### Instruction: In base $10,$ $44 \times 55$ does not equal $3506.$ In what base does $44 \times 55 = 3506$?### Response: This problem involves a different base system, where each digit represents a power of some base instead of $10.$ For example, in base $5,$ the number $123$ means $1 \times 5^2 + 2 \times 5^1 + 3 \times 5^0 = 38$ in base $10.$
To find the base that makes $44 \times 55 = 3506,$ I need to convert the digits of the product to their base $10$ equivalents using the same base.
Let me call the base $b.$ Then $44$ means $4 \times b^1 + 4 \times b^0 = 4b + 4$ and $55$ means $5 \times b^1 + 5 \times b^0 = 5b + 5.$
Multiplying these together, I get $(4b + 4)(5b + 5) = 20b^2 + 40b + 20.$
Similarly, $3506$ means $3 \times b^3 + 5 \times b^2 + 0 \times b^1 + 6 \times b^0 = 3b^3 + 5b^2 + 6.$
Setting these equal to each other, I get $20b^2 + 40b + 20 = 3b^3 + 5b^2 + 6.$
I want to find the value of $b$ that satisfies this equation.
I can simplify the equation by subtracting $20b^2 + 40b + 20$ from both sides, leaving me with $3b^3 - 15b^2 - 40b - 14 = 0.$
This is a cubic equation, which might be hard to solve in general, but I can use some intuition to guess some possible values of $b.$
Since $b$ is a base, it must be a positive integer greater than or equal to $2.$
I can try plugging in some small values of $b$ and see if they make the equation true.
If I try $b = 2,$ I get $3 \times 2^3 - 15 \times 2^2 - 40 \times 2 - 14 = 24 - 60 - 80 - 14 = -130,$ which is not zero.
If I try $b = 3,$ I get $3 \times 3^3 - 15 \times 3^2 - 40 \times 3 - 14 = 81 - 135 - 120 - 14 = -188,$ which is not zero either.
If I try $b = 4,$ I get $3 \times 4^3 - 15 \times 4^2 - 40 \times 4 - 14 = 192 - 240 - 160 - 14 = -222,$ still not zero.
If I try $b = 5,$ I get $3 \times 5^3 - 15 \times 5^2 - 40 \times 5 - 14 = 375 - 375 - 200 - 14 = -214,$ not zero.
If I try $b = 6,$ I get $3 \times 6^3 - 15 \times 6^2 - 40 \times 6 - 14 = 648 - 540 - 240 - 14 = -146,$ not zero.
If I try $b = 7,$ I get $3 \times 7^3 - 15 \times 7^2 - 40 \times 7 - 14 = 1029 - 735 - 280 - 14 = 0,$ which is zero!
So I have found a value of $b$ that satisfies the equation, and therefore makes $44 \times 55 = 3506$ in that base.
The base is $b = 7.$