Difference between revisions of "2003 AMC 10A Problems/Problem 25"
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− | == Problem == | + | ==Problem== |
Let <math>n</math> be a <math>5</math>-digit number, and let <math>q</math> and <math>r</math> be the quotient and the 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>? | Let <math>n</math> be a <math>5</math>-digit number, and let <math>q</math> and <math>r</math> be the quotient and the 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>? | ||
<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> | ||
− | == Solutions == | + | ==Solutions== |
− | === | + | ===Solution 1=== |
<math>11|(q+r)</math> implies that <math>11|(99q+q+r)</math> and therefore <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. | <math>11|(q+r)</math> implies that <math>11|(99q+q+r)</math> and therefore <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 2=== | |
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− | === Solution | ||
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>. Another way to think about it is the number of possible values of q when r, the remainder, is <math>0</math>. In this case, q itself has to be a multiple of <math>11</math>. <math>\left\lfloor \frac{999}{11} = 90 \right\rfloor</math>. Then we'll need to subtract <math>9</math> from <math>90</math> since we only want multiples of <math>11</math> greater than <math>100</math> <math>(90-9=81)</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>. Another way to think about it is the number of possible values of q when r, the remainder, is <math>0</math>. In this case, q itself has to be a multiple of <math>11</math>. <math>\left\lfloor \frac{999}{11} = 90 \right\rfloor</math>. Then we'll need to subtract <math>9</math> from <math>90</math> since we only want multiples of <math>11</math> greater than <math>100</math> <math>(90-9=81)</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 \ | + | 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>. |
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− | === Solution | + | ===Solution 3=== |
Let <math>n</math> equal <math>\overline{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, | ||
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<math>q+r=100a+10b+c+10d+e\equiv a-b+c-d+e\equiv 0\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: | + | The [[Divisibility rules#Divisibility Rule for Divisibility_rules#Divisibility_Rule_for_11|divisor trick for 11]] is as follows: |
"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 <math>11</math>." | "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 <math>11</math>." | ||
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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. | ||
− | + | ===Solution 4=== | |
− | + | 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 3, but it does not necessarily involve mods and so could potentially be faster.) | |
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− | === Solution | ||
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− | 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 | ||
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+ | ===Solution 5=== | ||
Defining <math>q</math> and <math>r</math> in terms of floor functions and <math>n</math> would yield: <math>q=\left \lfloor \frac{n}{100} \right \rfloor</math> and <math>r=n-100 \left \lfloor \frac{n}{100} \right \rfloor</math>. Since <math>q+r \equiv 0\pmod{11}</math>, <math>\left \lfloor \frac{n}{100} \right \rfloor + n-100 \left \lfloor \frac{n}{100} \right \rfloor \equiv 0\pmod{11}</math>. Simplifying gets us <math>n-99 \left \lfloor \frac{n}{100} \right \rfloor\equiv 0\pmod{11} \rightarrow n \equiv 0\pmod{11}</math> (<math>99 \left \lfloor \frac{n}{100} \right \rfloor\equiv 0\pmod{11}</math> is always true since floor function always yields an integer, and 99 is divisible by 11 w/o any remainder). After we come to this conclusion, it becomes easy to solve the rest of the problem (<math>\left \lfloor \frac{n}{99999} \right \rfloor - \left \lfloor \frac{n}{10000} \right \rfloor</math>). | Defining <math>q</math> and <math>r</math> in terms of floor functions and <math>n</math> would yield: <math>q=\left \lfloor \frac{n}{100} \right \rfloor</math> and <math>r=n-100 \left \lfloor \frac{n}{100} \right \rfloor</math>. Since <math>q+r \equiv 0\pmod{11}</math>, <math>\left \lfloor \frac{n}{100} \right \rfloor + n-100 \left \lfloor \frac{n}{100} \right \rfloor \equiv 0\pmod{11}</math>. Simplifying gets us <math>n-99 \left \lfloor \frac{n}{100} \right \rfloor\equiv 0\pmod{11} \rightarrow n \equiv 0\pmod{11}</math> (<math>99 \left \lfloor \frac{n}{100} \right \rfloor\equiv 0\pmod{11}</math> is always true since floor function always yields an integer, and 99 is divisible by 11 w/o any remainder). After we come to this conclusion, it becomes easy to solve the rest of the problem (<math>\left \lfloor \frac{n}{99999} \right \rfloor - \left \lfloor \frac{n}{10000} \right \rfloor</math>). | ||
~hw21 | ~hw21 | ||
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https://youtu.be/OpGHj-B0_hg?t=672 | https://youtu.be/OpGHj-B0_hg?t=672 | ||
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==Video Solution 2== | ==Video Solution 2== | ||
https://youtu.be/j5iM4V7ZT-Y | https://youtu.be/j5iM4V7ZT-Y | ||
− | + | ==See Also== | |
− | |||
− | == See Also == | ||
{{AMC10 box|year=2003|ab=A|num-b=24|after=Last Question}} | {{AMC10 box|year=2003|ab=A|num-b=24|after=Last Question}} | ||
{{AMC12 box|year=2003|ab=A|num-b=17|num-a=19}} | {{AMC12 box|year=2003|ab=A|num-b=17|num-a=19}} | ||
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:04, 5 January 2025
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
Solution 1
implies that and therefore , so . Then, can range from to for a total of numbers.
Solution 2
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 . Another way to think about it is the number of possible values of q when r, the remainder, is . In this case, q itself has to be a multiple of . . Then we'll need to subtract from since we only want multiples of greater than
Therefore, the number of possible values of such that is .
Solution 3
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.
Solution 4
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 3, but it does not necessarily involve mods and so could potentially be faster.)
Solution 5
Defining and in terms of floor functions and would yield: and . Since , . Simplifying gets us ( is always true since floor function always yields an integer, and 99 is divisible by 11 w/o any remainder). After we come to this conclusion, it becomes easy to solve the rest of the problem (). ~hw21
Video Solution 1
https://youtu.be/OpGHj-B0_hg?t=672
Video Solution 2
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.