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.

Our partners will collect data and use cookies for ad personalization and measurement. Learn how we and our ad partner Google, collect and use data. Agree & close