Can you simplify this otherwise tough problem??

Algebra Level 4

The number of distinct $4$ -tuples of rational numbers $(a,b,c,d)$ satisfying $a \log 2+ b \log 3 + c \log 5 + d \log 7=2012$ is equal to $k$ . Enter your answer as the sum of $k$ and all possible values of the $4$ -tuples.

Details : Logarithm is taken to base $10$ .

Example : If $k=2$ and the solutions are $(1,2,3,4), (5,6,7,8)$ then your answer should be $(1+2+3+4)+(5+6+7+8)+2.$

The answer is 4025.

1 solution

A Former Brilliant Member
Sep 16, 2014

This can be written as $\log\left(2^a3^b5^c7^d\right)=2012$ , which can be written as $2^a3^b5^c7^d=10^{2012}$ or $2^{2012}\times 5^{2012}$ , by comparing we get $a=c=2012$ and $b=d=0$ and $k=1$ , their sum is $2012+2012+0+0+1=4025.$

I did the same way! :)

Abhijith Asokan - 6 years, 9 months ago

If 3^b and 7^d can be some numbers or b and c can be some number other than 0, then the whole will come as irrational, this also need to be proven.

Kartik Sharma - 6 years, 8 months ago

a,b,c,d are already given to be rational!!

A Former Brilliant Member - 6 years, 8 months ago

if a=a1/a2,b=b1/b2,c=c1/c2,d=d1/d2, where a1,a2,b1,b2,c1,c2 are integers (a1,a2)=1,(b1,b2)=1,(c1,c2)=1,(d1,d2)=1, so we can transform to 2^a1 x 3^b1 x 5^c1 x 7^d1=10^(2012 x a2 x b2 x c2 x d2),so must be b1=0,d1=0,a1=2012a2,b1=2012b2 which means b=0,d=0,a=2012,b=2012 is unique solution and k=1. So a+b+c+d+k=2012+2012+1=4025=25x7x23

Nikola Djuric - 6 years, 6 months ago

I did the same way .

Since the base of log is 10, its only factors are 2 and 5. Hence a=c. 3 and 7 have to be eliminated. This is possible if they are 1 in the multiplication, that is exponents are 0,0...........

Niranjan Khanderia - 6 years, 8 months ago

Can you simplify this otherwise tough problem??

The answer is 4025.

1 solution

0 pending reports