# Prove .....

Consider the following results.

$$99^{1} = 99$$

$$99^{2} = 9801$$

$$99^{3} = 970299$$

$$99^{4} = 96059601$$

$$99^{5} = 9509900499$$

Prove that $$99^{n}$$ ends in 99 for odd n.

Note by Abdulrahman El Shafei
3 years, 10 months ago

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$$99\equiv(-1)\pmod{100} \Rightarrow 99^{n}\equiv(-1)^{n}\pmod{100}.$$But,whenever $$n$$ is an odd integer$$, (-1)^{n}=(-1) \Rightarrow 99^{n}\equiv(-1)\equiv(99)\pmod{100}$$ for all odd $$n.$$

- 3 years, 10 months ago

Seriously. And Obviously -_- This was easy. Welll Done Adarsh Kumar

- 3 years, 5 months ago

For this , let us consider two cases when $$n$$ is odd and when it is even ,

When $$n$$ is even :

$$99$$ could be expressed as $$100-1$$. So, we could write it as $$(100-1)^{n}$$. As in this case we have assumed $$n$$ to be even us let us write $$n=2k$$ . So we have $$(100-1)^{2k}$$ . Now , little bit expanding it , we have , $$10000+1-200)^{k}$$ . Now taking $$p=10000+1$$ and $$q=-200$$ , we have $$(p+q)^{n}$$ . Now let us prove one more result , for any $$(100+1)^{x}$$ , the last two digits would be $$01$$ . By expanding by binomial theory , $(100+1)^{x} = 100^{x}+a_{1}100^{x-1} \times 1 + . . . . a_{y}100 \times 1^{x-1} + 1^{x}$ So, expanding it will have $$x+1$$ terms. It is clear that the first $$x$$ terms would have $$00$$ in their last as they are multiples of some powers of $$100$$. But the last term will be 1. So , returning to our expansion we, would have , $p^{n} +a_{1}p^{n- 1} q^ {1} + . . . . . a_{y} p q^{n-1} + q ^{n}$ Now in this we have except $$p^{n}$$ all other terms would be a multiple of $$100$$ as $$q$$ is a motile of it . As we have earlier proove that the last term of $$p^{n}$$ would be 1 . And , as all other terms end in $$00$$ , therefore 3rd two digits of the whole computation would be $$01$$ . So we have thus profed that any power that is even would yield a result of 01 at the end when applied to $$100-1$$. Case 2 could be proved in a similar way and after proving it we would have proved the result so instead doing that long proof for case 2 also we could say that any odd number can be expressed $$n=2k+1$$ so just multiplying one more $$99$$ to any even power we would obtain the odd power. And multiplying the end two digits which are $$01$$ for any even power of 99 as we have proved , by 99 would give us 99 at the end. Hence , proved.

QED

- 3 years, 10 months ago

This is similar to prove that

$$(100-1)^n\equiv 99 \mod 100$$

For $$n$$ is an odd number.

Noticed that every term of the expansion is divisible by $$100$$ except for the last term, $$-1$$. Hence we get

$$(100-1)^n\equiv -1 \mod 100$$

$$(100-1)^n\equiv 99 \mod 100$$

- 3 years, 10 months ago

While for $$n$$ is an even number, a similar approach can prove that $$99^n$$ ends in $$01$$.

- 3 years, 10 months ago