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

(Alternate solution)
(formatting and notes)
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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 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 B</math>.  
  
 
=== Solution 2 ===
 
=== Solution 2 ===
−
Let <math>n</math> equal <math>\underline{abcde}</math>, where <math>a</math> through <math>e</math> are digits. Therefore,
+
Let <math>n</math> equal <math>\overline{abcde}</math>, where <math>a</math> through <math>e</math> are digits. Therefore,
  
−
<math>q=\underline{abc}=100a+10b+c</math>
+
<math>q=\overline{abc}=100a+10b+c</math>
  
−
<math>r=\underline{de}=10d+e</math>
+
<math>r=\overline{de}=10d+e</math>
  
 
We now take <math>q+r\bmod{11}</math>:
 
We now take <math>q+r\bmod{11}</math>:
Line 29: Line 29:
 
The divisor trick for 11 is as follows:
 
The divisor trick for 11 is as follows:
  
−
"Let <math>n=\underline{a_1a_2a_3\cdots a_x}</math> be an <math>x</math> digit integer. If <math>a_1-a_2+a_3-\cdots +(-1)^{x-1} a_x</math> is divisible by <math>11</math>, then <math>n</math> is also divisible by 11."
+
"Let <math>n=\overline{a_1a_2a_3\cdots a_x}</math> be an <math>x</math> digit integer. If <math>a_1-a_2+a_3-\cdots +(-1)^{x-1} a_x</math> is divisible by <math>11</math>, then <math>n</math> is also divisible by 11."
  
 
Therefore, the five digit number <math>n</math> is divisible by 11. The 5-digit multiples of 11 range from <math>910\cdot 11</math> to <math>9090\cdot 11</math>. There are <math>8181\Rightarrow \mathrm{(B)}</math> divisors of 11 between those inclusive.
 
Therefore, the five digit number <math>n</math> is divisible by 11. The 5-digit multiples of 11 range from <math>910\cdot 11</math> to <math>9090\cdot 11</math>. There are <math>8181\Rightarrow \mathrm{(B)}</math> divisors of 11 between those inclusive.
 +
 +
==== Notes ====
 +
 +
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>.
 +
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.
 +
 +
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 14:54, 24 January 2009

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

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.

Notes

The part labeled "divisor trick" actually follows from the same observation we made in the previous step: $10\equiv (-1)\pmod{11}$, therefore $10^{2k}\equiv 1$ and $10^{2k+1}\equiv (-1)$ for all $k$. For a $5-$digit number $\overline{abcde}$ we get $\overline{abcde}\equiv a\cdot 1 + b\cdot(-1) + c\cdot 1 + d\cdot(-1) + e\cdot 1 = a-b+c-d+e$, as claimed.

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 $+$, the result will have the same remainder modulo $11$ as the original number.

See Also

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Final Question
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All AMC 10 Problems and Solutions
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