could u Please help with this algebra 2 problem?

Question

A college student sends a post home with this message...

SE , ND
+MO , RE
____________
MON , EY

Each letter Represents a digit (0 to 9) and the calculation represent the sum. How much money was requested.

Please Explain THX SOOOO MUCH!!!

Answer 1

To find what each digit is we first need to determine at least one value. Now it would be a lot easier if we could assume that none of the columns has a carry over and/or if we were given a value for at least some like in the link on the previous example, but since we are not, we will start by assuming that M is 1 because no two single digit numbers (even with a carryover) is greater than 19, and therefore S + M + [maybe a carryover] = 10M + O means that M can only be a 1, assuming a number cannot start with a zero.

We know that none of the other numbers can be one and that :
S + 1 + [maybe a carryover] = O + 10

We will call this maybe carryover C3. C3 has two possible values, (0,1). If it is 1, we have S + 2 = O + 10. If it is 0, we have S + 1 = O + 10. From looking at this we can see that no matter what, S >= 8 (since O cannot be negative and S + 1 + C3 has to add to at least 10. So from this we can determine that S is either 8 or 9 and O is either 0 or 1, since the largest S value plus the largest C3 value = 11, and we are adding 10. Now since O has to be 0 or 1 and we have already used 1, O = 0;

So thus far we have:
S = (8,9)
E = ?
N = ?
D = ?
M = 1
O = 0
R = ?
Y = ?

The next thing to look at is trying to write some equations with the Carryover represented by C1, C2, C3:
D + E = Y + 10C1
N + R + C1 = E + 10C2
E + O + C2 = N + 10C3 or E + 0 + C2 = N + 10C3
S + 1 + C3 = O + 10 = 10

Since E + 0 is not N, we know C2 = 1.
Since E + 0 + C2 = N, this means E + 1 = N or E = N -1
And since N + R + C1 = E + 10C2
N + R + C1 = E + 10
N + R + C1 = 9 + N
Subtract N from both sides and get
R + C1 = 9, so R is either (8,9)

So since these are now our equations:
D + E = Y + 10C1
N + R + C1 = E + 10
E + 1 = N + 10C3
S + 1 + C3 = 10

So we need to figure out either S or R.
We know that both are either 8 or 9 so by solving one, we will have the other.
Since S + 1 + C3 = 10,
And E + 1 = N + 10C3, We can conclude that since the largest value that E can possibly be is 6 (8 and 9 are used already for S and R and N = E +1 so N <= 7) we know that E + 1 < 10. Therefore C3 = 0.
So:
S + 1 + C3 = S + 1 + 0 = 10
S = 9.
Since S= 9, R = 8
We now know that:
S = 9
E = ?
N = ?
D = ?
M = 1
O = 0
R = 8
Y = ?

So our equations are now:
D + E = Y + 10C1
N + 8 + C1 = E + 10
E + 1 = N
9 + 1 = 10

Since E = N - 1, we can use N + 8 + C1 = E + 10 to find C1.
N + 8 + C1 = E + 10
N + 8 + C1 = N - 1 + 10
8 + C1 = 9, so C1 = 1

So our equations are now:
D + E = Y + 10
N + 9 = E + 10
E + 1 = N

So we know that since we have used up 0,1,8,9 and E < N,
2 <= E <= 6
3 <= N <= 7
2 <= D <= 7
2 <= Y <= 7
Now since E + D >= 12 (because Y is at least 2 and we need to add 10) and D is at most 7, we can eliminate from E's possible values anything less than 5. So 5 <= E <= 6 or E = (5,6)

So since E = (5,6) then N = (6,7).
Since E + D >= 12, then when E = 5, D = 7 and when E = 6, then D = 7. But if E = 6 and D = 7 then there are no values left for N. So therefore:
E = 5
D = 7
N = 6

This means we have:
S = 9
E = 5
N = 6
D = 7
M = 1
O = 0
R = 8
Y = ?

To solve for Y we know that E + D = Y + 10 --> 5+ 7 = 12 = Y + 10,
so by subtracting 10 from both sides, Y = 2.

Now since we have all the values, we check:

_SE.ND
+MO.RE
=======
MON.EY

95.67
+10.85
=====
106.52

Checking this, you will see that it is actually a true statement.

Good luck!!

Answer 2

http://mathforum.org/library/drmath/view/57951.html