Elitmus
Exam
Logical Reasoning
Cryptography
H M K
* A V E
------------------------
A N X X
X AV H
M X V W
------------------
M A M V W X
Read Solution (Total 6)
-
- First we will see that which alphabets can be 0 or not.
(A,X,V,H,M,K,E)!=0 ; (W & N )=0,
Now , N can't be 0 bcoz E*K=X and E*M=X can't be possible it will definitely gives a carry. so N can't be 0 after adding something .
So, W==0.
Now, In the equation K*A=W, either K or A should be 5 to make W=0;
If we take K as 5 then in all the equation K*E ,and K* V always there will be 5 and 0 will be the result which is not possible.
So, A will be 5 for sure. And K will be 2,4,6 or 8.
As we can see that M=M in the last so X+X+A will not give any carry and A is 5. It can be only possible when X will be 2 so that the equation will be 2+2+carry(1) will be 5.
Now for the equation K*E=X; We will put 4 bcoz K now will be only 4,6 & 8.
So the equation will be 4*3=12 to make X =2. But it will not suite the whole equation multiple with because it will not give M a correct value so that X will come 2.
Now we will take K =6.
Now. the equation will be 6*7=42. The next equation M*E+carry=X as (4*7+4=32)
and the next equation H*E+carry=AN will be(8*7+3=59).
Now start putting all the values . The remaining alphabets will be clearly come out with an another value.
So our result be like as
8 4 6
*5 3 7
-------------------
5 9 2 2
2 5 3 8
4 2 3 0
------------------
4 5 4 3 0 2
- 9 years agoHelpfull: Yes(17) No(1)
- 846
*537
--------
5922
2538
4230
--------------
454302
- 9 years agoHelpfull: Yes(4) No(3)
- Just look the Double X in the equation ok
H M K
* A V E
------------------------
A N X X
X AV H
M X V W
------------------
M A M V W X
first write it as
H M K
* E
---------------
A N X X
just start substuting the values of X .. Start with 1 ok so to make X =1 we have to take E=7 and k =3 so we have 7(E)*3(k)=21 .2 will be carry and we have X=1 . Now we have one more X so we have the value of E=7 and we have to multiply with M so we can find a value that ends with 1 after adding the 2 (Carry from E*K=21). But if u go all the values from 0-9 you don't find any value for it.
so nw we will take X=2 .To make X=2 we take the value of E=2 and K=6 we get 2(E)*6(k)=12 we get X=2 but condition fails as we have E=2 we can't have the two alphabets with the same number .
after you go through some value u will find for E=7 and K=6 we get 42 ... 2 as the X and 4 carry .Now to make E*M =X we take M=4 .. then we will get E(7)*M(4)+ carry(4)=32 so X=2 and 3 will be carry.
nw replace all the X in the equation by 2.
H 4 6
* A V 7
------------------------
A N 2 2
2 AV H
M 2 V W
------------------
4 A 4 V W 2
we can see by adding 2+2=A(4)... but it cannot be 4 as we have M=4 .So there must be carry from previous addition.So lets tale A=5.Now start as
H 4 6
*5 V 7
---------------------
4 2 V W
Now 5(V)*6(k)=30.... W=0 and 3 will be carry now 5(A)*4(M)+carry=23 ..V=3 and 2 will be carry and then 5(A)*H=42 ... so the 8 will be the right value to get 42.
Now you can take down all the values .
- 9 years agoHelpfull: Yes(4) No(0)
- first of all split the question in three part
1) HMK
* E
------------
ANXX
---------
2) HMK
* V
----------
XAVH
----------
3) HMK
*A
-----------
MXVW
-----------
from part 1 result xx(check as 11,22,33,44)
if you think 11 then possible value for that is (3*7) or (7*3)
for 22 possible values are (2*6) ,(6*2),(4*3),(3*4),(8*4),(4*8),(7*6),(6*7)
take value for K=6 and E=7
so ,
846
*537
-----------
5922
2538
4230
-----------
454302
---------- - 9 years agoHelpfull: Yes(1) No(0)
- 846
*537
---------
5922
2538
4230
-----------
454302
- 9 years agoHelpfull: Yes(0) No(0)
- 846
537
______
5922
2538
4230
______
454302 - 9 years agoHelpfull: Yes(0) No(0)
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