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| − | ==Note==
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| − | <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 ==
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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>
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| − | \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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| − |
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| − | ==Solutions==
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| − | === 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.
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| − | 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>.
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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>.
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| − | 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>.
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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.
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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===
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| − | 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>.
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| − | ===Solution 3===
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| − | 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>
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| − |
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| − | ===Solution 4===
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| − | 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
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| − |
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| − | ===Solution 5===
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| − | 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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| − |
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| − | ~ Fasolinka
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| − |
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| − | === Solution 5===
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| − | 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
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| − | === Solution 6===
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| − | 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:
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| − | <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>
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| − |
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| − | -ConfidentKoala4
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| − |
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| − | ==Video Solutions==
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| − | https://youtu.be/EXWK7U8uXyk
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| − | https://www.youtube.com/watch?v=XHmKu-ZoRxI&feature=youtu.be
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| − |
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| − | == See Also ==
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| − |
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| − | {{AMC12 box|year=2001|num-b=11|num-a=13}}
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| − | {{AMC10 box|year=2001|num-b=25|after=Last Question}}
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| − | {{MAA Notice}}
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