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 Mental Arithmetic II (Posted on 2011-09-11)
Devise a procedure for mentally performing these divisions:

(I) A748A (base 14) by D (base 14)

(II) D57389 (base 16) by F (base 16)

 No Solution Yet Submitted by K Sengupta No Rating

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 Possible Solution Comment 1 of 1
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’.

 Posted by Harry on 2011-09-14 23:16:54

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