Solve: D O N A L D + G E R A L D = R O B E R T, D = 5, each letter lett er represents a unique digit from 0 to 9. _ _ D O N A L D +GERALD _ == == == == == == _ R O B E R T
1. Since D = 5, and D + D = T, T , then 5 + 5 = 10, so T is 0, and we carry a 1: _ _ _ _ _ _ 1 _ 5 O N A L 5 +GERAL5 _ == == == == == == _ R O B E R 0
2. Since O + E = O, and E cannot be zero (0 is already taken), ta ken), this addition can only be possible if E = 9. Then, O + 9 = O is only possible if we have a carry-over of 1 from the addition to its right, (N + R = B). We observe that O + 9 = O, which means that the result r esult is > 10, and we must have a carry-over carry- over of 1. We have: _ _ 1 _ _ _ 1 _ 5 O N A L 5 +G9RAL5 _ == == == == == == _ R O B 9 R 0 3. Since A + A = 9, we must have a carry-over from the previous column (from column L + L = R), and A must equal to 4 (i.e. 4 + 4 + 1 = 9). Thus, A=4. _ _ 1 _ _ 1 1 _ 5 O N 4 L 5 +G9R4L5 _ == == == == == == _ R O B 9 R 0 4. Since we carried over a 1 to accomplish O + 9 = O, then N + R = B implies that B > 10 (we·re (we ·re carrying the 1 over, and 0 is already taken by T, and a nd that is why B cannot equal e qual 10, so B > 10) Digits used up: { 0, 4, 5, 9 } Free: { 1, 2, 3, 6, 7, 8 } Since N + R > 10, then from the t he Free set we can only try to use combinations {3, 8}, {6, 7}, {6, 8}, or {7, 8}. We also observe that L + L + 1 = R, which implies that R must be odd. Then, from our 4 sets of possible assignments above, we can only use 3 and 7 as possible assignments to R, since they are the only odd digits occurring in the sets. Thus, R must be either 3 or 7. 7. We now try to figure f igure out if R is 3 or 7.
In the first column we see that
5 + G + 1 = R (and since we do not have any more digits left of R, i.e. we don·t carry over anything), R cannot be 3 (since 5 + G + 1 = 3 implies that G is negative). Thus, R must take value 7. We have:
_ _ 1 1 _ 1 1 _ 5 O N 4 L 5 +G974L5 _ == == == == == == _ 7 O B 9 7 0 5. Since 5 + G + 1 = 7, then G = 1. We have: _ _ 1 1 _ 1 1 _ 5 O N 4 L 5 +1974L5 _ == == == == == == _ 7 O B 9 7 0 6. Since L + L + 1 = 7, then we must have L = 3 or L = 8. We notice from 4 + 4 = 9 that we must be carrying car rying a 1 over from L + L. Since we carry a 1 from L + L, then we cannot have L = 3, and therefore L = 8: _ _ 1 1 _ 1 1 _ 5 O N 4 8 5 +197485 _ == == == == == == _ 7 O B 9 7 0 7. Digits used up: {1, 4, 5, 7, 8, 9, 9 , 0} Free: {2, 3, 6} We try to see which free digits match next. We have N + 7 = B, and we know that we must carry car ry a 1 after the t he addition, thus we can·t have N=2, because 2 + 7 < 10. If we let N = 3, then 3 + 7 = 10, and B cannot be zero (zero ( zero is already taken), therefore N cannot equal 3. The only other possibility is for N = 6. It immediately follows that B = 3 (since 6 + 7 = 13) _ _ 1 1 _ 1 1 _ 5 O 6 4 8 5 +197485 _ == == == == == == _ 7 O 3 9 7 0 8. The only remaining unused digit is 2. Therefore, let O = 2. Then 2 + 9 + 1 = 12, which works out. The final equation is then: _ _ 1 1 _ 1 1 _ 5 2 6 4 8 5
+197485 _ == == == == == == _ 7 2 3 9 7 0 Adding 526485 with197485 gives the result: 723970, therefore the t he addition is correct. The mapping is thus: 0 = T, 1 = G, 2 = O, 3 = B, 4 = A, 5 = D, 6 = N, 7 = R, 8 = L, 9 = E
Send+more=money steps.
Solving a cryptarithm by hand usually involves a mix of deductions and exhaustive tests of possibilities. possibilities. For instance, the following sequence of deductions so lves Dudeney's SEND + MORE = MONEY puzzle above (columns are numbered from right to left): 1. From column 5, M = 1 since it is the only carry-over possible from the sum of two single digit numbers in column 4. 2. To produce a carry from column 4 to column 5, S + M is at least 9, so S is 8 or 9, so S + M is 9 or 10, and so O is 0 or 1. But M = 1, so O = 0. 3. f there were a carry from column 3 to column 4 then E = 9 and so N = 0. But O = 0, so there is no carry, and S = 9. 4. f there were no carry from fro m column 2 to column 3 then E = N, which is impossible. Therefore there is a carry and N = E + 1. 5. If there were no carry from column 1 to column 2, then N + R = E mod 10, and N = E + 1, so E + 1 + R = E mod 10, so R = 9. But S = 9, so there must be a carry from column 1 to column 2 and R = 8. 6. To produce a carry from column 1 to column 2, we must have D + E = 10 + Y. As Y cannot be 0 o r 1, D + E is at least 12. As D is at most 7, then t hen E is at least 5. Also, N is at most 7, and N = E + 1. So E is 5 or 6. 7. If E were 6 then to make D + E at least 12, D would have to be 7. But N = E + 1, so N would also be 7, which is impossible. Therefore E = 5 and N = 6. 8. To make D + E at least 12 we must have D = 7, and so Y = 2. url for cross+ road = danger http://cryptarithms.awardspace.us/puzzle46.html
I assume no two characters are equal in any way. If we use numbers, R must be even cause at the end S + S = R ( Odd + Odd and Even + Even both equal Even, so R's got to be Even ) First conclusion: R is even ( Say, 2, 4, 6, 8, since solution set doesn't consist of any 0 ( i.e. 1-9 only, possibly ) ) Now CROSS +ROADS -----DANGER As you'd notice, D is extra, like something carried over in addition. So D's got to be 1 since any sum of 1-9 will can can give a carry of 1 alone. So, A will have a value of [2,3,4,5,6,7,8,9] cause D's already taken 1. First value found, D = 1 C+R = Set of numbers ( 12,13,14,15,16,17,18 ( 19 not possible as 10 is not allowed ) ) Which equal only to sums of: If A: C+R C+R Case 2: 3+9 || 9+3 4+8 || 8+4 5+7 || 7+5 ( Other combo's not possible cause C!=R ) Case 3: 4+9 || 9+4 5+8 || 8+5 6+7 || 7+6
Case 4: 5+9 || 9+5 6+8 || 8+6 Case 5: 6+9 || 9+6 7+8 || 8+7 Case 6: 7+9 || 9+7 Case 7: 8+9 || 9+8 Case 8: Not Possible as ( 9,9 ) is the only case and C!=R ( Assumed at the beginning of this puzzle solving ) Thus from the above equation, as R is even, we have R = Possiblity( 4, 6, 8 ) and C = Possiblity( 4, 5, 6, 7, 8, 9 ) So far, A = ( 2, 3, 4, 5, 7 ) [6's totally Odd for values of R] D=1 C = ( 4, 5, 6, 7, 8, 9 ) R = ( 4, 6, 8 ) Now since R is Even and definitely in [4, 6, 8] From R + O = N, we get R = N - O Thus, If R: N-O Case 4: 9-5 7-3 6-2 ( Other combo's not possible as in our assumption, we use R = 4 )
Case 6: 9-3 8-2 Case 8: Not possible as 1's already taken by D ( 9-1 being the only possible way ) Thus from the above, N = ( 6, 7, 8, 9 ) O = ( 2, 3, 5 ) A = ( 2, 3, 4, 5 ) D=(1) C = ( 4, 5, 6, 7, 8, 9 ) R = ( 4, 6 ) Also, A doesn't have 7 since C + R = A but R has only ( 4, 6 ) while C is ( 4, 5, 6, 7, 8, 9 ) ( Adding C+R will never yeild 17 ) Now similarly, G = O + A Thus, O = G - A If O: G-A Case 2: 7-5 6-4 5-3 Case 3: 8-5 7-4 5-2 ( 6-3 not Possible as O!=A, thus I G can't be 6 and A can't be 3 ) Case 5: 9-4 8-3 7-2
This gives us two conclusions, First, A is now ( 2, 4, 5 ) Second, G is ( 5, 7, 8, 9 ) Phew .. I need a break. Continuing, We now have: A = ( 2, 4, 5 ) D=(1) C = ( 4, 5, 6, 7, 8, 9 ) R = ( 4, 6 ) O = ( 2, 3, 5 ) N = ( 6, 7, 8, 9 ) G = ( 5, 7, 8, 9 ) Good so far? I hope so, else my entire evening is lifeless. :? Now since R = S + S and R has ( 4, 6 ) If R: S+S Case 4: 2+2 Case 6: 3+3 Therefore,, S = ( 2,3 ) Therefore Applying for E = S + D, we have: E = 2+1 = 3 or E = 3+1 = 4 Thus, E = ( 3,4 ) So this far,
D=(1) A = ( 2, 4, 5 ) C = ( 4, 5, 6, 7, 8, 9 ) R = ( 4, 6 ) N = ( 6, 7, 8, 9 ) O = ( 2, 3, 5 ) G = ( 5, 7, 8, 9 ) S = ( 2, 3 ) E = ( 3, 4 ) Not being able to proceed more as I did so long, I try to apply the smallest value sets to the solution of bigger b igger sets. If S is 2, R will be equal to 4 ( S + S ) and E will equal 3 ( S + D ) Thus, if R = 4 and E = 3 and S = 2 the only possible values of A will be ( 5 ) and O will be ( 5 ) too which makes this impossible since A cant equal O. Thus S will NOT be 2 and is surely 3 instead. Now if S is 3, we have: E=S+D=4 R=S+S=6 Thus, we now have possible values as: D=(1) A = ( 2, 5 ) C = ( 5, 7, 8, 9 ) R=(6) N = ( 6, 7, 8, 9 ) O = ( 2, 3, 5 ) G = ( 5, 7, 8, 9 ) S=(3) E=(4) Now A = C + R and has got to be either 12 or 15 since it gives D = 1 as carry. Thus only 9 + 6 satisfies it for A = 5 and 12 isn't possible with the remaining values.
Hence, A = ( 5 ) and C = ( 9 ) [Proof: C = A - R = 15 - 6 = 9 not considering the Carry character. This This proof is only additive to the above logical result] So again, so far, removing duplicates as we've been doing, D=(1) A=(5) C=(9) R=(6) S=(3) E=(4) N = ( 7, 8 ) O=(2) G = ( 7, 8 ) ( As numbers 3, 5 and 9 are already used up. ) Now N = R + O = 6 + 2 = 8 and hence G = 7 ( Odd one out ) [Proof for G: G = O + A = 2 + 5 = 7, again an additive additive proof.] Thus finally the solution is, D=1 A=5 N=8 G=7 E=4 R=6 C=9 O=2 S=3 Or in numeric order: D=1 O=2 S=3 E=4
A=5 R=6 G=7 N=8 C=9 Phew, took an hour. Hope am not wrong in my approach itself! If I am, I desperately need a life. :( P.s. Am attaching the TXT format in case the formatting I typed this in isn't showing well for reading here.
