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Difference between revisions of "2003 AMC 10A Problems/Problem 25"

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Since <math>q</math> is a quotient and <math>r</math> is a remainder when <math>n</math> is divided by <math>100</math>. So we have <math>n=100q+r</math>. Since we are counting choices where <math>q+r</math> is divisible by <math>11</math>, we have <math>n=99q+q+r=99q+11k</math> for some <math>k</math>. This means that <math>n</math> is the sum of two multiples of <math>11</math> and would thus itself be a multiple of <math>11</math>. Then we can count all the five digit multiples of <math>11</math> as in Solution 2. (This solution is essentially the same as Solution 2, but it does not necessarily involve mods and so could potentially be faster.)
 
Since <math>q</math> is a quotient and <math>r</math> is a remainder when <math>n</math> is divided by <math>100</math>. So we have <math>n=100q+r</math>. Since we are counting choices where <math>q+r</math> is divisible by <math>11</math>, we have <math>n=99q+q+r=99q+11k</math> for some <math>k</math>. This means that <math>n</math> is the sum of two multiples of <math>11</math> and would thus itself be a multiple of <math>11</math>. Then we can count all the five digit multiples of <math>11</math> as in Solution 2. (This solution is essentially the same as Solution 2, but it does not necessarily involve mods and so could potentially be faster.)
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==== Notes ====
 
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The part labeled "divisor trick" actually follows from the same observation we made in the previous step: <math>10\equiv (-1)\pmod{11}</math>, therefore <math>10^{2k}\equiv 1</math> and <math>10^{2k+1}\equiv (-1)</math> for all <math>k</math>.
 
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For a <math>5-</math>digit number <math>\overline{abcde}</math> we get <math>\overline{abcde}\equiv a\cdot 1 + b\cdot(-1) + c\cdot 1 + d\cdot(-1) + e\cdot 1 = a-b+c-d+e</math>, as claimed.
 
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Also note that in the "divisor trick" we actually want to assign the signs backwards - if we make sure that the last sign is a <math>+</math>, the result will have the same remainder modulo <math>11</math> as the original number.
 
  
 
== See Also ==
 
== See Also ==

Revision as of 17:11, 28 December 2020

Problem

Let $n$ be a $5$-digit number, and let $q$ and $r$ be the quotient and the remainder, respectively, when $n$ is divided by $100$. For how many values of $n$ is $q+r$ divisible by $11$?

$\mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090$

Solutions

Simple Solution

$11|q+r$ implies that $11|100q+r$, so $11|n$. Then, $n$ can range from $10010$ to $99990$ for a total of $\boxed{8181\Rightarrow \mathrm{(B)}}$ numbers.

Solution 1

When a $5$-digit number is divided by $100$, the first $3$ digits become the quotient, $q$, and the last $2$ digits become the remainder, $r$.

Therefore, $q$ can be any integer from $100$ to $999$ inclusive, and $r$ can be any integer from $0$ to $99$ inclusive.

For each of the $9\cdot10\cdot10=900$ possible values of $q$, there are at least $\left\lfloor \frac{100}{11} \right\rfloor = 9$ possible values of $r$ such that $q+r \equiv 0\pmod{11}$.

Since there is $1$ "extra" possible value of $r$ that is congruent to $0\pmod{11}$, each of the $\left\lfloor \frac{900}{11} \right\rfloor = 81$ values of $q$ that are congruent to $0\pmod{11}$ have $1$ more possible value of $r$ such that $q+r \equiv 0\pmod{11}$.

Therefore, the number of possible values of $n$ such that $q+r \equiv 0\pmod{11}$ is $900\cdot9+81\cdot1=8181 \Rightarrow B$.

Solution 2

Let $n$ equal $\overline{abcde}$, where $a$ through $e$ are digits. Therefore,

$q=\overline{abc}=100a+10b+c$

$r=\overline{de}=10d+e$

We now take $q+r\bmod{11}$:

$q+r=100a+10b+c+10d+e\equiv a-b+c-d+e\equiv 0\bmod{11}$

The divisor trick for 11 is as follows:

"Let $n=\overline{a_1a_2a_3\cdots a_x}$ be an $x$ digit integer. If $a_1-a_2+a_3-\cdots +(-1)^{x-1} a_x$ is divisible by $11$, then $n$ is also divisible by $11$."

Therefore, the five digit number $n$ is divisible by 11. The 5-digit multiples of 11 range from $910\cdot 11$ to $9090\cdot 11$. There are $8181\Rightarrow \mathrm{(B)}$ divisors of 11 between those inclusive.

Solution 3

Since $q$ is a quotient and $r$ is a remainder when $n$ is divided by $100$. So we have $n=100q+r$. Since we are counting choices where $q+r$ is divisible by $11$, we have $n=99q+q+r=99q+11k$ for some $k$. This means that $n$ is the sum of two multiples of $11$ and would thus itself be a multiple of $11$. Then we can count all the five digit multiples of $11$ as in Solution 2. (This solution is essentially the same as Solution 2, but it does not necessarily involve mods and so could potentially be faster.)

See Also

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Question
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 AMC 10 Problems and Solutions
2003 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 AMC 12 Problems and Solutions

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