Elitmus
Exam
Logical Reasoning
Cryptography
A P R
*O C T
-------------
P U R A
R O J R
R E C U
-------------
R A A J A A
What is the value of E ?
Read Solution (Total 13)
-
- E = 1
A P R
*O C T
-------------
P U R A
R O J R
R E C U
-------------
R A A J A A
==> R+R =A ; 2R=A
R+E+1=A ; R=e+1
U+U+J=J ; U= 0 or 5
U=0
APR
*O
----
RECU
ie
432
* 5
----
2160
====>E=1
APR
* T
----
PURA
432
* 7
----
3024
APR
* C
----
ROJR
432
* 6
----
2592
APR*OCT=RAAJAA
432*567 = 244944 - 10 years agoHelpfull: Yes(15) No(1)
- Its 432*567
We can simply judge as C*R=R but R+R=A hence R cannot be 5
Hence we have to start from 6*2 =_2 - 10 years agoHelpfull: Yes(3) No(1)
- i am not understanding it.. where can i get tutorial for such questions.. pls give the links in comment...
- 10 years agoHelpfull: Yes(2) No(0)
- E=1
APR
*OCT
-------
PURA
ROJR
RECU
-------
RAAJAA
HERE R+R=A=>2R=A then A must be{1,2,3,4}and R*C=R THEN R=2,C=6,A=4
4P2
O62
-----
PU24
2OJ2
2E6U
------
244J44
here 2+E+1=4 THEN E=1
SO ANS IS 1
- 10 years agoHelpfull: Yes(1) No(0)
- http://cryptarithmetic.wordpress.com/
- 10 years agoHelpfull: Yes(1) No(0)
- the value of E is 1.
- 10 years agoHelpfull: Yes(1) No(0)
- you can get such type of problems on cryptopress arithmetic problems by searching on google
- 10 years agoHelpfull: Yes(0) No(0)
- @ANN THERESSA, How u hav solved plzzz explain properly
- 10 years agoHelpfull: Yes(0) No(0)
- 432
*567
is ans nw solve for rest value - 10 years agoHelpfull: Yes(0) No(0)
- 432
*567
------
3024
2092
2160
------
244944 - 10 years agoHelpfull: Yes(0) No(0)
- 432
567
................
3024
2592
2160
................
244944 - 10 years agoHelpfull: Yes(0) No(0)
- 4 3 2
5 6 7
---------------
3 0 2 4
2 5 9 2
2 1 6 0
--------------------
2 4 4 9 4 4
--------------------- - 9 years agoHelpfull: Yes(0) No(0)
- first let R=2 if R=2 then A =4 then in first row
A P R * T =P U R A since R*T=A then it is clear that P=3
and proceed to find all the values likewise then
432
567
-----------
3024
2592
2160
------------
244944
is the correct answar. - 8 years agoHelpfull: Yes(0) No(0)
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