Elitmus
Exam
Logical Reasoning
Cryptography
i think this was the question
AGE
* RAM
----------------------
ASGR
SPBX
EXAS
-------------------------------------
EAGRBR
where S=1
Read Solution (Total 9)
-
- in last row there should be P in place of B.
463
*749
_____
4167
1852
3241
_____
346787 - 10 years agoHelpfull: Yes(11) No(5)
- A G E
R A M
------
ASGR
SPBX
EXAS
-------
EAGRPR
Here in final value B REPLACE P then its solve
we know s=1
so,
A G E
R
------
EXAS
R*E=s means S-1,so R=E=3,7 and also multiply E*M=R,if suppose we take R=7 and E=3 then S=1 and also in first row E*M=end with 7 then M=9,so we conform R=7,E=3,M=9
now first row G*M=end with G same number also carry generated from previous number,so i have carry 2,if G=6,then 6*9=54+carry 2=end with 6,so we have G=6,
M*A=end with S means 1,carry 5,so M(9)*A(4)=36+carry 5=41,so this line satisfy,A=4,
now
all value we know that
463
749
-----
4167
1852
3241
-----346787 - 10 years agoHelpfull: Yes(9) No(1)
- @aman And @Dilip both r wrng because ...
AGE
*RAM
----------
ASGR
SPBX*
EXAS**
-----------
EAGRBR
Both of u r given that B=5 in (SPBX) and B=8 in (EAGRBR)..
so how it possible ????
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- without using s=1 we can solve this ques,
the first thing m*g =g,but here any no will be add as carry which comes throug m*e.......so the m=9 g=6 and e=3
m*e=9*3=27 ,,,(here 2 as carry)(eq...1)
m*g=9*6=54...(54+2=56 )
there for we get the similar g here as 6
then we have to find a which is 4
we have to find the result that m*a=as,,the second digit is same
so m*a=9*4=36..(36+5=41 ..where 5 is carry when we multiply m*g)
from eq 1 we got the value of r as 7 ,,,
so we get all the values
463
749
-----
4167
1852
3241
--------
346787 - 10 years agoHelpfull: Yes(1) No(3)
- 237*329
if multiplication of two no. is m1
no. must be 3& 7
nothing else in this question - 10 years agoHelpfull: Yes(0) No(1)
- A G E
R A M
------
ASGR
SPBX
EXAS
-------
EAGRPR
Ans-346787 - 10 years agoHelpfull: Yes(0) No(0)
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