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Originally Posted by spindles
There goes an hour of my work day - I think this is not an overly complicated problem. The best method of trying to solve it is to start with the beginnings as shown by miriad people above and work out what each letter is not.
Given that p=1, and can't be anything else, r has to be 0, which makes h =9.
From this start, you can deduce that none of OCUS are 5 (as the resulting bottom line would be either 0 or 1, then O needs to be even, but not 6 or 8, because it invalidates "R=0"
and so on.
From here I did a bit of If this, then that, until I got an invalid result.
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h does not have to be 9, it could be 8 due to carryover from the previous column. You know that r is equal to 0 because you can't have a situation where there is a carryover and h is 9, because r cannot be 1. So if h can be 8 or 9, then none of the OCUS letters can be 9, because the result of two of these letters would result in 18 without the carryover and 19 with, but none of these answers would be possible because either you would use up both 8 and 9, which are the only possible values for h, or you would get 9 for two different letters. Also, as spindles mentioned none of the OCUS letters can be 5. It think there was some other deduction that I made, but I can't recall now.