Solving SEND + MORE = MONEYDate: 04/18/2002 at 21:11:51 From: George Hollinger Subject: Changing a letter problem into a math equation Hello, I have a problem that I cannot fiqure out. I have tried logical reasoning and can't get it. SEND + MORE ------- = MONEY I have to change the letters into numbers, and all the same letters have to be the same number. I hope you can help! Thanks, George, age 11 Date: 04/19/2002 at 02:59:50 From: Doctor Jeremiah Subject: Re: Changing a letter problem into a math equation Hi George. The secret is to notice that the answer has more letters (5 letters) than the question (4 letters). That M at the beginning of money is a carry from the thousands place, so M = 1. Now we have: SEND + 1ORE ------- = 1ONEY Now, in the thousands place there is a 1, so the only value for S that could cause a carry is S = 9 and that means O = 10. Now we have: 9END + 10RE ------- = 10NEY Now look at the hundreds place. If there were no carry from the tens place, E and N would be the same because E+0 = N, but E and N can't be the same, so there must be a carry from the tens place. Now we have: 1 1 <-- carry 9END + 10RE ------- = 10NEY Now the equation for the hundreds place is 1+E+0 = N or just 1+E = N. In the tens place we can have N+R = E+10 if there is no carry from the ones place, or we can have 1+N+R = E+10 if there is. First test: no carry from the ones place: N+R = E+10 and 1+E = N (1+E)+R = E+10 1+R = 10 R = 10-1 R = 9 But S = 9, so R cannot = 9. That means there is a carry from the ones place and we get: 1+N+R = E+10 and 1+E = N 1+(1+E)+R = E+10 2+R = 10 R = 10-2 R = 8 So now we have: 1 11 <-- carry 9END + 108E ------- = 10NEY N cannot be 0 or 1 because 0 and 1 are taken. N cannot be 2 because 1+2+8 = 11 and then E would equal 1, but it cannot equal 1 because 1 is taken. N could equal 3,4,5,6 or 7 but it cannot equal 8 or 9 because 8 and 9 are taken (and E must be 2,3,4,5 or 6 because it is 1 smaller than N). If E were 2, then for the ones place to carry D would have to be 8 or 9, and both 8 and 9 are taken, so E cannot be 2 (and N cannot be 3). If E were 3, then for the ones place to carry D would have to be 7,8, or 9, but D cannot be 7 because then Y would be 0, which is taken, so E cannot be 3 (and N cannot be 4). If E were 4, then for the ones place to carry D would have to be 6,7,8, or 9. D cannot be 6 because Y would be 0, but D cannot be 7 because Y would be 1, and 0 and 1 are both taken, so E cannot be 4 (and N cannot be 5). The only two possibilities for E now are 5 and 6. If E were 6, then N would be 7 and D would have to be 4 (which would make Y = 0), 5 (which would make Y = 1), 6 (which is taken by E), or 7 (which is taken by N). There are no solutions for E = 6, so E must be 5. So now we have: 1 11 <-- carry 956D + 1085 ------- = 1065Y Do the same reasoning for D and Y and get the answer. If you need more help check out this archived answer: Finding the Digits of SEND + MORE = MONEY http://www.mathforum.org/library/drmath/view/57968.html - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ |
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