so simple a 12-year-old made it!
~Elijahman~
Video Solution by OmegaLearn
https://youtu.be/7an5wU9Q5hk?t=574
Video Solution
~savannahsolver
See Also
| 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. 
