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’.