Difference between revisions of "2023 AMC 10B Problems/Problem 5"

Latest revision as of 14:48, 20 November 2023

Problem

Maddy and Lara see a list of numbers written on a blackboard. Maddy adds $3$ to each number in the list and finds that the sum of her new numbers is $45$ . Lara multiplies each number in the list by $3$ and finds that the sum of her new numbers is also $45$ . How many numbers are written on the blackboard?

$\textbf{(A) }10\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad\textbf{(E) }9$

Solution

Let there be $n$ numbers in the list of numbers, and let their sum be $S$ . Then we have the following

$S+3n=45$

$3S=45$

From the second equation, $S=15$ . So, $15+3n=45$ $\Rightarrow$ $n=\boxed{\textbf{(A) }10}.$

~Mintylemon66 (formatted atictacksh)

Solution 2

Let $x_1,x_2,x_3,...,x_n$ where $x_n$ represents the $n$ th number written on the board. Lara's multiplied each number by $3$ , so her sum will be $3x_1+3x_2+3x_3+...+3x_n$ . This is the same as $3\cdot (x_1+x_2+x_3+...+x_n)$ . We are given this quantity is equal to $45$ , so the original numbers add to $\frac{45}{3}=15$ . Maddy adds $3$ to each of the $n$ terms which yields, $x_1+3+x_2+3+x_3+3+...+x_n+3$ . This is the same as the sum of the original series plus $3 \cdot n$ . Setting this equal to $45$ , $15+3n=45 \Rightarrow n =\boxed{\textbf{(A) }10}.$

~vsinghminhas

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

@@ Line 1: / Line 1: @@
-== Problem ==
+==Problem==
-https://www.youtube.com/watch?v=lYBUbBu4W08
-How many positive integers not exceeding <math>2001</math> are multiples of <math>3</math> or <math>4</math> but not <math>5</math>?
-<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
-\qquad
-\textbf{(B) }801
-\qquad
-\textbf{(C) }934
-\qquad
-\textbf{(D) }1067
-\qquad
-\textbf{(E) }1167
-</math>
-==Solutions==
+<math>\textbf{(A) }10\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad\textbf{(E) }9</math>
-=== Solution 1===
-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>.
-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
-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>.
-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.
-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>
-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>
-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.
-This is correct.
-===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.
-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==
-===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>.
-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
-===Solution 4===
+==Video Solution 1 by SpreadTheMathLove==
-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>.
-~ mathmagical
+https://www.youtube.com/watch?v=SUnhwbA5_So
-===Solution 5===
+==Video Solution by Math-X (First understand the problem!!!)==
-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>.
+https://youtu.be/EuLkw8HFdk4?si=6dyj2QxkbBuNk6j7&t=951
-~ Fasolinka
+~Math-X
-=== Solution 5===
+==Video Solution==
-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
+https://youtu.be/-yk7ozNRrtQ
+~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
-=== Solution 6===
+==Video Solution by Interstigation==
-Similar to @above:
+https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L
-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:
+==See also==
-<cmath>667+500-166-133-100+33 = 1100-299 = 801.</cmath>
+{{AMC10 box|year=2023|ab=B|num-b=4|num-a=6}}
-Thus, the answer is <math>\boxed{\textbf{(B)}\ 801}.</math>
--ConfidentKoala4
-==Video Solutions==
-https://youtu.be/EXWK7U8uXyk
-https://www.youtube.com/watch?v=XHmKu-ZoRxI&feature=youtu.be
-== See Also ==
-{{AMC12 box|year=2001|num-b=11|num-a=13}}
-{{AMC10 box|year=2001|num-b=25|after=Last Question}}
 {{MAA Notice}}

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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