So im taking this class and extra credit was offered. I tried solving the following problem but no luck. Supposedly if you have a high IQ you can solve it. so here it is. Let's say D=5 Donald+Gerald=Robert Im guessing Donald is a 6 digit number and Gerald is a 6 digit number equaling Robert another 6 digit number. If any of you happen to know the answer please explain how you came up with it.
Each letter is assigned a different number, like 5 and d, and you get Robert by adding each into the other. Yeah, something like that.
D=5, T=0 because 5+5=10, then you solve it from there. 5ONAL5 +GERAL5 -------- ROBER0 D=5, O=2, N=6, A=4, L=8, G=1, E=9, R=7, B=3, T=0
Without additional clues, I am not sure this problem is solvable. But some letters can be inferred. Frist from "o + e = o" we know that either o or e could be 0. If e is 0, then it would be impossible for a + a = e to be correct. So o has to be 0, it follows e is 9, and a is 4, and l > 5. No much else I can come up at this moment.
I just subsituted the remaining numbers here and there and it eventually worked out, not the genius way of doing it, but it worked! If you didn't give me d=5 it would have taken forever, so I imagine there's an easier way of doing it rather than proccess of elimination. And I have a feeling that I've seen this type of puzzle somewhere before...
I did it so it is possible because i'm usually a single digit guy as far as IQ tests go Key was noticing there were ten letters...then looking for patterns. For instance.... o+e =o. e couldn't be zero since it was already taken, so there must have been a 1 carried forward. o+e+1=o. So....e=-1 -- which didn't make sense, so then i realized o+e+1 could equal o +10. suddenly E=9. And so on....(the 1's carried forward also helped decypher stuff) Clearly...i've got too much time on my hands today.
OK, now that I know we can assume 1) letters are case-insensitive and 2) each letter represents a unique digit. The problem is therefore completely solvable without random guessing. For me, the path starts from o + e = o Since we know neither o nor e can be 0, so 9 has to be involved. It only makes sense e is 9. It then follows a is 4, from a + a = e. Since 5 (D) + G = R, R is one of 6, 7, 8. Since L + L = R and it produces a carry-over, L is one of 6, 7, or 8, more over, R has to be an odd number because of the carry-over from 5 (D) + 5 (D) = 10. So R = 7. It follows that L = 8, G = 1. N + 7 (R) = B produces a carry-over. Thus, among the rest unpicked digits, 2, 3, 6, N has to be 6. It follows that B = 3, O =2.
Apparently, these things are called alphametics and there are more examples, I remembered these from somewhere so I searched donald+gerald=robert and found this page: http://www.geocities.com/Athens/Agora/2160/puzzlemenu.html Kind of addicting... I got the used+sex=words on the hard column without cheating.
