1) Two archaelogists, English and a French man come across a rock which has four letters carved on it. The english man thinks its an english word and the french thinks its a french word the clue for the word are as follows....
a) each alphabet has a number associated with it....for eg:a=1, b=2, c=3........ z=26.....
b) the sum of the 4 nums equals 20
c) the sum of any 3 nums of these 4 numbers, divided by the fourth one leaves a remainder zero....eg: if abcd is the word.....then a+b+c/d ,b+c+d/a ,c+d+a/b, d+a+b/c all leave a remainder zero.
Find the 4 letters (if possible the English word) with the help of the above clues...
3 comments:
Is the word JADE ?
I will just give my thought process. I am not sure if this is the most efficient one... Do let me know if there are other/better ways of doing this...
Since sum of any 3 is divisible by the 4th number, the sum of any 3 MUST be >= 4th number. Otherwise, it will not be divisible.
So, the highest possible number is 10. If there is any number above 10, the sum of the remaining 3 will be less than 10, and hence not divisible by 10.
So, from 1 to 10,
9 cannot be one of the numbers since the sum of remaining = 11,which is not divisible by 9.
8 cannot be one of the numbers since the sum of remaining = 12,which is not divisible by 8.
7 cannot be one of the numbers since the sum of remaining = 13,which is not divisible by 7.
6 cannot be one of the numbers since the sum of remaining = 14,which is not divisible by 6.
3 cannot be one of the numbers since the sum of remaining = 17,which is not divisible by 3.
That leaves us with 1, 2, 4, 5, 10
only 1, 4, 5, 10 make the total of 20... and also fulfils the other condition of divisibility.
These numbers correspond to A, D, E, J.
Couldnt find anything apart from JADE.
And is Deja the french word ?
Your answer is right..My logic was since a+b+c is divisible by d therefore (a+b+c)/d +1 = 20/d (a+b+c+d = 20 so dividing both sides by d, thats true for any one a,b or c). This means each number a,b,c & d is a factor of 20.
So there are only 6 numbers less than 20 which are its factors. 1,2,4,5,10,20. 20 is ruled out. So essentially it is just 1,4,5,10 which would add up to 20 and also would satisfy the divisibility criteria.
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