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Suppose we have $a_i, b_i, c_i \in \{0, 1, \dots , 9\}$ and $A=a_2a_1a_0, B=b_2b_1b_0, C=c_1c_0$. We want to perform the following operations with the restriction that only one digit operation is allowed to perform during the intermediate result

$(1 - 10)(A - B.10) + C.100$

My approach

  1. $-(B.10 - A) = -((b_2,b_1,b_0,0) - (a_2,a_1,a_0)) = -(b_2, d_1=(b_1-a_2), d_0=(b_0-a_1), -a_0)$

  2. $-((b_2,d_1,d_0,-a_0) + (b_2,d_1,d_0,-a_0,0)) = -(b_2, e_2=(b_2+d_1), e_1=(d_1+d_0), e_0=(d_0-a_0), -a_0)$

  3. $-((b_2,e_2,e_1,e_0,-a_0) - (c_1,c_0,0,0)) = -(b_2, (e_2-c_1), (e_1-c_0) , e_0, -a_0)$

After the above three steps now we can perform the borrow operation to obtain the correct result. But my above algorithms is not working correctly for example (1 - 10)(384 - 5780) + 6400

Where am I making mistake?

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As a start, I would first get rid of some of the minus signs, and write it as

(10-1)(10B-A)+100C.

The next step would be

10(10B-A)-(10B-A)+100C.

Then, find 10B-A and get the combined digits.

Are you allowed to write the first part as 9(10B_A)?

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  • $\begingroup$ Thank you and now I have solved the problem like your given idea $\endgroup$ – user110219 Mar 25 '15 at 10:54

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