Delhi Police Constable Exam Model Paper 7

GIVEN:
Number = 6300
FORMULA USED:
If factors of ‘N’ are ${\mathit{a}}^{\mathit{p}}\times {\mathit{b}}^{\mathit{q}}\times {\mathit{c}}^{\mathit{r}}\times \dots \dots .\left(\mathit{a}=2\right),$ then total number of odd factors $=\left(\mathit{q}+1\right)\times \left(\mathit{r}+1\right)$
CALCULATION
Factors of 6300 $=2\times 2\times 3\times 3\times 5\times 5\times 7=2^2\times 3^2\times 5^2\times 7$
Total odd factors = $6300=\left(2+1\right)\times \left(2+1\right)\times \left(1+1\right)$
$=3\times 3\times 2=18$
Points to remember
1) If factors of ‘N’ are ap $\times \mathit{b}\mathit{q}\times \mathit{c}\mathit{r}\left(\mathit{a}=2\right),$ then total number of factors $=\left(\mathit{p}+1\right)\times \left(\mathit{q}+1\right)\times \left(\mathit{r}+1\right)$
2) If factors of ‘N’ are ap × bq × cr (a = 2), then total number of odd factors = (q + 1) × (r + 1)
3) If factors of ‘N’ are $\mathit{a}\mathit{p}\times \mathit{b}\mathit{q}\times \mathit{c}\mathit{r}\left(\mathit{a}=2\right),$ then total number of even factors $=\mathit{p}\times \left(\mathit{q}+1\right)\times \left(\mathit{r}+1\right)$