Your calculator will hate this (2)

We have $7^4=(50-1)^2=2401\equiv 1\pmod{100}$ . Thus $7^{999}\equiv 7^3\equiv \boxed{43} \pmod{100}$

In the sequence 7^n for every 4 terms the combination of the last digits repeat. For example

7^1=7 7^2=49 7^3=343 7^4=2401 7^5=16807 And so on...

Therefore if we continue this sequence till the 999th term,then 999/4 would give a remainder of 3. For the 3rd term the last two digits are 43. So is it for the 999th term.

Slightly different solution ,

Let us split the question as ( $7^{3} * 7^{996}$ ).

We know that $7^4$ ends with 01 since $7^2=49$ and $49^2$ ends with $(50-49)^2 = 01$ .

$7^3 * (7^{4})^{249}$ = 343 * $01^{249}$ = 343 and hence last 2 digits are $\boxed{43}$

A mod-less approach:

${ 7 }^{ 999 }={ 7 }^{ 3 }{ \left( { 7 }^{ 4 } \right) }^{ 249 }=343{ \left( 2401 \right) }^{ 249 }=343{ \left( 100A+1 \right) }^{ 249 }=343\left( 100B+1 \right) =100C+43$

where $A,B,C$ are "some integers". Hence, $43$ are the last two digits.

Let $7^{999} = x$ . Now,

$7^{1000} = 7^{25\times 40} = 7^{25 \times \phi (100)} \equiv 1 \pmod{100}$

Now we know $\displaystyle 7x \equiv 1 \pmod{100}$ or $\displaystyle x\equiv 10a+3 \pmod{100}$ and we want to find $\displaystyle a$ such that $\displaystyle 7\times (10a+3) \equiv 1 \pmod{100}$ .

(The last digit of $\displaystyle x$ has to be $\displaystyle 3$ for getting $\displaystyle 1$ in unit's place of $\displaystyle 7x$ . Hence, the last two digits of $\displaystyle x$ were chosen to be $\displaystyle 10a+3$ . )

We get $\displaystyle a=4$ as the solution to the above congruence.

$\displaystyle \Rightarrow x \equiv \boxed{43} \pmod{100}$