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> | ||
− | == 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>. | 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, 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>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | Notice that <math>q+r=0\pmod{11}Rightarrow100q+r=0\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{99990-11010}{11}+1=8181</math> possible values. | ||
==See Also== | ==See Also== | ||
{{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}} | {{MAA Notice}} |
Revision as of 15:04, 23 December 2013
Contents
Problem
Let be a -digit number, and let and be the quotient and the remainder, respectively, when is divided by . For how many values of is divisible by ?
Solution 1
When a -digit number is divided by , the first digits become the quotient, , and the last digits become the remainder, .
Therefore, can be any integer from to inclusive, and can be any integer from to inclusive.
For each of the possible values of , there are at least possible values of such that .
Since there is "extra" possible value of that is congruent to , each of the values of that are congruent to have more possible value of such that .
Therefore, the number of possible values of such that is .
Solution 2
Notice that . This means that any number whose quotient and remainder sum is divisible by 11 must also be divisible by 11. Therefore, there are possible values.
See Also
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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