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Difference between revisions of "2023 AMC 10B Problems/Problem 15"

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m (→‎Solution 7 (Fastest Intuition))
 
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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?
 +
 +
<math>\textbf{(A) }30\qquad\textbf{(B) }30030\qquad\textbf{(C) }70\qquad\textbf{(D) }1430\qquad\textbf{(E) }1001</math>
 +
 
== Solution 1 ==
 
== Solution 1 ==
  
−
Consider 2,
+
We want <math>m\cdot2!\cdot3!\cdot4!\cdot\dots\cdot16!</math> to be a perfect square. Notice that we can rewrite and pair up certain elements:
−
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).
+
 
 +
<cmath>m\cdot2\cdot3!\cdot(4\cdot3!)\cdot5!\cdot(6\cdot5!)\cdot\dots\cdot15!\cdot(16\cdot15!)=m\cdot2\cdot4\cdot6\cdot\dots\cdot16\cdot(3!)^2(5!)^2\cdot\dots\cdot(15!)^2.</cmath>
  
−
There are even number of 3's in <math>2!\cdot3!\cdot4!\cdot5!...16!</math>
+
Note here that this is equivalent to simply <math>m\cdot2\cdot4\cdot\dots\cdot16</math> being a perfect square (this is intuitively obvious: i.e. if <math>a=bc</math> is a perfect square and so is <math>b</math>, then of course <math>c</math> must be a perfect square too). We can rewrite this as the following:
−
...etc,
 
  
 +
<cmath>m\cdot2^8\cdot(1\cdot2\cdot3\cdot\dots\cdot8)\equiv m\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\equiv m\cdot70.</cmath>
  
−
So, we original expression reduce to
+
The smallest <math>m</math> s.t. <math>70m</math> is a perfect square is, of course, <math>70</math> itself. QED.
−
<cmath>
 
−
\begin{align*}
 
−
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*}
 
−
</cmath>
 
  
−
~Technodoggo
+
~Technodoggo (rewritten as of 2024)
  
 
== Solution 2 ==
 
== 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:  
 
We can prime factorize the solutions:  
Line 33: Line 46:
  
 
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.
 
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)
+
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
 
7 appears in <math>7!</math> to <math>16!</math> and
−
14 appears in <math>7!</math> to <math>16!</math>.  
+
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>7!</math> to <math>16!</math>, 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>.
+
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
 
~aleyang
  
−
== Solution 3 ==
+
== Solution 4 ==
  
−
First, we note <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>.
+
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>.
  
−
~yourmomisalosinggame (a.k.a. Aaron) & ~Technodoggo (add more examples)
+
~Stead (a.k.a. Aaron) & ~Technodoggo (add more examples)
  
−
== Solution 4 (Bashy method) ==
+
== 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.
 
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.
  
Line 90: Line 103:
  
 
~arjken
 
~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 a minimum 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>
 +
 +
==Solution 9 (Braindead solution)==
 +
 +
List out 1 to 16:
 +
 +
<cmath>1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16.</cmath>
 +
 +
First, we remove all numbers that appear an even amount of times in the product:
 +
 +
<cmath>2, 4, 6, 8, 10, 12, 14, 16.</cmath>
 +
 +
Next, we remove all perfect squares:
 +
 +
<cmath>2, 6, 8, 10, 12, 14.</cmath>
 +
 +
Factoring each term gives:
 +
 +
<cmath>2^1, 2^1\cdot 3^1, 2^3, 2^1 \cdot 5^1, 2^2 \cdot 3^1, 2^1 \cdot 7^1.</cmath>
 +
 +
We multiply all the terms together and remove perfect squares to get <math>2^1 \cdot 5^1 \cdot 7^1 = \boxed{\text{ (C) }70}</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)
 +
 +
==Video Solution by MegaMath==
 +
 +
== https://www.youtube.com/watch?v=le0KSx3Cy-gt=28s ==
 +
 +
==See also==
 +
{{AMC10 box|year=2023|ab=B|num-b=14|num-a=16}}
 +
{{MAA Notice}}

Latest revision as of 19:55, 10 November 2024

Problem

What is the least positive integer $m$ such that $m\cdot2!\cdot3!\cdot4!\cdot5!...16!$ is a perfect square?

$\textbf{(A) }30\qquad\textbf{(B) }30030\qquad\textbf{(C) }70\qquad\textbf{(D) }1430\qquad\textbf{(E) }1001$

Solution 1

We want $m\cdot2!\cdot3!\cdot4!\cdot\dots\cdot16!$ to be a perfect square. Notice that we can rewrite and pair up certain elements:

\[m\cdot2\cdot3!\cdot(4\cdot3!)\cdot5!\cdot(6\cdot5!)\cdot\dots\cdot15!\cdot(16\cdot15!)=m\cdot2\cdot4\cdot6\cdot\dots\cdot16\cdot(3!)^2(5!)^2\cdot\dots\cdot(15!)^2.\]

Note here that this is equivalent to simply $m\cdot2\cdot4\cdot\dots\cdot16$ being a perfect square (this is intuitively obvious: i.e. if $a=bc$ is a perfect square and so is $b$, then of course $c$ must be a perfect square too). We can rewrite this as the following:

\[m\cdot2^8\cdot(1\cdot2\cdot3\cdot\dots\cdot8)\equiv m\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\equiv m\cdot70.\]

The smallest $m$ s.t. $70m$ is a perfect square is, of course, $70$ itself. QED.

~Technodoggo (rewritten as of 2024)

Solution 2

Perfect squares have all of the powers in their prime factorization even. To evaluate $2!\cdot3!\cdot4!\cdot5!...16!$ we get the following:

$(2^{15}) \times (3^{14}) \times ((2^2)^{13}) \times (5^{12}) \times ((2 \times 3)^{11}) \times (7^{10}) \times ((2^3)^{9}) \times ((3^2)^8) \times \\ ((5 \times 2)^7) \times (11^6) \times (((2^2) \times 3)^5) \times (13^4) \times ((7 \times 2)^3) \times ((3 \times 5)^2) \times ((2^4)^1)$.

Taking all powers $\mod 2$ we get:

$(2^1) \times ((2 \times 3)^1) \times (2^1) \times ((5 \times 2)^1) \times (3^1) \times ((7 \times 2)^1)$

Simplifying again, we finally get:

$(2^1) \times (5^1) \times (7^1)$

To make all the powers left even, we need to multiply by $(2 \times 5 \times 7)$ which is $\boxed{\text{(C) } 70}$.

~darrenn.cp

Solution 3

We can prime factorize the solutions: A = $2 \cdot 3 \cdot 5,$ B = $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13,$ C = $2 \cdot 5 \cdot 7,$ D = $2 \cdot 5 \cdot 11 \cdot 13,$ E = $7 \cdot 11 \cdot 13,$

We can immediately eliminate B, D, and E since 13 only appears in $13!, 14!, 15, 16!$, so $13\cdot 13\cdot 13\cdot 13$ 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 $7!$ to $16!$ and 14 appears in $14!$ to $16!$. So, there is an odd amount of 7's since there are 10 7's from $7!$ to $16!$ and 3 7's from $14!$ to $16!$ since 7 appears in 14 once, and $10+3=13$ 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 $\boxed{\text{C}}$.

~aleyang

Solution 4

First, we note that $3! = 2! \cdot 3$, $5! = 4! \cdot 5, ... 15! = 14! \cdot 15$. So, $2!\cdot3! ={2!}^2\cdot3 \equiv 3, 4!\cdot5!={4!}^2\cdot5 \equiv 5, ... 14!\cdot15!={14!}^2\cdot15\equiv15$. Simplifying the whole sequence and cancelling out the squares, we get $3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 15 \cdot 16!$. Prime factoring $16!$ and cancelling out the squares, the only numbers that remain are $2, 5,$ and $7$. Since we need to make this a perfect square, $m = 2 \cdot 5 \cdot 7$. Multiplying this out, we get $\boxed{\text{(C) } 70}$.

~Stead (a.k.a. Aaron) & ~Technodoggo (add more examples)

Solution 5 (Bashy method)

We know that a perfect square must be in the form $2^{2a_1}\cdot3^{2a_2}\cdot5^{2a_3}...p^{2a_n}$ where $a_1, a_2, a_3, ..., a_n$ are nonnegative integers, and $p$ is the largest and $nth$ prime factor of our square number.

Let's assume $r=m\cdot2!\cdot3!\cdot4!\cdot5!...16!$. We need to prime factorize $r$ 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 $m$. We can do this by finding the number of factors of $2$, $3$, $5$, $7$, $11$, and $13$.

Case 1: Factors of $2$

We first count factors of $2^1$ in each of the factorials. We know there is one factor of $2^1$ each in $2!$ and $3!$, two in $4!$ and $5!$, and so on until we have$8$ factors of $2^1$ in $16!$. Adding them all up, we have $1+1+2+2+...7+7+8=64$.

Now, we count factors of $2^2$ in each of the factorials. We know there is one factor of $2^2$ each in $4!$, $5!$, $6!$, and $7!$, two in $8!$, $9!$, $10$, and $11!$, and so on until we have $4$ factors of $2^1$ in $16!$. Adding them all up, we have $1+1+1+1+2+2+2+2+3+3+3+3+4=28$.

Now we count factors of $2^3$ in each of the factorials. Using a similar method as above, we have a sum of $1+1+1+1+1+1+1+1+2=10$.

Now we count factors of $2^4$ in each of the factorials. Using a similar method as above, we have a factor of $2^4$ in $16$, so there is $1$ factor of $2^4$.

Adding all the factors of $2$, we have $103$. Since $103$ is odd, $m$ has one factor of $2$.

Case 2: Factors of $3$

We use a similar method as in case 1. We first count factors of $3^1$. We obtain the sum $1+1+1+2+2+2+...4+4+4+5+5=50$.

We count factors of $3^2$. We obtain the sum $1+1+1+1+1+1+1+1=8$.

Adding all the factors of $3$, we have $58$. Since $58$ is even, $m$ has $0$ factors of $3$.

Case 3: Factors of $5$

We count the factors of $5^1$: $1+1+1+1+1+2+2+2+2+2+3+3=21$. Since $21$ is odd, $m$ has one factor of $5$.

Case 4: Factors of $7$

We count the factors of $7^1$: $1+1+1+1+1+1+1+2+2+2=13$. Since $13$ is odd, $m$ has one factor of $7$.

Case 5: Factors of $11$

We count the factors of $11^1$: $1+1+1+1+1+1=6$. Since $6$ is even, $m$ has $0$ factors of $11$.

Case 6: Factors of $13$

We count the factors of $13^1$: $1+1+1+1=4$. Since $4$ is even, $m$ has $0$ factors of $13$.

Multiplying out all our factors for $m$, we obtain $2\cdot5\cdot7=\boxed{\text{ (C) }70}$.

~arjken

Solution 6 (Answer Choices)

We see that all the answer choices are divisible by $10$ except for $1001$, and we also notice that the answer choices have $3$, $7$, $11$, or $13$ as a prime factor.

Testing, we see that $3$, $11$, $13$ have an even power in the product, so we have that all the other answer choices will not work.

Therefore we just have $\boxed{\text{ (C) }70}$.

Solution 7 (Fastest Intuition)

Notice that you can add the factor $1!$ to the expression to make it $m\cdot1!\cdot2!\cdot3!\cdot4!\cdot5!\cdot6!\cdot7!\cdot8!\cdot9!\cdot10!\cdot11!\cdot12!\cdot13!\cdot14!\cdot15!\cdot16!$

Every consecutive pair $n!\cdot(n+1)!$ can be simplified to $n+1$ (divide out $n!^2$).

Resulting is the product $m\cdot2\cdot4\cdot6\cdot8\cdot10\cdot12\cdot14\cdot16 = m\cdot2^8\cdot8!$

This is easily simplified to $m\cdot2\cdot5\cdot7 = m\cdot70$ when dividing out perfect squares.

This means that $m$ must have a minimum factor of $70$ which gives answer choice $\boxed{\text{ (C) }70}$.

~coolishu

Solution 8(Pretty Quick similar to solution 7)

$16!\left(15!\right)$ this can be written as $16\left(15!\right)\left(15!\right)$


Thus


$16\left(15!\right)^{2}$

Now lets try

$16!15!14!$

$16\left(15\right)\left(14!\right)15\left(14!\right)\left(14!\right)$


thus $16\left(15\right)^{2}\left(14\right)^{3}$


we can continue this pattern until one to get $m\cdot2!\cdot3!\cdot4!\cdot5!...16! = 16\left(15\right)^{2}\left(14\right)^{3}\left(13\right)^{4}\left(12\right)^{5}\left(11\right)^{6}....\left(1\right)^{15}$

we can remove all the even powers of expression above to get

$16\left(14\right)^{3}\left(12\right)^{5}\left(10\right)^{7}\left(8\right)^{9}\left(6\right)^{11}\left(4\right)^{13}\left(2\right)^{15}$


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 ${14}^3$

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

$\boxed{\text{ (C) }70}$.

Note: you can also just remove all pairs of two to get $16\left(14\right)\left(12\right)\left(10\right)\left(8\right)\left(6\right)\left(4\right)\left(2\right)$

Solution 9 (Braindead solution)

List out 1 to 16:

\[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16.\]

First, we remove all numbers that appear an even amount of times in the product:

\[2, 4, 6, 8, 10, 12, 14, 16.\]

Next, we remove all perfect squares:

\[2, 6, 8, 10, 12, 14.\]

Factoring each term gives:

\[2^1, 2^1\cdot 3^1, 2^3, 2^1 \cdot 5^1, 2^2 \cdot 3^1, 2^1 \cdot 7^1.\]

We multiply all the terms together and remove perfect squares to get $2^1 \cdot 5^1 \cdot 7^1 = \boxed{\text{ (C) }70}$.

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)

Video Solution by MegaMath

https://www.youtube.com/watch?v=le0KSx3Cy-gt=28s

See also

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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All AMC 10 Problems and Solutions

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