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− | == Problem == | + | ==Problem== |
− | https://www.youtube.com/watch?v=lYBUbBu4W08
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− | 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> | + | 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? |
− | \textbf{(A) }768
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− | \qquad
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− | \textbf{(B) }801
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− | \qquad
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− | \textbf{(C) }934
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− | \qquad
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− | \textbf{(D) }1067
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− | \qquad
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− | \textbf{(E) }1167
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− | </math> | |
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− | ==Solutions==
| + | <math>\textbf{(A) }10\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad\textbf{(E) }9</math> |
− | === 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.
| + | ==Solution== |
− | 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>.
| + | 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 |
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− | 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>S+3n=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>.
| + | <cmath>3S=45</cmath> |
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− | 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.
| + | From the second equation, <math>S=15</math>. So, <math>15+3n=45</math> <math>\Rightarrow</math> <math>n=\boxed{\textbf{(A) }10}.</math> |
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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.
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− | This is correct.
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− | ===Solution 2===
| + | ~Mintylemon66 (formatted atictacksh) |
− | 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>.
| + | ==Solution 2== |
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− | ===Solution 3===
| + | 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> |
− | 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>
| + | ~vsinghminhas |
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− | ===Solution 4===
| + | ==Solution 3== |
− | 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
| + | 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: |
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− | ===Solution 5===
| + | <cmath>3n + \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>.
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− | ~ Fasolinka
| + | <cmath>3 \sum_{i=1}^{n} a_i = 45</cmath> |
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− | === Solution 5===
| + | We can rewrite the first equation by multiplying both sides by <math>3</math>: |
− | Similar to some of the above solutions.
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− | 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>. | |
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− | ~ PlainOldNumberTheory
| + | <math>3(3n + \sum_{i=1}^{n} a_i) = 3(45)</math> |
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| + | <math>\Rightarrow 9n + 3 \sum_{i=1}^{n} a_i = 135</math> |
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− | === Solution 6===
| + | Now, subtract the second equation from the first: |
− | Similar to @above:
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− | 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:
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− | <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>
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− | Evaluating each of these we get:
| + | <cmath>(9n + 3 \sum_{i=1}^{n} a_i) - (3 \sum_{i=1}^{n} a_i) = 135 - 45</cmath> |
− | <cmath>667+500-166-133-100+33 = 1100-299 = 801.</cmath> | |
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− | Thus, the answer is <math>\boxed{\textbf{(B)}\ 801}.</math>
| + | <cmath>\Rightarrow 9n = 135 - 45</cmath> |
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− | -ConfidentKoala4
| + | <cmath>\Rightarrow 9n = 90</cmath> |
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− | ==Video Solutions== | + | <cmath>\Rightarrow n =\boxed{\textbf{(A) }10}</cmath> |
− | https://youtu.be/EXWK7U8uXyk
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− | https://www.youtube.com/watch?v=XHmKu-ZoRxI&feature=youtu.be
| + | ~ <math>Shalomkeshet</math> |
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− | == See Also == | + | ==Video Solution 1 by SpreadTheMathLove== |
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− | {{AMC12 box|year=2001|num-b=11|num-a=13}}
| + | https://www.youtube.com/watch?v=SUnhwbA5_So |
− | {{AMC10 box|year=2001|num-b=25|after=Last Question}} | + | |
| + | ==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== |
| + | {{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
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:
We can rewrite the first equation by multiplying both sides by :
Now, subtract the second equation from the first:
~
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.