Difference between revisions of "2018 AMC 8 Problems/Problem 21"
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==Solution 1== | ==Solution 1== | ||
| − | Looking at the values, we notice that <math>11-7=4</math>, <math>9-5=4</math> and <math>6-2=4</math>. This means we are looking for a value that is four less than a multiple of <math>11</math>, <math>9</math>, and <math>6</math>. The least common multiple of these numbers is <math>11\cdot3^{2}\cdot2=198</math>, so the numbers that fulfill this can be written as <math>198k-4</math>, where <math>k</math> is a positive integer. This value is only a three digit integer when <math>k</math> is <math>1, 2, 3, 4</math> or <math>5</math>, which gives <math>194, 392, 590, 788,</math> and <math>986</math> respectively. Thus we have <math>5</math> values, so our answer is <math>\boxed{\textbf{(E) }5}</math> | + | Looking at the values, we notice that <math>11-7=4</math>, <math>9-5=4</math> and <math>6-2=4</math>. This means we are looking for a value that is four less than a multiple of <math>11</math>, <math>9</math>, and <math>6</math>. The least common multiple of these numbers is <math>11\cdot3^{2}\cdot2=198</math>, so the numbers that fulfill this can be written as <math>198k-4</math>, where <math>k</math> is a positive integer. This value is only a three-digit integer when <math>k</math> is <math>1, 2, 3, 4</math> or <math>5</math>, which gives <math>194, 392, 590, 788,</math> and <math>986</math> respectively. Thus, we have <math>5</math> values, so our answer is <math>\boxed{\textbf{(E) }5}</math>. |
==Solution 2== | ==Solution 2== | ||
| − | Let us create the equations: <math>6x+2 = 9y+5 = 11z+7</math>, and we know <math>100 \leq 11z+7 <1000</math>, it gives us <math>9 \leq z \leq 90</math>, which is the range of the value of z. Because of <math>6x+2=11z+7</math>, then <math>6x=11z+5=6z+5(z+1)</math>, so (z+1) must be | + | Let us create the equations: <math>6x+2 = 9y+5 = 11z+7</math>, and we know <math>100 \leq 11z+7 <1000</math>, it gives us <math>9 \leq z \leq 90</math>, which is the range of the value of z. Because of <math>6x+2=11z+7</math>, then <math>6x=11z+5=6z+5(z+1)</math>, so <math>(z+1)</math> must be a mutiple of 6. Because of <math>9y+5=11z+7</math>, then <math>9y=11z+2=9z+2(z+1)</math>, so <math>(z+1)</math> must also be a mutiple of <math>9</math>. Hence, the value of <math>(z+1)</math> must be a common multiple of <math>6</math> and <math>9</math>, which means multiples of <math>18 (LCM \text{ of }\ 6, 9)</math>. So, let's say <math>z+1 = 18p</math>; then, <math>9 \leq z = 18p-1 \leq 90</math>, so <math>1 \leq p \leq 91/18\ or \ 1 \leq p \leq 5</math>. Thus, the answer is <math>\boxed{\textbf{(E) }5}</math>. |
| + | |||
| + | ~LarryFlora | ||
| + | |||
==Solution 3== | ==Solution 3== | ||
By the [[Chinese Remainder Theorem]], we have that all solutions are in the form <math>x=198k+194</math> where <math>k\in \mathbb{Z}.</math> Counting the number of values, we get <math>\boxed{\textbf{(E) }5}.</math> | By the [[Chinese Remainder Theorem]], we have that all solutions are in the form <math>x=198k+194</math> where <math>k\in \mathbb{Z}.</math> Counting the number of values, we get <math>\boxed{\textbf{(E) }5}.</math> | ||
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~mathboy282 | ~mathboy282 | ||
| − | ==Video Solution== | + | ==Solution 4== |
| − | https://youtu.be/CPQpkpnEuIc - | + | We can use modular arithmetic. Set up the equations: <math>x \equiv 2 \mod 6,</math> <math>x \equiv 5 \mod 9,</math> and <math>x \equiv 7 \mod 11.</math> These equations can also be written as <math>x+4 \equiv 0 \mod 6,</math> <math>x+4 \equiv 0 \mod 9,</math> and <math>x+4 \equiv 0 \mod 11.</math> Since <math>x+4</math> is congruent to numbers <math>6, 9,</math> and <math>11,</math> then it must also be congruent to their LCM. Thus, <math>x+4 \equiv 0 \mod 198,</math> since 198 is the LCM of <math>6, 9,</math> and <math>11.</math> Since these numbers have to be three digits, they can only be <math>194, 392, 590, 788,</math> and <math>986.</math> This gives us the answer of <math>\boxed{\textbf{(E) }5}.</math> |
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| + | ~ethancui0529 | ||
| + | |||
| + | ==Solution 5== | ||
| + | Let <math>N</math> be the three digit positive integer. <math>N = 6a + 2 = 9b + 5 = 11c + 7</math>. Then, we add four to all sides and write <math>N + 4 = 6(a+1) = 9(b+1) = 11(c+1)</math>. Now, we know that <math>N + 4</math> is divisible by 6, 9, and 11. The LCM of 6, 9, and 11 is equal to 198, so <math>N = 198k - 4</math>. From this, we can figure out that <math>N</math> can be 5 different three digit numbers -- <math>194, 392, 590, 788,</math> and <math>986</math>. <math>\therefore</math>, the answer is <math>\boxed{\textbf{(E) }5}.</math> | ||
| + | |||
| + | ~DY | ||
| + | |||
| + | == Video Solution by Pi Academy (Fast and Easy) == | ||
| + | |||
| + | https://youtu.be/7xlBdxcsP3I?si=lfZZBB7i_6tOlKr- | ||
| + | |||
| + | |||
| + | ==Video Solution (CREATIVE THINKING!!!)== | ||
| + | https://youtu.be/kb_0r9fEyEI | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/7an5wU9Q5hk?t=939 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | ==Video Solutions== | ||
| + | https://youtu.be/CPQpkpnEuIc | ||
| + | |||
| + | ~ Happytwin | ||
| + | |||
| + | https://youtu.be/hoCdk8AC-0c | ||
| − | + | ~savannahsolver | |
| − | https:// | + | https://www.youtube.com/watch?v=PjYwbGm_2aM ~David |
==See Also== | ==See Also== | ||
Latest revision as of 18:18, 23 November 2024
Contents
Problem
How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
Solution 1
Looking at the values, we notice that 
, 
and 
. This means we are looking for a value that is four less than a multiple of 
, 
, and 
. The least common multiple of these numbers is 
, so the numbers that fulfill this can be written as 
, where 
is a positive integer. This value is only a three-digit integer when is 
or 
, which gives 
and 
respectively. Thus, we have values, so our answer is 
.
Solution 2
Let us create the equations: 
, and we know 
, it gives us 
, which is the range of the value of z. Because of 
, then 
, so 
must be a mutiple of 6. Because of 
, then 
, so must also be a mutiple of . Hence, the value of must be a common multiple of and , which means multiples of 
. So, let's say 
; then, 
, so 
. Thus, the answer is .
~LarryFlora
Solution 3
By the Chinese Remainder Theorem, we have that all solutions are in the form 
where 
Counting the number of values, we get 
~mathboy282
Solution 4
We can use modular arithmetic. Set up the equations: 
and 
These equations can also be written as 
and 
Since 
is congruent to numbers 
and 
then it must also be congruent to their LCM. Thus, 
since 198 is the LCM of and 
Since these numbers have to be three digits, they can only be and 
This gives us the answer of
~ethancui0529
Solution 5
Let 
be the three digit positive integer. 
. Then, we add four to all sides and write 
. Now, we know that 
is divisible by 6, 9, and 11. The LCM of 6, 9, and 11 is equal to 198, so 
. From this, we can figure out that can be 5 different three digit numbers -- and . 
, the answer is
~DY
Video Solution by Pi Academy (Fast and Easy)
https://youtu.be/7xlBdxcsP3I?si=lfZZBB7i_6tOlKr-
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/7an5wU9Q5hk?t=939
~ pi_is_3.14
Video Solutions
~ Happytwin
~savannahsolver
https://www.youtube.com/watch?v=PjYwbGm_2aM ~David
See Also
| 2018 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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. 
