Mistakes give rise to Problems- 13

Algebra Level 4

In maths, we do $a\times b=ab$ . But if you do that while there is the log function, $\color{#D61F06}{\log (a) \times \log (b) = \log (a\times b)}$ then that will be a big mistake!

But for some pairs of integers $(a,b)$ , for which $\log (a)$ and $\log (b)$ are also integers, the above property is true. Find the sum of all $a$ and $b$ in these pairs.

If you get $n$ pairs $(a_1,b_1),(a_2,b_2),\ldots,(a_n,b_n)$ , then answer should be reported as $\displaystyle \sum_{k=1}^n \biggl(a_k+b_k \biggr)$

Details and assumptions :

Assume we take the log in base 10.
We only consider $a$ and $b$ as integers, and so are $\log(a)$ and $\log(b)$

This problem is a part of the set Mistakes Give rise to problems .

The answer is 202.

1 solution

Aditya Raut
Aug 7, 2014

If you call $\log (a)= x$ and $\log (b)=y$ , problem is actually to find the number of $\textbf{integer solutions}$ of $xy=x+y$

because we know that $\log (a) + \log (b) = \log (a\times b)$ , means $RHS$ is equal to $x+y$ .

This is simple to do,

$xy=x+y$

$xy-x-y=0$

$xy-x-y+1=1$

$(x-1)(y-1)=1$

This has integer solutions only at $x=y=0$ and $x=y=2$

Thus we have $\log (a)= \log(b)=0$ or $\log (a)=\log (b)=2$

This give us $a=b=10^0$ or $a=b=10^2$

Giving the pairs $(a,b)= (1,1) , (100,100)$ and hence the answer is $\boxed{202}$

Really? Level 4? Should be Level 2.

Sharky Kesa - 6 years, 7 months ago

Did it the same way...

敬全钟 - 6 years, 10 months ago

I know that the value of 5,4are less than 1 but if take log 20 and log9 are not equal.hence i found that LHS=RHS. am i right. pl comment.

amar nath - 6 years, 10 months ago

Nobody asked $\log(a+b)$ ? where comes the question of $\log(9)$ ?

And FYI , $\log(a) \times \log(b) \neq \log(a+b)$ , so no1 out of $LHS$ and $RHS$ will become $\log(9)$

Aditya Raut - 6 years, 10 months ago

suppose log 5xlog4=log5+log4 on left hand side1.3010=R.H.S=1.3020 log20=log5+log4 are same. please comment

amar nath - 6 years, 10 months ago

Well. even without calculation, i can tell your comment is wrong. $\log (4) <1$ and also $\log (5) <1$ And hence LHS is $\log (4) \times \log (5)$ always less than 1. And $\log(20)>1$ obviously. You are wrong with your calculations.

And $\log (a) + \log(b) = \log(ab)$ is an identity , what you saw is approximate values. But actually, they are equal.

Aditya Raut - 6 years, 10 months ago

I am the biggest fool. I tried this prob, did it correct added the no's up to 202 and typed 402.... Kept wondering why my answer got wrong and I couldn't even notice it!! Revisited the prob today,and got it right I felt like crying for loosing my ratings...Aargh

Pankaj Joshi - 6 years, 10 months ago

Mistakes give rise to Problems- 13

This problem is a part of the set Mistakes Give rise to problems .

The answer is 202.

1 solution

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