SOLUTION: What is 453 base 7 times 4 base 7?

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Question 609908: What is 453 base 7 times 4 base 7?
Found 2 solutions by jsmallt9, LisaDrapeau:
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
This problem can be done two ways:
  • Convert the base 7 numbers to base ten, multiply and then convert back to base 7; or
  • Try to do base 7 multiplication.
I'm going to do the second one. The key is to carry when you get 7 or more instead of 10 or more when you multiply:

  453
x   4
-----

4 x 3 is twelve which is a seven with 5 left over:

   1
  453
x   4
-----
    5

4 x 5 is twenty which is two sevens with 6 left over:

  21
  453
x   4
-----
   65

4 x 4 is sixteen which is 2 sevens with 2 left over:

 221
  453
x   4
-----
 2265

So 453 base 7 times 4 base 7 is 2265 base 7

Answer by LisaDrapeau(7) About Me  (Show Source):
You can put this solution on YOUR website!

The other poster's previously stated solution is NOT CORRECT. Here is the proper math:


 453
  x 4
 -----
     5  (3x4 is 12 which is ONE 7 with remainder of 5, so carry the ONE)


   1
 453
  x 4
 -----
   05 (5x4 is 20, then plus 1 is 21 which is THREE 7s with 0 left over, so carry the THREE)


 31
 453
  x 4
 -----
  505 (4x4 is 16, then plus 3 equals 19 which is TWO 7s with 5 left over, so carry the TWO)


231
  453
   x 4
  -----
  2505 (you can imagine a zero in front of the "453" so 4x0 is 0 plus 2 is 2 which is ZERO 7s with 2 left over)


Notice the pattern? The process of multiplication is no different regardless of base. The only thing different is in what is carried. In this case it is sets of 7 instead of the sets of 10 you are used to.


The correct answer is 2505 in base 7.


This can be checked by converting the numbers to base 10, then multiplying, and converting back to base 7:


453 base 7 = (4 sets of 7^2) + (5 sets of 7^1) + (3 sets of 7^0)
                 =     (4*49)      +      (5*7)      +     (3*1)
                 =       196        +        35         +       3
                 = 234


4 base 7 = (4 sets of 7^0)
             =     (4*1)
             = 4


234 x 4 = 936 and that converted to base 7 is:


936/7=133 with remainder of 5 so put 5 in rightmost position and
133/7=19 with remainder of 0 so put 0 in position to left of the 5.... 05 and
19/7=2 with remainder of 5 so put 5 in position to left of the 05... 505 and
2 is less than 7 so put the 2 in position to left of the 505... 2505


So to sum up (carried numbers in both cases are in the top row):


   Base 7    =    Base 10
-----------------------------
    231               11
     453              234
      x 4              x 4
   ----------------------
     2505             936


I hope that helps :)