Find the product of $6_8 \cdot 7_8.$ Express your answer in base $8.$
I notice that the problem is asking me to multiply two numbers in base $8.$ I remember that base $8$ means that each digit represents a power of $8,$ so $6_8 = 6 \cdot 8^0 = 6$ and $7_8 = 7 \cdot 8^0 = 7.$ I also remember that to multiply numbers in different bases, I can use the same algorithm as for base $10,$ but with a different base. I set up the multiplication as follows:

   6_8
x  7_8
------
  ?_8 I start by multiplying the ones digits, $6_8$ and $7_8.$ I know that $6 \cdot 7 = 42$ in base $10,$ but I need to convert this to base $8.$ I recall that to convert a number from base $10$ to base $8,$ I can repeatedly divide by $8$ and write the remainders as the digits in reverse order. I divide $42$ by $8$ and get a quotient of $5$ and a remainder of $2.$ I write the remainder $2$ as the ones digit of the product, and carry the quotient $5$ to the next place. 6_8
x  7_8
------
  2_8
 5 I move on to multiplying the tens digits, $0_8$ and $7_8.$ I know that $0 \cdot 7 = 0$ in any base, so I write a zero as the tens digit of the product, and add the carry $5.$ 6_8
x  7_8
------
  2_8
 50_8
+ I get $52_8$ as the final product. 