You visited us 1 times! Enjoying our articles? Unlock Full Access!

Binomial Theorem for Any Index

If 10m divide...

Question

If $10^{m}$ divides $101^{100} - 1$ , then the greatest value of $m$ is

Open in App

Solution

$101^{100} - 1 = (1 + 100)^{100} - 1 = 1 +^{100} C_{1} (100) +^{100} C_{2} (100)^{2} + \dots +^{100} C_{100} (100)^{100} - 1 =^{100} C_{1} (100) +^{100} C_{2} (100)^{2} + \dots +^{100} C_{100} (100)^{100} = 100^{2} (1 + k) = 10^{4} (1 + k)$
Where $k$ is multiple of $10$

Hence, the greatest value of $m$ is $4$

Suggest Corrections

Similar questions

10^{m}

101^{100} - 1

m

101^{100} - 1

(- 24)^{- 1}

- \frac{1}{m}

3^{- 1}

m

{(1 + x)}^{n} = n \sum r = 0 a_{r} x^{r} a n d b_{r} = 1 + \frac{a_{r}}{a_{r - 1}} a n d n \prod r = 1 b_{r} = \frac{{(101)}^{100}}{100!}

f (x) = \frac{1}{4} x^{4} - \frac{2}{3} x^{3} - \frac{3}{2} x^{2} + 2; x \in [- 2, 4]

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program