2003 AMC 12B Problems/Problem 18: Difference between revisions

Revision as of 11:37, 21 February 2010

Problem

Let $n$ be a 5-digit number, and let $q$ and $r$ be the quotient and 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

Suppose $n = 100\cdot q + r = 99\cdot q + (q+r)$

Since $11|(q+r)$ and $11|99q$ , $11|n$

$10000 \leq n \leq 99999$ , so there are $\left\lfloor\frac{99999}{11}\right\rfloor-\left\lceil\frac{10000}{11}\right\rceil+1 = \boxed{8181}$ values of $q+r$ that are divisible by $11 \Rightarrow {B}$ .

@@ Line 1: / Line 1: @@
-suppose n=a*10^4+b*10^3+c*10^2+d*10+e,
+== Problem ==
-where a,b,c,d,e are integers between [0,9]
+Let <math>n</math> be a 5-digit number, and let <math>q</math> and <math>r</math> be the quotient and remainder, respectively, when <math>n</math> is divided by <math>100</math>. For how many values of <math>n</math> is <math>q+r</math> divisible by <math>11</math>?
-then q=(a*10^2+b*10+c)%11, r=d*10+e, and
-n%11=(a*100+b*10+c)%11
+<math> \mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090 </math>
-(q+r)%11 = n%11, since 10000<=n<=99999
-there are 9090-910+1=8181 n values that are multiples of 11,
+== Solution ==
-thus there are 8181 (q+r) values that are multiples of 11
+Suppose <math>n = 100\cdot q + r = 99\cdot q + (q+r)</math>
+Since <math>11|(q+r)</math> and <math>11|99q</math>, <math>11|n</math>
+<math>10000 \leq n \leq 99999</math>, so there are <math>\left\lfloor\frac{99999}{11}\right\rfloor-\left\lceil\frac{10000}{11}\right\rceil+1 = \boxed{8181}</math> values of <math>q+r</math> that are divisible by <math>11 \Rightarrow {B}</math>.

Page

Toolbox

Search

2003 AMC 12B Problems/Problem 18: Difference between revisions

Revision as of 11:37, 21 February 2010

Problem

Solution