How to find the last two digits of $299^{33}$. Is there any trick?

Find the last two digits of the number $N=299^{33}$

Is there any trick of finding the last two digits of such a big number. If yes, then kindly share it with me

This is a GMAT exam question.

Since your question is tagged with GMAT, I am going to assume you are not a mathematician... I am not either, and all this talk of Euler's theorem and modulus will give me a headache ;)

So, simply put... the trick is that only the rightmost two digits of each number being multiplied can have any effect on the rightmost two digits of the answer. So you can just drop the "$2$" from $299$ because $299$ squared ends with same $2$ digits as $99$ squared.

$2$nd power: $99 \times 99 = 9801, \ldots$ now you can drop the "$98$" because it has no effect on the rightmost two digits of the answer.

$3$rd power: $01 \times 99 = 99$

$4$th power: $99 \times 99 = 9801, \ldots$ drop the "$98$" again

$5$th power: $01 \times 99 = 99$

etc... See the pattern? Even powers will end with "$01$" and odd powers will end with "$99$"

Hint:

$$299 = 300 -1$$

Hence $N \equiv (-1)^{33} \pmod {100}$

Edit:

The following page could be helpful.

modular arithmetic wikipedia page

Let $a, q, b, r \in \mathbb{Z}$, and $$a=bq+r$$

then we say $a \equiv r \mod q.$

Using modular arithmetic we are able to generally solve some elementary number theory problem by focusing on what we are interested. For example in your case, we can focus on the last two digits, or if your interest is whether a number is odd or even, just deal with $\mod 2$.

Some operations that we can do include:

if $a \equiv b \pmod q$ and $c \equiv d \pmod q$ then we have $$a+c \equiv b+d \pmod q$$

and $$ac \equiv bd \pmod q$$

and $$a^k \equiv b^k \pmod q$$

Since $$299 = 3\times 100 -1,$$

we have $$299 \equiv -1 \pmod{100}$$

$$299^{33} \equiv (-1)^{33} = -1 \pmod{100}$$

Note that $$99 \equiv -1 \pmod{100}$$

Hence the last two digits is $99$

Note: $$299^{33}=(300-1)^{33}=300^{33}-33\cdot 300^{32}\cdot 1+\cdots +33\cdot 300\cdot 1^{32}-1^{33}=M\cdot 100-1=\cdots99.$$

You can rewrite $299^{33}$ as $(300-1)^{33}$. Last $2$ digits of sum of all terms in binomial expansion except the last term is $00$. Because they can be divided by $300$. When we add the last term $-1$ to $00$, answer is $99$.