2003 AMC 10A Problems/Problem 25: Difference between revisions

Revision as of 09:44, 6 March 2008

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 $\lfloor \frac{100}{11} \rfloor = 9$ 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 $\lfloor \frac{900}{11} \rfloor = 81$ 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 $\underline{abcde}$ , where $a$ through $e$ are digits. Therefore,

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

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

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 $n=\underline{a_1a_2a_3\cdots a_x}$ 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.

@@ Line 5: / Line 5: @@
 == Solution ==
+=== 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>.
@@ Line 14: / Line 15: @@
 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 ===
+Let <math>n</math> equal <math>\underline{abcde}</math>, where <math>a</math> through <math>e</math> are digits. Therefore,
+<math>q=\underline{abc}=100a+10b+c</math>
+<math>r=\underline{de}=10d+e</math>
+We now take <math>q+r\bmod{11}</math>:
+<math>q+r=100a+10b+c+10d+e\equiv a-b+c-d+e\equiv 0\bmod{11}</math>
+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."
+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.
 == See Also ==

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