Source: Many sources.
Hi Ken,
Here is the solution to the current problem.
Thanks again for the page,
David Berthold
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SEND
+MORE
-----
MONEY
Assuming M is not a leading zero, M<>0.
The thousand's digit sum must generate a carry.
Since the carry cannot be >1 then M=1.
S+M+[0,1]=10+O; S=[8,9]+O;
Since S<=9 then O=0or1; So O=0.
Then S=[8,9]
For E+O+[0,1] to yield N then the ten's digit sum generates a carry.
Therefore E+1=N+[0,10] yields E+1=N or E=N+9.
Since E=N+9 only allows N=0 then E+1=N and the hundreds generate no carry.
S+1+carry=10; Therefore S=9.
N+R+[0,1]=10+E and substituting E+1=N yields R+[0,1]=9;
So R=8 and the one's digit must have a carry.
So D+E=10+Y; substituting E+1=N again yields D+N=Y+11.
The maximum D+N is now 6+7 therefore Y<=2; So Y=2.
So D+N=6+7
N=E+1; If N=7 then D and E both must be 6, so N=6.
So E=5 and D=7.
That leaves only one solution!
Check: 9567
+1085
-----
10652
(by the way if m=0 there are 24 other solutions)
============================================================================
SEND
+MORE
+GOLD
-----
MONEY
Assuming M is not a leading zero, M<>0.
The thousand's digit sum must generate a carry of 1 or 2.
S+M+G could be as high as 9+2+8=19+[1,2]=[20,21]
That would make O=[0,1]
If O=[0,1] then the maximum hundred's digit carry is 1.
Then S+M+G+1 is a maximum of 20.
Then O=0. If O=0 then hundred's carry is 0.
Therefore the thousand's carry is a maximum of 1 and M=1.
Since her parents assume the minimum, choose S+G=9.
S+G=[2+7,3+6,4+5]
Then the hundred's carry is 0 and O=0.
There must be a ten's digit carry so E+[1,2]=N
Again, to minimize choose the ten's carry=1; So E+1=N
To minimize, choose the one's carry=0.
Then N+R+L=E+10
Substituting N=E+1 yields R+L=9
And Y=2D+E
Y D E
- - -
9 2 5
8 2 4
8 3 2
7 2 3
Remember that S+G=[2+7,3+6,4+5]
All combinations of D and E above eliminate two of these pairs.
Therefore R+L and S+G cannot both be equal to 9.
So the ten's carry is not 0.
So the minimum ten's carry must be 1.
To minimize, choose the one's carry=1.
Then N+R+L+1=E+10
Substituting N=E+1 yields R+L=8
Remember that S+G=[2+7,3+6,4+5]
And R+L=[2+6,3+5]
This leaves the following choices:
S+G R+L
--- ---
2+7 3+5
3+6 none
4+5 2+6
Either way 2 and 5 must be used in the combination.
And 2D+E=Y+10 yields the following possibilities:
E D Y
- - -
2 6 4 Not since 2 is used
2 7 6 Not since 2 is used
3 6 5 Not since 5 is used
3 8 9 =
4 7 8
5 4 3 Not since 5 is used
5 6 7 Not since N=E+1
5 7 9 Not since 5 is used
6 3 2 Not since 2 is used
7 4 5 Not since 5 is used
7 6 9
8 3 4
8 4 6
9 2 3 Not since 2 is used
9 3 5 Not since 5 is used
9 4 7 Not since N=E+1
That leaves the following:
E D N Y
- - - -
3 8 4 9 No combination left for S+G and R+L =
4 7 5 8 No combination left for S+G and R+L
7 6 8 9 No combination left for S+G and R+L
8 3 9 4 No combination left for S+G and R+L
8 4 9 6
So E=8, N=9, D=4, and Y=6
That eliminates S+G=4+5 and R+L=2+6.
So S+G=2+7 (in either order)
And R+L=3+5 (in either order)
That leaves 4 solutions but the MONEY remains the same.
2894 2894 7894 7894
+1038 +1058 +1038 +1058
+7054 +7034 +2054 +2034
----- ----- ----- -----
10986 10986 10986 10986
(The possible sums are 10986, 13580, 15326, 15762, 16740, 17508)
(by the way if m=1 there are 28 solutions)
(and if m=0 there are 20 other solutions)
============================================================================
So the second student asked for at least 10986-10652 = $334 more!
1) SEND + MORE = MONEY
We can check easily that M,O and S take only the respective values 1,0 and
9, so we have to solve 9END + 10RE = 10NEY with N = E+1
Two cases:
1) D+E=Y
then N+R=10+E => R=9 impossible
2) D+E=Y+10
then N+R+1=10+E => R=8
the sole possible couple (D,Y) is 7,2
hence the solution is 9567 + 1085 = 10652
2) SEND + MORE + GOLD = MONEY
We have again one possible value for M, i.e. M=1
With the equation 2D+E=Y or 2D+E=Y+10 or 2D+E=Y+20,we can identify 42
possible set of 3 elements (D,E,Y) and for each of them,we have to solve the
following equations:
N+R+L+(0 or 1 or 2) = E+(0 or 1 or 2)
2O+E +(0 or 1 or 2) = N+(0 or 1 or 2)
S+G = O+(7 or 8 or 9)
Without computer but with a lot of patience, we determine 3 families od
solutions:
1) 7894 + 1058 + 2034 = 10986 S and G can be permuted, R and L too
2) 6054 + 1720 + 9734 = 17508 same conditions as above on S&G,R&L
3) 9876 + 1458 + 3426 = 14760 same conditions as above on S&G,R&L
Globally there are 12 different solutions.