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Another CryptArithmatic Problem

Replace letters with numbers assuming numbers can't be repeated. 
 
     SEND
  + MORE
  ----------
 = MONEY
 
Replace Numbers with letters - Maths Puzzle

Process of decryption is here! 
 

Decrypting The CryptArithmatic Problem


What was the problem? 

Let's take a look at the equation once again.
       S  E N D
  +   M O R E
  -----------------
 = M O N E Y
 
Now letter M must be representing the carry generated & it must be 1. 
 
And if M = 1 then S must be 9 or 8 if carry is generated from hundreds place. 
In any case, O can be either 0 or 1. But can't be 1 as M = 1 hence O = 0.
 
If O = 0 then E + 0 = N i.e. E = N if there is no carry from tens place. 
Hence, N = E + 1.
 
Let's turn towards tens place now. With no carry from units place N + R = 10 + E .
Putting N =  E + 1, we get, R = 9. And with carry from units place, we have,
1 + N + R = 10 + E, gives us R = 8. Hence either R = 9 and S = 8 or R = 8 or S = 9.
 
For a moment, let's assume R = 9 and S = 8 with no carry from units place, then, 
 
   8 E N D
+ 1 0 9  E
=========
1 0 N E Y 
 
For this to be correct, we need carry at thousands place generated from hundreds 
place. That's only possible if E = 9 and carry is generated from tens place
forwarded to hundreds place. Since 9 is already being used for R this combination
is just impossible.
 
Hence, R = 8 and S = 9 with carry 1 from units place.
 
Now it looks like,
 
   9 E N D
+ 1 0 8 E
========
1 0 N E Y 
 
This means D + E >= 10 i.e. D + E = 10 + Y.  And numbers left are 2,3,4,5,6,7.
 
If E = 2 then D must be 8 or 9 for D + E >= 10.
Since 8 and 9 already taken, this is not possible.
 
If E = 3 then D can be 7 but Y would be 0 in the case. 
Since O = 0 already taken this value of E is also not valid. 
For any other value of D, D + E < 10 .
 
If E = 4 then D = 7 or 6 and Y = 0 or 1.
Both are taken hence this value of E in invalid.
Also,D <= 5 in the case gives  D + E < 10 .
 
If E = 6, N = 7 then, D <= 5. 
With D = 5 or 4, Y = 0 or 1, both are used for O and M already.
And for D = 2 or 3,D + E < 10 .
In short, this value of E is also not valid.   
 
So only value of E left is 5. Hence, N = 6 and D = 7. That gives, Y = 2.
 
Maths Puzzle -  Correct numbers for letters
 
To conclude,
 
  9 5 6 7
+ 1 0 8 5
=========
1 0 6 5 2 

Burn The Strings

You have two strings whose only known property is that when you light one end of either string it takes exactly one hour to burn. The rate at which the strings will burn is constant and strings are identical.

How do you measure 45 minutes?


That's how to measure 45 minutes!

Timer Using Burning Strings


What was the challenge? 

1. Burn both the ends of 1st string and 1 end of other string.

2. After 30 minutes, 1st string will be burnt out and half of the other string will be burnt out.So far we have counted 30 minutes.

3. Now burn the other end of 2nd end. It will take 15 minute to burnt out totally. This way, we have counted 30 + 15 = 45 minutes using 2 strings.

Logical Puzzle of burning strings
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