Here's a question......!
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Decode the following :
SEND
+
MORE
.......
....
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Answers
Step-by-step explanation:
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.
Answer:
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