Elitmus
Exam
Logical Reasoning
Cryptography
G A E
M R A
-----------------------------------------------------
X B R X
A B K S
B R P R
-------------------------------------------------------
G X G X P X
Read Solution (Total 7)
-
- GAE
MRA
----------
XBRX
ABKS
BRPR
----------
GXGXPX
1) A & R take values between (1,2,3,4...9)
2) Hence we can conclude that maximum carry that can be generated will be 1.
So,
B+1=G
3) Now, M*G=B_
4) We form the table for B & G
B|G
-----
1|2
2|3
3|4
5|6
6|7
5) We check all possible combination with the help of equations in 2 & 3.
6) For B=5, M&G should be high, hence we choose M=9,G=6.. which also fits the condition mentioned in the table.
7) We form other equations like
A+R=X-----(1)
X+B+P=G---(2)
B+K+R=X---(3)
R+S=P-----(4)
keeping a CARRY of 1 or 2 in mind.
8) As mentioned in point 1 that carry 1 is generated...
So, values of A+R can be...11,12,......,19.
keeping a Carry of 1 or 2 in mind we select the values of A & R, which comes to be 4 & 7.
9) now its easy to figure out values by placing the values we have got in proper place. - 10 years agoHelpfull: Yes(7) No(10)
- 643
*974
-----
2572
4501
5787
--------
626282 - 10 years agoHelpfull: Yes(5) No(4)
- 6 4 3
9 7 4
.................................
2 5 7 2
4 5 0 1
5 7 8 7
...........................
6 2 6 2 8 2
** **
here no logic
assume s=1, bcz here we will get more value i.e R & E values
only possiblity 7 * 3 or 3 * 7
then R=7,E=3 and p=8
then, m=9 and A will become 4 then x=2
and G =6 then B will be 5 - 9 years agoHelpfull: Yes(2) No(1)
- 643*974=626282
- 10 years agoHelpfull: Yes(1) No(3)
- 643
974
------------
2572
4501
5787
------------
626282 - 10 years agoHelpfull: Yes(0) No(2)
- 643
x975
2572
4501
5787
626282 - 10 years agoHelpfull: Yes(0) No(4)
- 6 4 3
9 7 4
.................................
2 5 7 2
4 5 0 1
5 7 8 7
...........................
6 2 6 2 8 2 - 9 years agoHelpfull: Yes(0) No(1)
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