Suppose $A$ and $B$ are digits for which the product of the two-digit numbers $\underline{2A}$ and $\underline{AB}$ is 1971. Find the sum of $A$ and $B$.
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Suppose A and B are digits for which the product of the two-digit numbers 2A and AB is 1971. Find A+B.
2A*AB = (20+A)(10A+B) = 1971
1971 = 3^3*73
So, its only two-digit factors are
27 and 73
A+B=10
Suppose A and B are digits for which the product of the two-digit numbers 2A and AB is 1971. Find A+B.
2A*AB = (20+A)(10A+B) = 1971
1971 = 3^3*73
So, its only two-digit factors are
27 and 73
A+B=10
Well, to solve this problem, let's just unleash the clown within me.
Ahem, ladies and gentlemen, welcome to the Math Circus! Today, we have a magical two-digit number show for you!
Now, let's first multiply $\underline{2A}$ and $\underline{AB}$.
To do that, we take the first digit of $\underline{2A}$, which is $2$, and multiply it by the first digit of $\underline{AB}$, which is $A$. This gives us $2A^2$.
Next, we take the last digit of $\underline{2A}$, which is just $A$, and multiply it by the last digit of $\underline{AB}$, which is $B$. This gives us $AB$.
Now, let's add these two magical products together: $2A^2 + AB = 1971$.
Hmm, let me take a sip of my clown coffee and think about this for a moment.
Since $1971$ doesn't really give us any clues, let's try out some possible values for $A$ and $B$ and see what happens, shall we?
We know that $A$ and $B$ are digits, which means they can only take on values from $0$ to $9$. So let's try them all, one by one!
If $A = 0$, then $2A^2 + AB = B$. But wait, that doesn't equal $1971$.
Okay, let's try $A = 1$. Then we have $2A^2 + AB = 2 + B$. But that still doesn't equal $1971$.
Time to put on my thinking clown hat...let's keep trying!
If $A = 2$, then $2A^2 + AB = 8 + B$. But that's still nowhere close to $1971$.
Ah, I see something funny here! If $A = 3$, then $2A^2 + AB = 18 + B$.
Wait, wait, hold your applause, my clown friends! This is getting interesting. If $B = 7$, then we have $2A^2 + AB = 18 + 7 = 25$.
Oh, dear audience, do you see what's happening here? $25$ is a special number, and it's special because...because...well, $5 \times 5 = 25$!
So, finally, the magical solution to our problem is $A = 3$ and $B = 5$. And guess what, folks? The sum of $A$ and $B$ is $3 + 5 = \boxed{8}$!
Ta-da! Thank you for joining me at the Math Circus today. I hope you had a clown-tastic time!
To find the sum of $A$ and $B$ in the given problem, we need to find the two-digit numbers $\underline{2A}$ and $\underline{AB}$, whose product is 1971.
We can start by considering the factorization of 1971. The prime factorization of 1971 is $3 \times 17 \times 39$. Since we are looking for two-digit numbers, we can ignore the factor of 39 (because it would make the product more than 100). Thus, we can think of the product as $3 \times 17$.
Now, let's consider the two-digit number $\underline{2A}$ as a product of its digits. The tens digit is 2, and the units digit is $A$. So, we can write $\underline{2A}$ as $10 \times 2 + A = 20 + A$.
Similarly, let's consider the two-digit number $\underline{AB}$. The tens digit is $A$, and the units digit is $B$. So, $\underline{AB}$ can be written as $10 \times A + B = 10A + B$.
Since the product of $\underline{2A}$ and $\underline{AB}$ is $3 \times 17$, we can write the equation as $(20 + A)(10A + B) = 3 \times 17$.
Multiplying out the left side of the equation gives $200A + 20B + 10A^2 + AB = 51$.
Looking at the equation, we can see that $A$ and $B$ must be single-digit integers. Since the product includes $10A^2$ and $200A$, $A$ cannot be 0.
We can now try different values of $A$ and $B$ to see if any of them satisfy the equation. We have the following pairs:
1. $A = 1, B = 9$
2. $A = 3, B = 1$
3. $A = 4, B = 7$
Checking each of these pairs in the equation, we find that only the pair $A = 3$ and $B = 1$ satisfies the equation. Thus, the sum of $A$ and $B$ is $3 + 1 = 4$.
Hence, the sum of $A$ and $B$ is 4.
Suppose A and B are digits for which the product of the two-digit numbers 2A and AB is 1971. Find A+B.
2A*AB = (20+A)(10A+B) = 1971
1971 = 3^3*73
So, its only two-digit factors are
27 and 73
A+B=10