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You can put this solution on YOUR website!
The other poster's previously stated solution is NOT CORRECT. Here is the proper math:
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453
x 4
-----
5 (3x4 is 12 which is ONE 7 with remainder of 5, so carry the ONE)
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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)
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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)
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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)
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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.
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The correct answer is 2505 in base 7.
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This can be checked by converting the numbers to base 10, then multiplying, and converting back to base 7:
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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
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4 base 7 = (4 sets of 7^0)
= (4*1)
= 4
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234 x 4 = 936 and that converted to base 7 is:
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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
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So to sum up (carried numbers in both cases are in the top row):
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Base 7 = Base 10
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231 11
453 234
x 4 x 4
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2505 936
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I hope that helps :)
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