How can we solve these problems withing using brute force?


  • $\begingroup$ What is brute force? Is it using logic to reduce the number of possible cases, then testing those? $\endgroup$ – The Chaz 2.0 Nov 3 '11 at 14:53
  • $\begingroup$ It also looks like they are intended for 5th graders. I'm working on applying some logic to it now. Brute force would be writing a computer program in order to solve this or (shudder) manually trying every combination. I think we can do better.. $\endgroup$ – foodals Nov 3 '11 at 14:53
  • $\begingroup$ this can be done by applying systems of equations with several variables $\endgroup$ – Peđa Terzić Nov 3 '11 at 15:06
  • $\begingroup$ pedja, how? can you show an example? $\endgroup$ – foodals Nov 3 '11 at 15:11
  • $\begingroup$ These are known as "cryptarithms" or "verbal arithmetic". You can get started in Wikipedia's page on solving them. As the page notes, one usually uses a mix of logic to eliminate possibilities and reduce to a small number, and then a bit of trial-and-error. $\endgroup$ – Arturo Magidin Nov 3 '11 at 16:31

More commonly, when problems like this are presented, you get the intermediate products (one digit of the multiplier times the multiplicand) which are added to get the final answer. Then you have many more clues to work with. But for MIX * SIX = SCARCE, for example, the fact that you carry the S means M has to be large. E has to be 1,4, or 9 because it is the last digit of a square. We could have X=8, E=4; X=9, E=1; or X=3, E=9. But I am afraid there will be a bunch of brute force in these.




So you have to calculate left hand side.You will get expression similar to the right hand side. Use fact that equality is true if corresponding coefficients are equal.Applying this condition you should get $5$ equations with $8$ unknown variables.

  • $\begingroup$ Mind taking it one step further? Give me one of the equations? $\endgroup$ – foodals Nov 3 '11 at 15:59
  • $\begingroup$ $A\cdot N=S$ or $D \cdot T=S$ for example... $\endgroup$ – Peđa Terzić Nov 3 '11 at 16:07
  • $\begingroup$ I'm not sure how this helps, since you don't have as many equations as unknowns you can't solve the system directly. $\endgroup$ – Dan is Fiddling by Firelight May 2 '12 at 18:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.