Latest Elitmus Aptitude Question SOLUTION: W H Y * N U T ------------ O O N P O Y P Y O U H A ------------------- O N E P O P

The basic logic for solving crypt arithmetic questions can be put in three steps

(1) Understanding the hints from addition
(2) understanding hints from last digits of multiplication
(3) understanding hints from first digits of multiplication

(1) One can track 9's and 0's and carries by spending a few seconds on additions. But please understand we can track it every time, for example there are no hints of addition in this particular example.
(2) understanding logics of multiplication by last digit
This is simply observing what is happening in multiplication tables
I am just listing some observations for explanation sake
There is no 3 ending in 3 table, similarly no 7 ending in 7 table and no 9 ending in 9 table
The entire 5 table ends with 5's or 0's
2,4 and 8 when multiplied by 6 ends with 2, 4 and 6 respectively

So, as far as multiplication is concerned first check for 1's and 0's, which you rarely get for ELITMUS level
Second check for 5's and 6's, which can be found a lot of times in ELITMUS

Now lets see how to implement that,
In the second row,
W H Y * U = O Y P Y
As both of them are ending with Y, this can happen only with 5 or 6( as explained above)
So, either Y must be 5 or U must be 6
Y cannot be 5 for the reason that the entire 5 table ends with 5's and 0's, but if we observe the problem
WHY * T is ending with P
WHY * U is ending with Y
WHY * N is ending with A

So, Y cannot be 5, so U must be 6

Now lets move to the last step, checking the hints from the first digits of multiplication

All the three results are starting with O's
If we can the multiplication tables this will happen in 2 table, lot of results start with 1 and in 3 table where it starts with 1 and 2 and in 4 table. But it will not happen after 4 table

Whatever the Hints, Cryptarithmetic always involves a little bit of trail and error, here the trail and error is O can be 1 or 2, we have to try that

All the hints try to reducee the amount of trail and error, these steps has resulted zero trail and error in some particular problems

I hope it helped, if any further doubts are there post it

2 3 4
*
7 6 5
-------
1 1 7 0
1 4 0 4
1 6 3 8
-------------
1 7 9 0 1 0
------------

2 3 4
*
7 6 5
-------
1 1 7 0
1 4 0 4
1 6 3 8
-------------
1 7 9 0 1 0
------------

check this:
http://engineerscircle.files.wordpress.com/2012/12/elitmus-guide_-guide-to-solve-cryptic-multiplication-question.pdf

step 1: go for 2nd multiplication means (WHY*U)
now u ll observe Y*U=y, means tat after multiplying any number with y in
return u r getting y only.so we hv only 4 sets of logic which fits over
here. i.e(Y*U=Y)
1.2*6=12(in this last digit represent y)
2.4*6=24(")
3.5*3=15(")
4.8*6=48(") try with this logic one after 1. but second will give perfect combination(4*6)so put Y=4 nd U=6.
step 2:take O as 1. as hit nd trial.

mahesh please xplain how u have taken the numbers?

can anybody help me .... what is the basic logic to solve these type of question
????

Sorry question is wrong.....