Fair Distribution Of Water

In Sahara desert , 3 men found a big 24L Jar is full of water. Since there is shortage of water so they decided to distribute the water among themselves such that they all have equal amounts of it. But they only have a 13L, a 5L and an 11 liter Jar.

How do they do it? 

Here is how to do it!

Source 

Equal Distribution Of Water

What was the challenge?

1. First pour 24L into 13L and 11L jar.There will be no water in 24L jar.

2. Now pour 13L jar into 5L jar till 5L is filled. So 8L of water will be left in 13L jar.

3. This 8L of water from 13L jar is emptied in empty 24L jar.This will leave 13L jar empty.

4. Now pour 11L water from 11L jar into 13L of jar. There is still space for 2L of water in 13L jar.

5. Pour 5L jar into 13L of jar which had space for 2L only. So 3L of water will be left in 5L of jar.

6. This 3L of water is emptied in 11L jar.

7. A 13L jar full of water is again poured into 5L jar leaving behind 8L of water in it.

8. A 5L jar full of water is finally poured in 11L jar already having 3L of water. 

9. This way, 24L, 11L and 13L of jar would have 8L of water each.

Another CryptArithmatic Problem

Replace letters with numbers assuming numbers can't be repeated. 

     SEND
  + MORE
  ----------
 = MONEY

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.

To conclude,

  9 5 6 7
+ 1 0 8 5
=========
1 0 6 5 2