Find the highest power of 12 in 10² × 11² ×12² × 13² ×..........× 30² × 31²

(Q). Find the highest power of 12 in 10² × 11² ×12² × 13² ×……….× 30² × 31².

Solution:

10² × 11² ×12² × 13² ×……….× 30² × 31²
= (10 × 11 × 12 × 13 ×……….× 30 × 31)²

= ${\left( {\frac{{31!}}{{9!}}} \right)^2}$
so we have to find the highest power of 12 in ${\left( {\frac{{31!}}{{9!}}} \right)^2}$

now ⇒ 12 = 2² × 3
⇒

∴ Highest power of 2 in $\frac{{31!}}{{9!}}$ = 26 – 7 = 19
Highest power of 2 in ${\left( {\frac{{31!}}{{9!}}} \right)^2}$ = (19) × 2 = 38
Highest power of 2² in ${\left( {\frac{{31!}}{{9!}}} \right)^2}$ = $\frac{{38}}{2}$ = 19

Now highest power of 3 in

∴ Highest power of 3 in $\left( {\frac{{31!}}{{9!}}} \right)$ = 14 – 4 = 10
Highest power of 3 in ${\left( {\frac{{31!}}{{9!}}} \right)^2}$ = 10 + 10 = 20

Hence highest power of 12 in ${\left( {\frac{{31!}}{{9!}}} \right)^2}$
= Minimum of (Highest power of 2² in ${\left( {\frac{{31!}}{{9!}}} \right)^2}$ & Highest power of 3 in ${\left( {\frac{{31!}}{{9!}}} \right)^2}$ )

= 19 Answer

Find the highest power of 12 in 10² × 11² ×12² × 13² ×……….× 30² × 31²

Submit a Comment Cancel reply

Recent Posts

Recent Comments