# 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, \[\color{Red}{\log (a) \times \log (b) = \log (a\times b)}\] 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 \(b\) asintegers, and so are \(\log(a)\) and \(\log(b)\)