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| Mental Arithmetic II (Posted on 2011-09-11) |
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Devise a procedure for mentally performing these divisions:
(I) A748A (base 14) by D (base 14)
(II) D57389 (base 16) by F (base 16)
Possible Solution
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Comment 1 of 1
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May we assume that these divisions have no remainder? If so, then...
(I) Working in base 14:
Let A748A divided by D be tsrqp.
Thus: A748A = D * tsrqp
= (14 – 1)*tsrqp
= tsrqp0 - tsrqp
Therefore: A748A + tsrqp = tsrqp0, and dealing with each
‘column’ in this addition, from right to left gives:
A + p = 0 (mod 14) => p = 4
8 + q + 1(carry) = 4 (mod 14) => q = 9
4 + r + 1(carry) = 9 (mod 14) => r = 4
7 + s = 4 (mod 14) => s = B
A + t + 1(carry) = B (mod 14) => t = 0 so we have finished.
Answer: tsrqp = B494 (base 14)
(II) Working in base 16:
Let D57389 divided by F be utsrqp. Then, using the same approach
as above, it follows that
D57389 + utsrqp = utsrqp0
and the column calculations needed are:
9 + p = 0 (mod 16) => p = 7
8 + q + 1(carry) = 7 (mod 16) => q = E
3 + r + 1(carry) = E (mod 16) => r = A
7 + s = A (mod 16) => s = 3
5 + t = 3 (mod 16) => t = E
D + u + 1(carry) = E (mod 16) => u = 0 so we have finished
Answer: utsrqp = E3AE7 (base 16)
On a good day, with practice, I might be able to do these ‘mentally’.
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Posted by Harry
on 2011-09-14 23:16:54 |
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