In Need of Math Help said:

My niece just sent me this problem and it is difficult for me to understand. Please show you work if you are able to help me because I must tell my niece how I got to this answer and show her how to do it. The question is-

The story goes that a young man away at college needed some extra cash. He sent his mother this plea. He wanted her to send the amont indicated by the following sum: **SEND + MORE = MONEY** Each letter stands for a different digit, 0 through 9. No two letters stand for the same digit. How much money did the young man want.

Thank you so much for your help

Method 1

1--M is obviously 1 as the sum of S and N, with, or without, a carryover (c.o.) from E + O cannot possibly equal 20 or more.

2--As for S, S + 1 = 10 + O or S - O = 9 or S + 1 + c.o. = 10 + O or S - O = 9 - c.o..

3--SInce the c.o. can also be no more than 1 from E + O, S - O equals either 8 or 9.

4--S + M can result only in 10 or 11, but 1 is already used, thus, O must be 0 and S = 8 or 9.

5--If S = 8, E must then be 9 making N = 0 which cannot be as O = 0 thus making S = 9.

So where are we?

.......9END

....+ 10RE

-----------------

.....10NEY

6--SInce 9 + 1 = 10, E + 0 = N has no c.o.

7--Since E + 0 = N, there must be a c.o. from N + R making E + 1 = N.

8--To get the c.o. for E + 0, N + R must be greater than 9 or N + R = E + 10 for no c.o. from D + E or N + R = E

+ 9 if there is a c.o.

9--Substiruting N = E + 1 into both expressions for N + R, we have E + 1 + R = E + 10 and E + 1 + R = E + 9.

10--Thus, R = 8 or 9.

11--Since S already = 9, R must be 8 and E + D = 12 or more.

12--The only possibilities for E, D, and N are 5, 6, or 7. Since N = E + 1 and E + D = 12 or more, E = 5, N = 6, D = 7, and Y = 2.

Method 2

1--M must equal 1 since MONEY is the sum of two 4-digit numbers and must be less than 20,000.

2--S must equal 9 and O must equal 0 as S + 1 = 1(O).

3--E + 1 = N because E + 0 = N can't have E = N meaning we have a c.o. from the previous column.

4--R must be 8 since N + R = E requires us to carry a 10 and since E + 1 = N this implies that R = 9. Since S = 9,

R must be 8 and there is a 1 c.o. from the previous column.

5--E must be 5 and N must be 6 because E + 1 = N and D + E = Y forces us to carry a 10. The only numbers left are 2, 3, 4, 5, 6, and 7. Only E = 5, N = 6, D = 7, and Y = 2 is the only solution that satisfies both expressions.