Elitmus
Exam
Logical Reasoning
Cryptography
Q Z F
R Q A
------------------
S A X V
S W F S
X S A X
- ----------------------
X X R Q Q V
Read Solution (Total 12)
-
- Its very simple.... its clear that S cant be more than 4 as it will lead to a carry which is not feasible...
so start putting values for S=1,2,3,4
For S=1
we can take Q=3 and F=7 or reverse as only multipication of 7 with 3 and 9 with 9 results to unit digit of 1
when s=1, x=2
so we have to take R in a way whose multipication with 7 results in a unit digit of 2 that is possible with 6 only so take R=6
Now here S=1 and X=2 so Q comes as 3
If you proceed with this u will find that addition of X and S is not generating any carry therefore to make unit digit of addition A, F and X as 3 we need to take A as 4
When A=4, V=8
Solve further u will easily get z=5 and w=0 - 9 years agoHelpfull: Yes(24) No(1)
- hint is to find out that s= 1,2,3,4, bcas we have s+s+carry=x
so by taking s=1 ill start my procedure.
nw we can have carry=0,1,2
so ill start with carry=0
with s=1 and carry=0
1+1+0=> x=2
we have s=1,x=2 now
and we have a hint here that x+s=Q
so 2+1=> Q=3
nowe have s=1 Q=3 x=2
in multiplication it is simple to solve from 2nd row bcas we have repetition of letter S
QZF*Q=SWFS
=>3ZF*3=1WF1
no to get last digit 1 we have to sub F=7
=>3Z7*3=1W71
=> no we must get F=7 on RHS
so go on substituting z=4,5,6...
at z=5 it will satisfy our equation
so => 357*3=1071
itsclear that z=5 and w=0
no we have Q=3, s=1,x=2, Z=5, W=0, F=7
now we have in third column form right hand side that A+F+X=carry+ Q
=>A+7+2=carry+Q
=>9+A=carry+Q
=>to get Q=3 we must have A=4
=>so A=4
now ill take 1st row to get A
QZF*A=SAXV
=> 357*4=142V
=> from this equation to satisfy this we must have v=8
now we have v=8,A=4
now from 4th column from right hand side S+W+A+1(carry)= R
=> 1+0+4+1=6
soR=6
i think i got it now
357
* 634
---------
1428
1071
2142
----------
226338
----------
- 9 years agoHelpfull: Yes(10) No(1)
- 3 5 7
6 3 4
--------------------
1 4 2 8
1 0 7 1
2 1 4 2
--------------------
2 2 6 3 3 8 - 9 years agoHelpfull: Yes(7) No(0)
- 357
*634
---------------
1428
1071
2142
--------------------------
226338
- 9 years agoHelpfull: Yes(2) No(0)
- please explain it
- 9 years agoHelpfull: Yes(1) No(0)
- try to find clue..
nd here 2S + Carry= X
now carry, can be 0,1,2;
so, probable values of S=1,2,3,4;
and X=2,3,4,5,6,7,8,9;
now you can easily optimize the value of R,Q,F corresponding to each value of X and S;
lets assume when carry =0;
then S= 1,2,3,4
nd corresponding value of X= 2,4,6,8;
now suppose S=1; then X= 2;
now see the nest clue
R * F=X and
Q * F=S;
now if S =1;then Q=3 or 7 and F=7 or 3 so R=6 or 4;
now proceed with Q = 3,f =7 , R =6;
Q Z F
R Q A
------------------
S A X V
S W F S
X S A X
- ----------------------
X X R Q Q V
then you will find Z must be 5;
now the solution is:
3 5 7
* 6 3 4
---------------------------------
1 4 2 8
1 0 7 1
2 1 4 2
--------------------------
2 2 6 3 3 8
and the worst part is i cudnt solve it in the xam hall..but yes juss after in the bus i crackd it
- 9 years agoHelpfull: Yes(1) No(0)
- s
- 9 years agoHelpfull: Yes(1) No(0)
- do we get options in these type of problems?? plz help...
- 9 years agoHelpfull: Yes(1) No(0)
- S cant be more than 4,SINCE THERE IS NO CARRY(x=x)
now for S=1,X will be 2 and Q=3
NOW
WE HAVE 3ZF*3=1WF1 => so F must be 7
So- 3Z7*3=1W71 by checking we will get Z=5 and W=0
therefore 357*3=1071
now 357*R=21A2 SO must be 6
Hence 357*6=2142(A=4)
so finally-
357
*634
1428
1071
2142
226338 - 9 years agoHelpfull: Yes(1) No(0)
- 357
*634
-------------
1428
1071
2132
--------------
226338 - 9 years agoHelpfull: Yes(0) No(0)
- Q#3, Z#5, F#7 , R#6 ,A#4 ,S#1 ,W#0 ,X#2 , V#8.
- 9 years agoHelpfull: Yes(0) No(0)
- USE HIT AND TRIAL FOR F & A starting from 2 to 9 hint is
QZF
* Q
------
SWFS
-------
ANS:- 357
*634
-----------------------
1428
1071
2142
---------------
226338 - 9 years agoHelpfull: Yes(0) No(0)
Elitmus Other Question