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Difference between revisions of "2003 AMC 12A Problems/Problem 18"

(This question was in amc10a 2003)
(→‎Solution 1)
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Therefore, <math>q</math> can be any integer from <math>100</math> to <math>999</math> inclusive, and <math>r</math> can be any integer from <math>0</math> to <math>99</math> inclusive.  
 
Therefore, <math>q</math> can be any integer from <math>100</math> to <math>999</math> inclusive, and <math>r</math> can be any integer from <math>0</math> to <math>99</math> inclusive.  
  
−
For each of the <math>9\cdot10\cdot10=900</math> possible values of <math>q</math>, there are at least <math>\lfloor \frac{100}{11} \rfloor = 9</math> possible values of <math>r</math> such that <math>q+r \equiv 0\pmod{11}</math>.  
+
For each of the <math>9\cdot10\cdot10=900</math> possible values of <math>q</math>, there are at least <math>\left\lfloor \frac{100}{11} \right\rfloor = 9</math> possible values of <math>r</math> such that <math>q+r \equiv 0\pmod{11}</math>.  
  
−
Since there is <math>1</math> "extra" possible value of <math>r</math> that is congruent to <math>0\pmod{11}</math>, each of the <math>\lfloor \frac{900}{11} \rfloor = 81</math> values of <math>q</math> that are congruent to <math>0\pmod{11}</math> have <math>1</math> more possible value of <math>r</math> such that <math>q+r \equiv 0\pmod{11}</math>.  
+
Since there is <math>1</math> "extra" possible value of <math>r</math> that is congruent to <math>0\pmod{11}</math>, each of the <math>\left\lfloor \frac{900}{11} \right\rfloor = 81</math> values of <math>q</math> that are congruent to <math>0\pmod{11}</math> have <math>1</math> more possible value of <math>r</math> such that <math>q+r \equiv 0\pmod{11}</math>.  
  
 
Therefore, the number of possible values of <math>n</math> such that <math>q+r \equiv 0\pmod{11}</math> is <math>900\cdot9+81\cdot1=8181 \Rightarrow\boxed{(B)} </math>.
 
Therefore, the number of possible values of <math>n</math> such that <math>q+r \equiv 0\pmod{11}</math> is <math>900\cdot9+81\cdot1=8181 \Rightarrow\boxed{(B)} </math>.

Revision as of 12:59, 11 March 2019

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 $11$?

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

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 $900\cdot9+81\cdot1=8181 \Rightarrow\boxed{(B)}$.

Solution 2

Notice that $q+r\equiv0\pmod{11}\Rightarrow100q+r\equiv0\pmod{11}$. This means that any number whose quotient and remainder sum is divisible by 11 must also be divisible by 11. Therefore, there are $\frac{99990-10010}{11}+1=8181$ possible values. The answer is $\boxed{(B)}$.

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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