Latest Elitmus Aptitude Question SOLUTION: F Y H E W T ___________ Z E T F E K Z Z Y Z T X ___________ Y F Y W E F

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(#M40034221) ELITMUS Logical Reasoning question Cryptography Keep an EYE

F Y H
E W T
___________
Z E T F
E K Z Z
Y Z T X
___________
Y F Y W E F

Asked In Elitmus Ankur Aggarwal (10 years ago)

Unsolved Read Solution (8)

Is this Puzzle helpful?   (11)   (6) Submit Your Solution Cryptography

Read Solution (8) :

amit kumar  (10 years ago)

see,FYH*T=ZETF.IT means Y*T=T.THIS HAPPENS ONLY WHEN T=3 AND Y=4(REMEMBER IT.AS I THINK).BCOZ H*T WOULD HAVE GIVEN A CARRY OF 1 AND THIS CARRY WOULD BE ADDED. i,e 4*3+1=13.
F4H
EW3
ZETF


now look ABOVE for H. TO GET A CARRY OF 1 FROM H*3,H CAN BE 5 OR 6. NOW 5 WILL CONTRADICT SINCE IT WILL GIVE F=5(AS H=5). SO H=6
F46
EW3
ZE3F
EKZZ
4ZTX
4F4WEF
NOW 6*3=18.SO B=8
WRITE ALL THE FOUND DIGITS AGAIN AND SOLVE..FROM HERE ON ITS EASY

Is this solution Helpfull? Yes (16) | No (3)  

Ankur Aggarwal  (10 years ago)

F=8
Y=4
H=6
E=5
W 7
T=3
Z=2
K=9
X=0

I have used hit & Trial, using F Y H * W = E K Z Z. It took around half hour to do it. Any other lead to solve this problem?

Is this solution Helpfull? Yes (11) | No   

tausif  (10 years ago)

The Answer is
846
573*
---------
2538
5922*
4230**
---------
484758

Is this solution Helpfull? Yes (2) | No   

varun indra  (10 years ago)

how y*t=t means y=4,t=3 ?
it could also mean that y=6 and t=2/4/8

Is this solution Helpfull? Yes (2) | No   

kumar sarvjeet  (10 years ago)

FYH*T=ZETF,=>Y*T=T, IN THIS CASE T=3 & Y=4(WHENEVER this type of case happens)
put H=6,BY HIT & TRIAL , we get F=8,
now putting these values, we get, X=0,Z=2,E=5,K=9,W=7,

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Abhishek kUNDU  (9 years ago)

from where did you get the value of alphabets ? please explain that one.

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Mohammad Danish Khan  (6 years ago)

for those who can't understand t is not 5 or 6
fyh*t=zetf in this let's see y*t=t. if t=5 then there should be no carry at tens place from units place so h=1 but h cannot be 1 as units place is h*t=f and if h=1 it becomes 1*t=f which doesn't satisfy.

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ROHAN KUMAR SRIVASTAVA  (6 years ago)

Since E+Z should not give carry, we have a possible value of F is 9,8,7,6,5,4,3.
(1)Let's first consider F=9 and solve.
FYH
xT
------
ZETF ..............(1)
------
the product of two no can only be 9 when we take 3*3=9 and 7*7=49.We cannot take these no in consideration because two letters cannot have same no.
So Fcannot be 9....
(2)Let's take F=8,and solve.
2*4=8,2*9=18,3*6=18,4*7=28,6*8=48.
(a)considering first(2*4=8)
T=2 and H=8
putting this in equation (1).
we get,Y=6,Z=1,E=7.
FYH
xW
----
EKZZ .................(2)
----
Z=1 while having H=8(we cannot get such combination) is not possible from equation(2).
same is the case with H=2,T=8.
(b)Now,considering second(2*9=18)
putting this in equation (1) we cannot get any soution from equation(you can check).
(c)Now,lets consider third(3*6=18)
ie.T=3,H=6.
putting this in equation(1) we get,
Y=4,Z=2,E=5.
using these results in equation(2) we get....
W=7,k=9.
using these all solution in the question we get..
F=8,Y=4,T=3,H=6,Z=2,E=5,W=7,K=9,X=0 ANS.

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