2016 AMC 8 Problems/Problem 5: Difference between revisions
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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 1== | ||
From the second bullet point, we know that the second digit must be <math>3</math>. | 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: | ||
<math>9(1)=9\\ | <math>9(1)=9\\ | ||
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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>. | ||
~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== | ==Video Solution== | ||
https:// | 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 | |||
==Solution 3== | |||
We can set up a system of modular congruences | |||
n = 1 (mod 9) | |||
n = 3 (mod 10) | |||
9x + 1 = 3 (mod 10) | |||
-1 -1 | |||
9x = 2(mod 10) | |||
+70 | |||
9x = 72(mod 10) | |||
/9 /9 | |||
x = 8 (mod 10) | |||
10f + 8 | |||
9(10f+8)+1= n | |||
90f + 72 + 1 = n | |||
90f + 73 = n | |||
n = 73 (mod 90) | |||
73/11 = 6 r7 | |||
- timi821 | |||
==Video Solution== | |||
https://youtu.be/aKWQl7kEMy0 | |||
~savannahsolver | |||
Latest revision as of 09:32, 9 November 2025
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
Solution 3
We can set up a system of modular congruences
n = 1 (mod 9) n = 3 (mod 10)
9x + 1 = 3 (mod 10)
-1 -1
9x = 2(mod 10)
+70
9x = 72(mod 10) /9 /9
x = 8 (mod 10)
10f + 8 9(10f+8)+1= n 90f + 72 + 1 = n 90f + 73 = n n = 73 (mod 90)
73/11 = 6 r7
- timi821
Video Solution
~savannahsolver
