Elitmus
Exam
Logical Reasoning
Cryptography
AIM
PRJ
------
KJJK
KIKN
KTKA
---------
KKNKIK
Read Solution (Total 12)
-
- 253
467
--------
1771
1518
1012
-----------
118151 - 10 years agoHelpfull: Yes(3) No(0)
- here we know that
k+t=k hence it means t=0;
we know that there is no carry at the k from the left end so it means
k+t - 9 years agoHelpfull: Yes(3) No(2)
- 253
*467
---------
1771
1518
1012
-----------
118151
- 10 years agoHelpfull: Yes(0) No(0)
- Plz Explain above solutions
- 10 years agoHelpfull: Yes(0) No(0)
- first see t+k =k means t =0 now start putting the value of k from 1 to 9 u will get the solution
- 10 years agoHelpfull: Yes(0) No(1)
- can u clearly explain this
- 10 years agoHelpfull: Yes(0) No(0)
- First take KJJK and verify which replaces J and K . It can be easily known by starting from 1. For every number of J , K must ends in less than or equal to J-1. By this we can easily find KJJK as 1771. and we already know that T =0.
- 9 years agoHelpfull: Yes(0) No(0)
- in above problem j+k+A+c1=k where c1 means carry; and 1 st row j*a=k_
so j+a+0/1=10
j can take values A can take values
j------6 means a------4 /3 j*A= 6*4 = 2__
j-------7 means a------3/2 7*3=2__ 7*2=1___
j-------8 means a--------2/1 8*2=1_
k can take values either 1 or 2;
putting k=1 and verifying we can get the ans; - 9 years agoHelpfull: Yes(0) No(0)
- 253
467
------------
1771
1518
1012
-------------
108151
here- K+T=K ( from this T=0)
put T =0 in place of T and after that put the value between 1 to 9 in place of K and check for it by hit and trial method. - 9 years agoHelpfull: Yes(0) No(0)
- 2 5 3
*4 6 7
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1 7 7 1
1 5 1 8
1 0 1 2
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1 7 7 0 8 - 9 years agoHelpfull: Yes(0) No(0)
- sory printing mistake in previous on
2 5 3
* 4 6 7
--------------
1 7 7 1
1 5 1 8
1 0 1 2
------------
1 1 8 1 5 - 9 years agoHelpfull: Yes(0) No(0)
- a-2,i-5,m-3,j-7,r-6,p-4,k-1,n-8,t-0
- 9 years agoHelpfull: Yes(0) No(0)
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