Elitmus
Exam
Numerical Ability
Log and Antilog
(logA + logB +logC)/log6=6; find how many possible solution exist.
Read Solution (Total 10)
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- Rajnikanth the way u done is correct but
log(A*B*C) = 6*log 6 then log(A*B*C) = log(6^6)
so A*B*C = 6*6*6*6*6*6
so no of possibilities are
(6*6,6*6,6*6) = 1
(6*6*6,6*6,6) = 6
(6*6*6*6,6,6) = 3
so total 10 ways i think - 10 years agoHelpfull: Yes(32) No(2)
- (log A + log B + log C)/log 6 = 6
=>(log (A*B*C))/log 6 = 6
=>(log (A*B*C))=6*log 6
=>(log (A*B*C))=log 36
Therefore A*B*C=36
So no.of possible solution will be 8
1*1*36
1*2*18
1*3*12
1*4*9
1*6*6
2*2*9
2*3*6
3*3*4 - 10 years agoHelpfull: Yes(10) No(36)
- (logA+logB+logC)/ log6 =6
log(ABC)=6*log6
ABC=6^6
we can factorize 6 as (1*2*3) or(1*1*6)
case 1: when solutions are different (at least two)
possible combinations (1,2,3)= 6 solutions
or (1,1,6)= 3 solutions
case 2: when all solutions are equal i.e. (A*B*C)=(6*6, 6*6, 6*6)= 1 solution
hence total no. of solutions= 6+3+1= 10
- 10 years agoHelpfull: Yes(10) No(2)
- 6*log6=36.....????
- 10 years agoHelpfull: Yes(3) No(3)
- Ans will be 14! / (6! 6! 2!)
Explanation:
ABC = 2^6 * 3^6
The question now boils down to dividing 6 no. of 2's and 6 no. of 3's among A, B & C.
The arrangement is equivalent to possible arrangement of followings -
0 0 0 0 | 0 0 * * | * * * * (this is similar to integral solution question i.e. A + B + C = 12 etc.)
No. of possible arrangement is 14! / (6! 6! 2!) - 9 years agoHelpfull: Yes(2) No(1)
- Rajnikant bhai.sachmuch rajnikant ban gaya..6log6= 36....how is it possible rajnikant jee??
- 9 years agoHelpfull: Yes(1) No(1)
- I have a query:
Why we considering only combinations in powers of 6
Can we write A*B*C as 2^6 * 3^6
So we can also have A = 1, B = 2 , C= 2^5*3^6?????
And so we can form different combinations with distributed powers of 2 and 3. - 9 years agoHelpfull: Yes(1) No(3)
- A*B*C = 12
A,B,C are different numbers so > 1*4*3 = 12 , 2*6*1 = 12, and both can be arranged by 3 ways so ans must be 6. - 10 years agoHelpfull: Yes(0) No(7)
- log(A*B*C) = 6*log 6 then log(A*B*C) = log(6^6)
so A*B*C = 6*6*6*6*6*6
no of ways of dividing 6 sixes among 3 persons a,b,c is n+r-1 C r-1
6+3-1 C 3-1==>8C2==>28
so 28 solutions are possible
- 8 years agoHelpfull: Yes(0) No(1)
- log(A*B*C)=6*log(6)
=log(6^6)
=log(6*6*6*6*6*6)
A*B*C=(2^6)*(3^6)
if a^n * b^m then no of solution is (n+1)*(m+1)
so here ans is 7*7=49
no of solution =49 - 7 years agoHelpfull: Yes(0) No(0)
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