A) Y + 2N = Y, --> N = 0 B) T + 2E = 10 + T, --> E = 5 C) 1 + R + 2T must be > 20 because W + 2 has to be > 10 in order for their to be a remainder of 1 to add to F in order to get S. After making that observation it is just a matter of trial and error substituting large enough values for T and R to satisfy condition C while making sure that S = F + 1.
@lisa this is the best solution.....without any fucking hit n trial
(digit 0) 2N = 0 (mod 10)
(digit 1) 2E = 0 (mod 10) no carry from digit 0 possible Therefore N=0 and E=5. Then O=9 and I=1 requiring two carries. Further S=F+1. Digit 2 no gives the equation R + 2T + 1 = 20 + X The smallest digit left is 2. So X>=2 Therefore R + 2T >= 21. As not both R and T can be 7 one of then must be larger that is 8. We are left with some case checkings. Case R=8: Then T>=6.5 that is T=7 and X=3. There are no consecutive numbers left for F and S. Case T=8: Then R>=5 and as 5 is in use R>=6. Case R=6: As X=3 there are no consecutive numbers left. Case R=7: As X=4 we get F=2 and S=3. The remaining digit 6 will be Y. Therefore we get a unique solution.
hmm is it possible that your riddle is wrong? :s
ReplyDeleteor am i misreading it?
because:
Y+N = Y
and T+E=T
would make that both N and E are equal to 0 if i'm correct?
might be messed up with missing spaces or something?
kind regards
29786
ReplyDelete850
+ 850
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31486
A) Y + 2N = Y, --> N = 0
B) T + 2E = 10 + T, --> E = 5
C) 1 + R + 2T must be > 20 because
W + 2 has to be > 10 in order for their to be a remainder of 1 to add to F in order to get S. After making that observation it is just a matter of trial and error substituting large enough values for T and R to satisfy condition C while making sure that S = F + 1.
@Lisa
ReplyDeleteThat's exactly how I did it. :-)
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ReplyDelete@lisa
ReplyDeletethis is the best solution.....without any fucking hit n trial
(digit 0) 2N = 0 (mod 10)
(digit 1) 2E = 0 (mod 10)
no carry from digit 0 possible
Therefore N=0 and E=5.
Then O=9 and I=1 requiring two carries. Further S=F+1.
Digit 2 no gives the equation
R + 2T + 1 = 20 + X
The smallest digit left is 2. So X>=2
Therefore R + 2T >= 21. As not both R and T can be 7
one of then must be larger that is 8.
We are left with some case checkings.
Case R=8:
Then T>=6.5 that is T=7 and X=3. There are no consecutive
numbers left for F and S.
Case T=8:
Then R>=5 and as 5 is in use R>=6.
Case R=6: As X=3 there are no consecutive numbers left.
Case R=7: As X=4 we get F=2 and S=3. The remaining digit 6 will be Y.
Therefore we get a unique solution.
29786
+850
+850
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31486
thts wat i hav got lik lisa
ReplyDelete