Difference between revisions of "2003 AMC 10A Problems/Problem 25"
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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 == | + | == Solutions == |
+ | === Simple Solution === | ||
+ | <math>11|q+r</math> implies that <math>11|100q+r</math>, so <math>11|n</math>. Then, <math>n</math> can range from <math>10010</math> to <math>99990</math> for a total of <math>\boxed{8181\Rightarrow \mathrm{(B)}}</math> numbers. | ||
+ | |||
=== Solution 1 === | === 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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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>. | 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 === | ||
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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. | ||
− | === | + | ==== Note ==== |
− | |||
− | |||
− | |||
− | |||
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>. | 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>. | ||
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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. | 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. | ||
+ | |||
+ | === Solution 3 === | ||
+ | |||
+ | Since <math>q</math> is a quotient and <math>r</math> is a remainder when <math>n</math> is divided by <math>100</math>. So we have <math>n=100q+r</math>. Since we are counting choices where <math>q+r</math> is divisible by <math>11</math>, we have <math>n=99q+q+r=99q+11k</math> for some <math>k</math>. This means that <math>n</math> is the sum of two multiples of <math>11</math> and would thus itself be a multiple of <math>11</math>. Then we can count all the five digit multiples of <math>11</math> as in Solution 2. (This solution is essentially the same as Solution 2, but it does not necessarily involve mods and so could potentially be faster.) | ||
== See Also == | == See Also == | ||
− | {{AMC10 box|year=2003|ab=A|num-b=24|after= | + | {{AMC10 box|year=2003|ab=A|num-b=24|after=Last Question}} |
− | + | {{AMC12 box|year=2003|ab=A|num-b=17|num-a=19}} | |
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 17:12, 28 December 2020
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 ?
Solutions
Simple Solution
implies that , so . Then, can range from to for a total of numbers.
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
Let equal , where through are digits. Therefore,
We now take :
The divisor trick for 11 is as follows:
"Let be an digit integer. If is divisible by , then is also divisible by ."
Therefore, the five digit number is divisible by 11. The 5-digit multiples of 11 range from to . There are divisors of 11 between those inclusive.
Note
The part labeled "divisor trick" actually follows from the same observation we made in the previous step: , therefore and for all . For a digit number we get , 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 as the original number.
Solution 3
Since is a quotient and is a remainder when is divided by . So we have . Since we are counting choices where is divisible by , we have for some . This means that is the sum of two multiples of and would thus itself be a multiple of . Then we can count all the five digit multiples of as in Solution 2. (This solution is essentially the same as Solution 2, but it does not necessarily involve mods and so could potentially be faster.)
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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