Question

If $N=$ $2^a\times 3^b\times 5^c$, how many numbers(in terms of $N$) are less than $N$ and are co prime to it?

• A. $\frac{2}{15}$ $N$
• B. $\frac{4}{15}$ $N$
• C. $\frac{8}{15}$ $N$
• D. $\frac{2}{5}$ $N$

Answer 1

The correct option is B $\frac{4}{15}$ $N$

$N[1-\frac{1}{p}][1-\frac{1}{q}][1-\frac{1}{r}]$----,
where $N=p^a{q}^b{r}^c$-----& $p,q,r$ are primes.
$N=2^a{3}^b{5}^c$
$=N[1-\frac{1}{2}][1-\frac{1}{3}][1-\frac{1}{5}]$
$=N\times \frac{1}{2}\times \frac{2}{3}\times \frac{4}{5}$
$=\frac{4}{15}N$