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Difference between revisions of "2003 AMC 12A Problems/Problem 18"
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<math> \mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090 </math>
 
<math> \mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090 </math>
  
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== Solution ==
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== Solution 1 ==
  
 
When a <math>5</math>-digit number is divided by <math>100</math>, the first <math>3</math> digits become the quotient, <math>q</math>, and the last <math>2</math> digits become the remainder, <math>r</math>.  
 
When a <math>5</math>-digit number is divided by <math>100</math>, the first <math>3</math> digits become the quotient, <math>q</math>, and the last <math>2</math> digits become the remainder, <math>r</math>.  
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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.  
  
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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>.  
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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>.  
  
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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>.  
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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>.
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== Solution 2 ==
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Notice that <math>q+r\equiv0\pmod{11}\Rightarrow100q+r\equiv0\pmod{11}</math>. This means that any number whose quotient and remainder sum is divisible by 11 must also be divisible by 11. Therefore, there are <math>\frac{99990-10010}{11}+1=8181</math> possible values. The answer is <math>\boxed{\textbf{(B) }8181}</math>.
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== Solution 3 ==
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Let <math>abcde</math> be the five digits of <math>n</math>. Then <math>q = abc</math> and <math>r = de</math>. By the divisibility rules of <math>11</math>, <math>q = a - b + c \pmod{11}</math> and <math>r = -d + e \pmod{11}</math>, so <math>q + r = a - b + c - d + e = abcde = n \pmod{11}</math>. Thus, <math>n</math> must be divisble by <math>11</math>. There are <math>\frac{99990 - 10010}{11} + 1 = 8181</math> five-digit multiples of <math>11</math>, so the answer is <math>\boxed{\textbf{(B) }8181}</math>.
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==Video Solution 1==
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https://youtu.be/OpGHj-B0_hg?t=672
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 +
~IceMatrix
  
 
==See Also==
 
==See Also==
 +
{{AMC10 box|year=2003|ab=A|num-b=24|after=Last Question}}
 
{{AMC12 box|year=2003|ab=A|num-b=17|num-a=19}}
 
{{AMC12 box|year=2003|ab=A|num-b=17|num-a=19}}
 +
{{MAA Notice}}

Latest revision as of 17:20, 18 July 2022

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 $\boxed{\textbf{(B) }8181}$.

Solution 3

Let $abcde$ be the five digits of $n$. Then $q = abc$ and $r = de$. By the divisibility rules of $11$, $q = a - b + c \pmod{11}$ and $r = -d + e \pmod{11}$, so $q + r = a - b + c - d + e = abcde = n \pmod{11}$. Thus, $n$ must be divisble by $11$. There are $\frac{99990 - 10010}{11} + 1 = 8181$ five-digit multiples of $11$, so the answer is $\boxed{\textbf{(B) }8181}$.

Video Solution 1

https://youtu.be/OpGHj-B0_hg?t=672

~IceMatrix

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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