Latest Elitmus Aptitude Question SOLUTION: (logA + logB +logC)/log6=6; find how many possible solution exist.

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(#M40052223) ELITMUS Numerical Ability question log Keep an EYE

(logA + logB +logC)/log6=6; find how many possible solution exist.

Asked In Elitmus jitu (9 years ago)

Unsolved Read Solution (10)

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Read Solution (10) :

arun  (9 years ago)

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

Is this solution Helpfull? Yes (32) | No (2)  

Rajnikant  (9 years ago)

(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

Is this solution Helpfull? Yes (10) | No (36)  

Mithun Kumar Kesari  (9 years ago)

(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

Is this solution Helpfull? Yes (10) | No (2)  

Amit kumar rana  (9 years ago)

6*log6=36.....????

Is this solution Helpfull? Yes (3) | No (3)  

anu radha  (8 years ago)

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!)

Is this solution Helpfull? Yes (2) | No (1)  

Abhishek kUNDU  (9 years ago)

Rajnikant bhai.sachmuch rajnikant ban gaya..6log6= 36....how is it possible rajnikant jee??

Is this solution Helpfull? Yes (1) | No (1)  

Nupur  (9 years ago)

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.

Is this solution Helpfull? Yes (1) | No (3)  

swati  (9 years ago)

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.

Is this solution Helpfull? Yes | No (7)  

krishna mudhiaj  (7 years ago)

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

Is this solution Helpfull? Yes | No (1)  

pinki kumawat  (7 years ago)

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

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