Difference between revisions of "2023 AMC 10B Problems/Problem 15"
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==Problem== | ==Problem== | ||
What is the least positive integer <math>m</math> such that <math>m\cdot2!\cdot3!\cdot4!\cdot5!...16!</math> is a perfect square? | What is the least positive integer <math>m</math> such that <math>m\cdot2!\cdot3!\cdot4!\cdot5!...16!</math> is a perfect square? | ||
− | |||
− | Consider 2 | + | <math>\textbf{(A) }30\qquad\textbf{(B) }30030\qquad\textbf{(C) }70\qquad\textbf{(D) }1430\qquad\textbf{(E) }1001</math> |
+ | ==Video Solution by MegaMath== | ||
+ | |||
+ | https://www.youtube.com/watch?v=le0KSx3Cy-g&t=28s | ||
+ | |||
+ | == Solution 1 == | ||
+ | |||
+ | Consider 2 -- | ||
there are odd number of 2's in <math>2!\cdot3!\cdot4!\cdot5!...16!</math> (We're not counting 3 2's in 8, 2 3's in 9, etc). | there are odd number of 2's in <math>2!\cdot3!\cdot4!\cdot5!...16!</math> (We're not counting 3 2's in 8, 2 3's in 9, etc). | ||
− | |||
− | |||
− | So, original expression | + | There are even number of 3's in |
+ | ...etc, | ||
+ | |||
+ | |||
+ | So, we can reduce our original expression to | ||
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
− | m \cdot 2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 16 &= | + | m \cdot 2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 16 &\equiv m \cdot 2^8 \cdot (1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8)\\ |
+ | &\equiv m \cdot (2 \cdot 3) \cdot (2 \cdot 2) \cdot 5 \cdot (2 \cdot 3) \cdot 7 \cdot 2 \cdot (2 \cdot 2)\\ | ||
+ | &\equiv m \cdot 2 \cdot 5 \cdot 7\\ | ||
+ | m &= 2 \cdot 5 \cdot 7 = 70 | ||
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
+ | |||
+ | ~Technodoggo ~minor edits by lucaswujc &sravya_m18 | ||
+ | |||
+ | == Solution 2 == | ||
+ | Perfect squares have all of the powers in their prime factorization even. To evaluate <math>2!\cdot3!\cdot4!\cdot5!...16!</math> we get the following: | ||
+ | |||
+ | <math>(2^{15}) \times (3^{14}) \times ((2^2)^{13}) \times (5^{12}) \times ((2 \times 3)^{11}) \times (7^{10}) \times ((2^3)^{9}) \times ((3^2)^8) \times \\ ((5 \times 2)^7) \times (11^6) \times (((2^2) \times 3)^5) \times (13^4) \times ((7 \times 2)^3) \times ((3 \times 5)^2) \times ((2^4)^1)</math>. | ||
+ | |||
+ | Taking all powers <math>\mod 2</math> we get: | ||
+ | |||
+ | <math>(2^1) \times ((2 \times 3)^1) \times (2^1) \times ((5 \times 2)^1) \times (3^1) \times ((7 \times 2)^1)</math> | ||
+ | |||
+ | Simplifying again, we finally get: | ||
+ | |||
+ | <math>(2^1) \times (5^1) \times (7^1)</math> | ||
+ | |||
+ | To make all the powers left even, we need to multiply by <math>(2 \times 5 \times 7)</math> which is <math>\boxed{\text{(C) } 70}</math>. | ||
+ | |||
+ | ~darrenn.cp | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | We can prime factorize the solutions: | ||
+ | A = <math>2 \cdot 3 \cdot 5,</math> | ||
+ | B = <math>2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13, </math> | ||
+ | C = <math>2 \cdot 5 \cdot 7, </math> | ||
+ | D = <math>2 \cdot 5 \cdot 11 \cdot 13, </math> | ||
+ | E = <math>7 \cdot 11 \cdot 13, | ||
+ | </math> | ||
+ | |||
+ | We can immediately eliminate B, D, and E since 13 only appears in <math>13!, 14!, 15, 16!</math>, so <math>13\cdot 13\cdot 13\cdot 13</math> is a perfect square. | ||
+ | Next, we can test if 7 is possible (and if it is not we can use process of elimination). | ||
+ | 7 appears in <math>7!</math> to <math>16!</math> and | ||
+ | 14 appears in <math>14!</math> to <math>16!</math>. | ||
+ | So, there is an odd amount of 7's since there are 10 7's from <math>7!</math> to <math>16!</math> and 3 7's from <math>14!</math> to <math>16!</math> since 7 appears in 14 once, and <math>10+3=13</math> which is odd. So we need to multiply by 7 to get a perfect square. Since 30 is not a divisor of 7, our answer is 70 which is <math>\boxed{\text{C}}</math>. | ||
+ | |||
+ | ~aleyang | ||
+ | |||
+ | == Solution 4 == | ||
+ | |||
+ | First, we note that <math>3! = 2! \cdot 3</math>, <math>5! = 4! \cdot 5, ... 15! = 14! \cdot 15</math>. So, <math>2!\cdot3! ={2!}^2\cdot3 \equiv 3, 4!\cdot5!={4!}^2\cdot5 \equiv 5, ... 14!\cdot15!={14!}^2\cdot15\equiv15</math>. Simplifying the whole sequence and cancelling out the squares, we get <math>3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 15 \cdot 16!</math>. Prime factoring <math>16!</math> and cancelling out the squares, the only numbers that remain are <math>2, 5,</math> and <math>7</math>. Since we need to make this a perfect square, <math>m = 2 \cdot 5 \cdot 7</math>. Multiplying this out, we get <math>\boxed{\text{(C) } 70}</math>. | ||
+ | |||
+ | ~Stead (a.k.a. Aaron) & ~Technodoggo (add more examples) | ||
+ | |||
+ | == Solution 5 (Bashy method) == | ||
+ | We know that a perfect square must be in the form <math>2^{2a_1}\cdot3^{2a_2}\cdot5^{2a_3}...p^{2a_n}</math> where <math>a_1, a_2, a_3, ..., a_n</math> are nonnegative integers, and <math>p</math> is the largest and <math>nth</math> prime factor of our square number. | ||
+ | |||
+ | Let's assume <math>r=m\cdot2!\cdot3!\cdot4!\cdot5!...16!</math>. We need to prime factorize <math>r</math> and see which prime factors are raised to an odd power. Then, we can multiply one factor each of prime number with an odd number of factors to <math>m</math>. We can do this by finding the number of factors of <math>2</math>, <math>3</math>, <math>5</math>, <math>7</math>, <math>11</math>, and <math>13</math>. | ||
+ | |||
+ | Case 1: Factors of <math>2</math> | ||
+ | |||
+ | We first count factors of <math>2^1</math> in each of the factorials. We know there is one factor of <math>2^1</math> each in <math>2!</math> and <math>3!</math>, two in <math>4!</math> and <math>5!</math>, and so on until we have<math>8</math> factors of <math>2^1</math> in <math>16!</math>. Adding them all up, we have <math>1+1+2+2+...7+7+8=64</math>. | ||
+ | |||
+ | Now, we count factors of <math>2^2</math> in each of the factorials. We know there is one factor of <math>2^2</math> each in <math>4!</math>, <math>5!</math>, <math>6!</math>, and <math>7!</math>, two in <math>8!</math>, <math>9!</math>, <math>10</math>, and <math>11!</math>, and so on until we have <math>4</math> factors of <math>2^1</math> in <math>16!</math>. Adding them all up, we have <math>1+1+1+1+2+2+2+2+3+3+3+3+4=28</math>. | ||
+ | |||
+ | Now we count factors of <math>2^3</math> in each of the factorials. Using a similar method as above, we have a sum of <math>1+1+1+1+1+1+1+1+2=10</math>. | ||
+ | |||
+ | Now we count factors of <math>2^4</math> in each of the factorials. Using a similar method as above, we have a factor of <math>2^4</math> in <math>16</math>, so there is <math>1</math> factor of <math>2^4</math>. | ||
+ | |||
+ | Adding all the factors of <math>2</math>, we have <math>103</math>. Since <math>103</math> is odd, <math>m</math> has one factor of <math>2</math>. | ||
+ | |||
+ | Case 2: Factors of <math>3</math> | ||
+ | |||
+ | We use a similar method as in case 1. We first count factors of <math>3^1</math>. We obtain the sum <math>1+1+1+2+2+2+...4+4+4+5+5=50</math>. | ||
+ | |||
+ | We count factors of <math>3^2</math>. We obtain the sum <math>1+1+1+1+1+1+1+1=8</math>. | ||
+ | |||
+ | Adding all the factors of <math>3</math>, we have <math>58</math>. Since <math>58</math> is even, <math>m</math> has <math>0</math> factors of <math>3</math>. | ||
+ | |||
+ | Case 3: Factors of <math>5</math> | ||
+ | |||
+ | We count the factors of <math>5^1</math>: <math>1+1+1+1+1+2+2+2+2+2+3+3=21</math>. Since <math>21</math> is odd, <math>m</math> has one factor of <math>5</math>. | ||
+ | |||
+ | Case 4: Factors of <math>7</math> | ||
+ | |||
+ | We count the factors of <math>7^1</math>: <math>1+1+1+1+1+1+1+2+2+2=13</math>. Since <math>13</math> is odd, <math>m</math> has one factor of <math>7</math>. | ||
+ | |||
+ | Case 5: Factors of <math>11</math> | ||
+ | |||
+ | We count the factors of <math>11^1</math>: <math>1+1+1+1+1+1=6</math>. Since <math>6</math> is even, <math>m</math> has <math>0</math> factors of <math>11</math>. | ||
+ | |||
+ | Case 6: Factors of <math>13</math> | ||
+ | |||
+ | We count the factors of <math>13^1</math>: <math>1+1+1+1=4</math>. Since <math>4</math> is even, <math>m</math> has <math>0</math> factors of <math>13</math>. | ||
+ | |||
+ | Multiplying out all our factors for <math>m</math>, we obtain <math>2\cdot5\cdot7=\boxed{\text{ (C) }70}</math>. | ||
+ | |||
+ | ~arjken | ||
+ | |||
+ | ==Solution 6 (Answer Choices)== | ||
+ | We see that all the answer choices are divisible by <math>10</math> except for <math>1001</math>, and we also notice that the answer choices have <math>3</math>, <math>7</math>, <math>11</math>, or <math>13</math> as a prime factor. | ||
+ | |||
+ | Testing, we see that <math>3</math>, <math>11</math>, <math>13</math> have an even power in the product, so we have that all the other answer choices will not work. | ||
+ | |||
+ | Therefore we just have <math>\boxed{\text{ (C) }70}</math>. | ||
+ | |||
+ | ==Solution 7 (Fastest Intuition)== | ||
+ | Notice that you can add the factor <math>1!</math> to the expression to make it <math>m\cdot1!\cdot2!\cdot3!\cdot4!\cdot5!\cdot6!\cdot7!\cdot8!\cdot9!\cdot10!\cdot11!\cdot12!\cdot13!\cdot14!\cdot15!\cdot16!</math> | ||
+ | |||
+ | Every consecutive pair <math>n!\cdot(n+1)!</math> can be simplified to <math>n+1</math> (divide out <math>n!^2</math>). | ||
+ | |||
+ | Resulting is the product <math>m\cdot2\cdot4\cdot6\cdot8\cdot10\cdot12\cdot14\cdot16 = m\cdot2^8\cdot8!</math> | ||
+ | |||
+ | This is easily simplified to <math>m\cdot2\cdot5\cdot7 = m\cdot70</math> when dividing out perfect squares. | ||
+ | |||
+ | This means that <math>m</math> must have at minimun a factor of <math>70</math> which gives answer choice <math>\boxed{\text{ (C) }70}</math>. | ||
+ | |||
+ | ~coolishu | ||
+ | |||
+ | ==Solution 8(Pretty Quick similar to solution 7)== | ||
+ | <math>16!\left(15!\right)</math> | ||
+ | this can be written as | ||
+ | <math>16\left(15!\right)\left(15!\right)</math> | ||
+ | |||
+ | |||
+ | Thus | ||
+ | |||
+ | |||
+ | <math>16\left(15!\right)^{2}</math> | ||
+ | |||
+ | Now lets try | ||
+ | |||
+ | <math>16!15!14!</math> | ||
+ | |||
+ | <math>16\left(15\right)\left(14!\right)15\left(14!\right)\left(14!\right)</math> | ||
+ | |||
+ | |||
+ | thus | ||
+ | <math>16\left(15\right)^{2}\left(14\right)^{3}</math> | ||
+ | |||
+ | |||
+ | we can continue this pattern until one to get | ||
+ | <math>m\cdot2!\cdot3!\cdot4!\cdot5!...16! = 16\left(15\right)^{2}\left(14\right)^{3}\left(13\right)^{4}\left(12\right)^{5}\left(11\right)^{6}....\left(1\right)^{15}</math> | ||
+ | |||
+ | we can remove all the even powers of expression above to get | ||
+ | |||
+ | <math>16\left(14\right)^{3}\left(12\right)^{5}\left(10\right)^{7}\left(8\right)^{9}\left(6\right)^{11}\left(4\right)^{13}\left(2\right)^{15}</math> | ||
+ | |||
+ | |||
+ | we can see that all off the numbers are even and multiples of 2,3,5,7 | ||
+ | |||
+ | the biggest possible number we should get is 2 times 3 times 5 times 7 | ||
+ | which is 210 | ||
+ | which means out anwer should be A or C | ||
+ | 30 or 70. | ||
+ | |||
+ | Then look all the stuff with 7 which would only consist of <math>{14}^3</math> | ||
+ | |||
+ | which needs an extra 7 to give it company, so put answer must be divisble by 7 and less than 210 | ||
+ | |||
+ | which leaves 70 or C as our answer | ||
+ | |||
+ | <math>\boxed{\text{ (C) }70}</math>. | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | Note: you can also just remove all pairs of two to get <math>16\left(14\right)\left(12\right)\left(10\right)\left(8\right)\left(6\right)\left(4\right)\left(2\right)</math> | ||
+ | |||
+ | ==Video Solution 2 by OmegaLearn== | ||
+ | https://youtu.be/fB81j1vbwdM | ||
+ | |||
+ | ==Video Solution 3 by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=sqVY5-h4vfo | ||
+ | |||
+ | ==Video Solution 4 by paixiao== | ||
+ | https://www.youtube.com/watch?v=EvA2Nlb7gi4&t=238s | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/eW9eBpalm7I | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==See also== | ||
+ | {{AMC10 box|year=2023|ab=B|num-b=14|num-a=16}} | ||
+ | {{MAA Notice}} |
Latest revision as of 07:24, 7 July 2024
Contents
- 1 Problem
- 2 Video Solution by MegaMath
- 3 Solution 1
- 4 Solution 2
- 5 Solution 3
- 6 Solution 4
- 7 Solution 5 (Bashy method)
- 8 Solution 6 (Answer Choices)
- 9 Solution 7 (Fastest Intuition)
- 10 Solution 8(Pretty Quick similar to solution 7)
- 11 Video Solution 2 by OmegaLearn
- 12 Video Solution 3 by SpreadTheMathLove
- 13 Video Solution 4 by paixiao
- 14 Video Solution
- 15 See also
Problem
What is the least positive integer such that
is a perfect square?
Video Solution by MegaMath
https://www.youtube.com/watch?v=le0KSx3Cy-g&t=28s
Solution 1
Consider 2 --
there are odd number of 2's in (We're not counting 3 2's in 8, 2 3's in 9, etc).
There are even number of 3's in ...etc,
So, we can reduce our original expression to
~Technodoggo ~minor edits by lucaswujc &sravya_m18
Solution 2
Perfect squares have all of the powers in their prime factorization even. To evaluate we get the following:
.
Taking all powers we get:
Simplifying again, we finally get:
To make all the powers left even, we need to multiply by which is
.
~darrenn.cp
Solution 3
We can prime factorize the solutions:
A =
B =
C =
D =
E =
We can immediately eliminate B, D, and E since 13 only appears in , so
is a perfect square.
Next, we can test if 7 is possible (and if it is not we can use process of elimination).
7 appears in
to
and
14 appears in
to
.
So, there is an odd amount of 7's since there are 10 7's from
to
and 3 7's from
to
since 7 appears in 14 once, and
which is odd. So we need to multiply by 7 to get a perfect square. Since 30 is not a divisor of 7, our answer is 70 which is
.
~aleyang
Solution 4
First, we note that ,
. So,
. Simplifying the whole sequence and cancelling out the squares, we get
. Prime factoring
and cancelling out the squares, the only numbers that remain are
and
. Since we need to make this a perfect square,
. Multiplying this out, we get
.
~Stead (a.k.a. Aaron) & ~Technodoggo (add more examples)
Solution 5 (Bashy method)
We know that a perfect square must be in the form where
are nonnegative integers, and
is the largest and
prime factor of our square number.
Let's assume . We need to prime factorize
and see which prime factors are raised to an odd power. Then, we can multiply one factor each of prime number with an odd number of factors to
. We can do this by finding the number of factors of
,
,
,
,
, and
.
Case 1: Factors of
We first count factors of in each of the factorials. We know there is one factor of
each in
and
, two in
and
, and so on until we have
factors of
in
. Adding them all up, we have
.
Now, we count factors of in each of the factorials. We know there is one factor of
each in
,
,
, and
, two in
,
,
, and
, and so on until we have
factors of
in
. Adding them all up, we have
.
Now we count factors of in each of the factorials. Using a similar method as above, we have a sum of
.
Now we count factors of in each of the factorials. Using a similar method as above, we have a factor of
in
, so there is
factor of
.
Adding all the factors of , we have
. Since
is odd,
has one factor of
.
Case 2: Factors of
We use a similar method as in case 1. We first count factors of . We obtain the sum
.
We count factors of . We obtain the sum
.
Adding all the factors of , we have
. Since
is even,
has
factors of
.
Case 3: Factors of
We count the factors of :
. Since
is odd,
has one factor of
.
Case 4: Factors of
We count the factors of :
. Since
is odd,
has one factor of
.
Case 5: Factors of
We count the factors of :
. Since
is even,
has
factors of
.
Case 6: Factors of
We count the factors of :
. Since
is even,
has
factors of
.
Multiplying out all our factors for , we obtain
.
~arjken
Solution 6 (Answer Choices)
We see that all the answer choices are divisible by except for
, and we also notice that the answer choices have
,
,
, or
as a prime factor.
Testing, we see that ,
,
have an even power in the product, so we have that all the other answer choices will not work.
Therefore we just have .
Solution 7 (Fastest Intuition)
Notice that you can add the factor to the expression to make it
Every consecutive pair can be simplified to
(divide out
).
Resulting is the product
This is easily simplified to when dividing out perfect squares.
This means that must have at minimun a factor of
which gives answer choice
.
~coolishu
Solution 8(Pretty Quick similar to solution 7)
this can be written as
Thus
Now lets try
thus
we can continue this pattern until one to get
we can remove all the even powers of expression above to get
we can see that all off the numbers are even and multiples of 2,3,5,7
the biggest possible number we should get is 2 times 3 times 5 times 7 which is 210 which means out anwer should be A or C 30 or 70.
Then look all the stuff with 7 which would only consist of
which needs an extra 7 to give it company, so put answer must be divisble by 7 and less than 210
which leaves 70 or C as our answer
.
Note: you can also just remove all pairs of two to get
Video Solution 2 by OmegaLearn
Video Solution 3 by SpreadTheMathLove
https://www.youtube.com/watch?v=sqVY5-h4vfo
Video Solution 4 by paixiao
https://www.youtube.com/watch?v=EvA2Nlb7gi4&t=238s
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2023 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.