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Mental Arithmetic II (Posted on 2011-09-11) Difficulty: 3 of 5
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    
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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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