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| − | ==Note== | + | ==Problem== |
| − | <math>
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| − | \textbf{That is just a trick. The following question is from 2001 AMC10 Problem 25.}
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| − | </math>
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| | | | |
| − | == Problem ==
| + | Maddy and Lara see a list of numbers written on a blackboard. Maddy adds <math>3</math> to each number in the list and finds that the sum of her new numbers is <math>45</math>. Lara multiplies each number in the list by <math>3</math> and finds that the sum of her new numbers is also <math>45</math>. How many numbers are written on the blackboard? |
| − | How many positive integers not exceeding <math>2001</math> are multiples of <math>3</math> or <math>4</math> but not <math>5</math>?
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| | | | |
| − | <math> | + | <math>\textbf{(A) }10\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad\textbf{(E) }9</math> |
| − | \textbf{(A) }768 | |
| − | \qquad | |
| − | \textbf{(B) }801 | |
| − | \qquad | |
| − | \textbf{(C) }934 | |
| − | \qquad | |
| − | \textbf{(D) }1067 | |
| − | \qquad | |
| − | \textbf{(E) }1167 | |
| − | </math> | |
| | | | |
| − | ==Solutions==
| + | ==Solution== |
| − | === Solution 1===
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| | | | |
| − | Out of the numbers <math>1</math> to <math>12</math> four are divisible by <math>3</math> and three by <math>4</math>, counting <math>12</math> twice.
| + | Let there be <math>n</math> numbers in the list of numbers, and let their sum be <math>S</math>. Then we have the following |
| − | Hence <math>6</math> out of these <math>12</math> numbers are multiples of <math>3</math> or <math>4</math>.
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| | | | |
| − | The same is obviously true for the numbers <math>12k+1</math> to <math>12k+12</math> for any positive integer <math>k</math>.
| + | <cmath>S+3n=45</cmath> |
| | | | |
| − | Hence out of the numbers <math>1</math> to <math>60=5\cdot 12</math> there are <math>5\cdot 6=30</math> numbers that are divisible by <math>3</math> or <math>4</math>.
| + | <cmath>3S=45</cmath> |
| − | Out of these <math>30</math>, the numbers <math>15</math>, <math>20</math>, <math>30</math>, <math>40</math>, <math>45</math> and <math>60</math> are divisible by <math>5</math>.
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| − | Therefore in the set <math>\{1,\dots,60\}</math> there are precisely <math>30-6=24</math> numbers that satisfy all criteria from the problem statement.
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| | | | |
| − | Again, the same is obviously true for the set <math>\{60k+1,\dots,60k+60\}</math> for any positive integer <math>k</math>.
| + | From the second equation, <math>S=15</math>. So, <math>15+3n=45</math> <math>\Rightarrow</math> <math>n=\boxed{\textbf{(A) }10}.</math> |
| | | | |
| − | We have <math>1980/60 = 33</math>, hence there are <math>24\cdot 33 = 792</math> good numbers among the numbers <math>1</math> to <math>1980</math>. At this point we already know that the only answer that is still possible is <math>\boxed{\textbf{(B)}}</math>, as we only have <math>20</math> numbers left.
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| | | | |
| − | By examining the remaining <math>20</math> by hand we can easily find out that exactly <math>9</math> of them match all the criteria, giving us <math>792+9=\boxed{\textbf{(B) }801}</math> good numbers.
| + | ~Mintylemon66 (formatted atictacksh) |
| − | This is correct.
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| | | | |
| − | ===Solution 2===
| + | ==Solution 2== |
| − | We can solve this problem by finding the cases where the number is divisible by <math>3</math> or <math>4</math>, then subtract from the cases where none of those cases divide <math>5</math>. To solve the ways the numbers divide <math>3</math> or <math>4</math> we find the cases where a number is divisible by <math>3</math> and <math>4</math> as separate cases. We apply the floor function to every case to get <math>\left\lfloor \frac{2001}{3} \right\rfloor</math>, <math>\left\lfloor \frac{2001}{4} \right\rfloor</math>, and <math>\left\lfloor \frac{2001}{12} \right\rfloor</math>. The first two floor functions were for calculating the number of individual cases for <math>3</math> and <math>4</math>. The third case was to find any overlapping numbers. The numbers were <math>667</math>, <math>500</math>, and <math>166</math>, respectively. We add the first two terms and subtract the third to get <math>1001</math>. The first case is finished.
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| − | The second case is more or less the same, except we are applying <math>3</math> and <math>4</math> to <math>5</math>. We must find the cases where the first case over counts multiples of five. Utilizing the floor function again on the fractions <math>\left\lfloor \frac{2001}{3\cdot5} \right\rfloor</math>, <math>\left\lfloor \frac{2001}{4\cdot5} \right\rfloor</math>, and <math>\left\lfloor \frac{2001}{3\cdot4\cdot5} \right\rfloor</math> yields the numbers <math>133</math>, <math>100</math>, and <math>33</math>. The first two numbers counted all the numbers that were multiples of either four with five or three with five less than <math>2001</math>. The third counted the overlapping cases, which we must subtract from the sum of the first two. We do this to reach <math>200</math>. Subtracting this number from the original <math>1001</math> numbers procures <math>\boxed{\textbf{(B)}\ 801}</math>.
| + | Let <math>x_1,x_2,x_3,...,x_n</math> where <math>x_n</math> represents the <math>n</math>th number written on the board. Lara's multiplied each number by <math>3</math>, so her sum will be <math>3x_1+3x_2+3x_3+...+3x_n</math>. This is the same as <math>3\cdot (x_1+x_2+x_3+...+x_n)</math>. We are given this quantity is equal to <math>45</math>, so the original numbers add to <math>\frac{45}{3}=15</math>. Maddy adds <math>3</math> to each of the <math>n</math> terms which yields, <math>x_1+3+x_2+3+x_3+3+...+x_n+3</math>. This is the same as the sum of the original series plus <math>3 \cdot n</math>. Setting this equal to <math>45</math>, <math>15+3n=45 \Rightarrow n =\boxed{\textbf{(A) }10}.</math> |
| | | | |
| − | ===Solution 3===
| + | ~vsinghminhas |
| − | First find the number of such integers between 1 and 2000 (inclusive) and then add one to this result because 2001 is a multiple of <math>3</math>.
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| | | | |
| − | There are <math>\frac45\cdot2000=1600</math> numbers that are not multiples of <math>5</math>. <math>\frac23\cdot\frac34\cdot1600=800</math> are not multiples of <math>3</math> or <math>4</math>, so <math>800</math> numbers are. <math>800+1=\boxed{\textbf{(B)}\ 801}</math>
| + | ==Solution 3== |
| | | | |
| − | ===Solution 4===
| + | If the list of numbers written on the board is <math>a_1, a_2, a_3, \ldots, a_n</math>, then we can formulate two equations: |
| − | Take a good-sized sample of consecutive integers; for example, the first <math>25</math> positive integers. Determine that the numbers <math>3, 4, 6, 8, 9, 12, 16, 18, 21,</math> and <math>24</math> exhibit the properties given in the question. <math>25</math> is a divisor of <math>2000</math>, so there are <math>\frac{10}{25}\cdot2000=800</math> numbers satisfying the given conditions between <math>1</math> and <math>2000</math>. Since <math>2001</math> is a multiple of <math>3</math>, add <math>1</math> to <math>800</math> to get <math>800+1=\boxed{\textbf{(B)}\ 801}</math>.
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| | | | |
| − | ~ mathmagical
| + | <cmath>3n + \sum_{i=1}^{n} a_i = 45</cmath> |
| | | | |
| − | ===Solution 5===
| + | <cmath>3 \sum_{i=1}^{n} a_i = 45</cmath> |
| − | By PIE, there are <math>1001</math> numbers that are multiples of <math>3</math> or <math>4</math> and less than or equal to <math>2001</math>. <math>80\%</math> of them will not be divisible by <math>5</math>, and by far the closest number to <math>80\%</math> of <math>1001</math> is <math>\boxed{\textbf{(B)}\ 801}</math>.
| |
| | | | |
| − | ~ Fasolinka
| + | We can rewrite the first equation by multiplying both sides by <math>3</math>: |
| | | | |
| − | === Solution 5===
| + | <math>3(3n + \sum_{i=1}^{n} a_i) = 3(45)</math> |
| − | Similar to some of the above solutions.
| |
| − | We can divide <math>2001</math> by <math>3</math> and <math>4</math> to find the number of integers divisible by <math>3</math> and <math>4</math>. Hence, we find that there are <math>667</math> numbers less than <math>2001</math> that are divisible by <math>3</math>, and <math>500</math> numbers that are divisible by <math>4</math>. However, we will need to subtract the number of multiples of <math>15</math> from 667 and that of <math>20</math> from <math>500</math>, since they're also divisible by 5 which we don't want. There are <math>133</math> + <math>100</math> = <math>233</math> such numbers. Note that during this process, we've subtracted the multiples of <math>60</math> twice because they're divisible by both <math>15</math> and <math>20</math>, so we have to add <math>33</math> back to the tally (there are <math>33</math> multiples of <math>60</math> that does not exceed <math>2001</math>). Lastly, we have to subtract multiples of both <math>3</math> AND <math>4</math> since we only want multiples of either <math>3</math> or <math>4</math>. This is tantamount to subtracting the number of multiples of <math>12</math>. And there are <math>166</math> such numbers. Let's now collect our numbers and compute the total: <math>667</math> + <math>500</math> - <math>133</math> - <math>100</math> + <math>33</math> - <math>166</math> = <math>\boxed{\textbf{(B)}\ 801}</math>.
| |
| | | | |
| − | ~ PlainOldNumberTheory
| + | <math>\Rightarrow 9n + 3 \sum_{i=1}^{n} a_i = 135</math> |
| | | | |
| | + | Now, subtract the second equation from the first: |
| | | | |
| − | === Solution 6===
| + | <cmath>(9n + 3 \sum_{i=1}^{n} a_i) - (3 \sum_{i=1}^{n} a_i) = 135 - 45</cmath> |
| − | Similar to @above:
| |
| − | Let the function <math>M_{2001}(n)</math> return how many multiples of <math>n</math> are there not exceeding <math>2001</math>. Then we have that the desired number is:
| |
| − | <cmath>M_{2001}(3)+M_{2001}(4)-M_{2001}(3\cdot 4)-M_{2001}(3 \cdot 5) - M_{2001}(4 \cdot 5)+M_{2001}(3 \cdot 4 \cdot 5)</cmath> | |
| | | | |
| − | Evaluating each of these we get:
| + | <cmath>\Rightarrow 9n = 135 - 45</cmath> |
| − | <cmath>667+500-166-133-100+33 = 1100-299 = 801.</cmath> | |
| | | | |
| − | Thus, the answer is <math>\boxed{\textbf{(B)}\ 801}.</math>
| + | <cmath>\Rightarrow 9n = 90</cmath> |
| | | | |
| − | -ConfidentKoala4
| + | <cmath>\Rightarrow n =\boxed{\textbf{(A) }10}</cmath> |
| | | | |
| − | ==Video Solutions==
| + | ~ShalomKeshet |
| − | https://youtu.be/EXWK7U8uXyk
| |
| | | | |
| − | https://www.youtube.com/watch?v=XHmKu-ZoRxI&feature=youtu.be
| + | ==Video Solution 1 by SpreadTheMathLove== |
| | | | |
| − | == See Also == | + | https://www.youtube.com/watch?v=SUnhwbA5_So |
| | | | |
| − | {{AMC12 box|year=2001|num-b=11|num-a=13}}
| + | ==Video Solution by Math-X (First understand the problem!!!)== |
| − | {{AMC10 box|year=2001|num-b=25|after=Last Question}} | + | https://youtu.be/EuLkw8HFdk4?si=6dyj2QxkbBuNk6j7&t=951 |
| | + | |
| | + | ~Math-X |
| | + | |
| | + | ==Video Solution== |
| | + | |
| | + | https://youtu.be/-yk7ozNRrtQ |
| | + | |
| | + | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) |
| | + | |
| | + | ==Video Solution by Interstigation== |
| | + | https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L |
| | + | |
| | + | ==See also== |
| | + | {{AMC10 box|year=2023|ab=B|num-b=4|num-a=6}} |
| | {{MAA Notice}} | | {{MAA Notice}} |
Problem
Maddy and Lara see a list of numbers written on a blackboard. Maddy adds 
to each number in the list and finds that the sum of her new numbers is 
. Lara multiplies each number in the list by and finds that the sum of her new numbers is also . How many numbers are written on the blackboard?
Solution
Let there be 
numbers in the list of numbers, and let their sum be 
. Then we have the following
![\[S+3n=45\]](//latex.artofproblemsolving.com/4/e/e/4ee46f0c0e92dd3ac54e1c12065ec6b94c453721.png)
![\[3S=45\]](//latex.artofproblemsolving.com/a/5/b/a5bf5591b7e535363264a8943dd7c4a9f53036ea.png)
From the second equation, 
. So, 
~Mintylemon66 (formatted atictacksh)
Solution 2
Let 
where 
represents the th number written on the board. Lara's multiplied each number by , so her sum will be 
. This is the same as 
. We are given this quantity is equal to , so the original numbers add to 
. Maddy adds to each of the terms which yields, 
. This is the same as the sum of the original series plus 
. Setting this equal to , 
~vsinghminhas
Solution 3
If the list of numbers written on the board is 
, then we can formulate two equations:
![\[3n + \sum_{i=1}^{n} a_i = 45\]](//latex.artofproblemsolving.com/d/c/1/dc14031cfd91a0958f35d55ed589018e33650cf6.png)
![\[3 \sum_{i=1}^{n} a_i = 45\]](//latex.artofproblemsolving.com/5/a/b/5ab0629d2b29952b98e7107167e978beab07eb39.png)
We can rewrite the first equation by multiplying both sides by :
Now, subtract the second equation from the first:
![\[(9n + 3 \sum_{i=1}^{n} a_i) - (3 \sum_{i=1}^{n} a_i) = 135 - 45\]](//latex.artofproblemsolving.com/6/e/6/6e6fecab8c887b46f9e9e99cc20300cbe95397bb.png)
![\[\Rightarrow 9n = 135 - 45\]](//latex.artofproblemsolving.com/9/0/f/90f231a54aa30e6b93d39e77dd14f50ae0e7133a.png)
![\[\Rightarrow 9n = 90\]](//latex.artofproblemsolving.com/d/1/2/d129e9531572d33225c45b7d7d2db5bdcab9b88a.png)
![\[\Rightarrow n =\boxed{\textbf{(A) }10}\]](//latex.artofproblemsolving.com/d/4/4/d44cb07ee9c739fe3c20233273836e2f824921fc.png)
~ShalomKeshet
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=SUnhwbA5_So
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/EuLkw8HFdk4?si=6dyj2QxkbBuNk6j7&t=951
~Math-X
Video Solution
https://youtu.be/-yk7ozNRrtQ
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by Interstigation
https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L
See also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
to each number in the list and finds that the sum of her new numbers is 
. Lara multiplies each number in the list by 
numbers in the list of numbers, and let their sum be 
. Then we have the following
![\[S+3n=45\]](http://latex.artofproblemsolving.com/4/e/e/4ee46f0c0e92dd3ac54e1c12065ec6b94c453721.png)
![\[3S=45\]](http://latex.artofproblemsolving.com/a/5/b/a5bf5591b7e535363264a8943dd7c4a9f53036ea.png)
. So, 
where 
represents the 
. This is the same as 
. We are given this quantity is equal to 
. Maddy adds 
. This is the same as the sum of the original series plus 
. Setting this equal to 
, then we can formulate two equations:
![\[3n + \sum_{i=1}^{n} a_i = 45\]](http://latex.artofproblemsolving.com/d/c/1/dc14031cfd91a0958f35d55ed589018e33650cf6.png)
![\[3 \sum_{i=1}^{n} a_i = 45\]](http://latex.artofproblemsolving.com/5/a/b/5ab0629d2b29952b98e7107167e978beab07eb39.png)
![\[(9n + 3 \sum_{i=1}^{n} a_i) - (3 \sum_{i=1}^{n} a_i) = 135 - 45\]](http://latex.artofproblemsolving.com/6/e/6/6e6fecab8c887b46f9e9e99cc20300cbe95397bb.png)
![\[\Rightarrow 9n = 135 - 45\]](http://latex.artofproblemsolving.com/9/0/f/90f231a54aa30e6b93d39e77dd14f50ae0e7133a.png)
![\[\Rightarrow 9n = 90\]](http://latex.artofproblemsolving.com/d/1/2/d129e9531572d33225c45b7d7d2db5bdcab9b88a.png)
![\[\Rightarrow n =\boxed{\textbf{(A) }10}\]](http://latex.artofproblemsolving.com/d/4/4/d44cb07ee9c739fe3c20233273836e2f824921fc.png)
