Calculate the last 2 digits of the following expression

Question

Find last 2 digits of the following expression with modular arithmetic.

$$\LARGE 7^{7^{7^{7}}} - 7^{7^{7}} - 7^7 - 7$$

I have tried taking remainders of all terms $\mod 100$ then subtracting , eventually realising it can depend on more digits.

Note: 1st no. is 7^7^7^7 , it wasn't showing it clearly in $\LaTeX$.

One thing I struggle with is how to come with an approach to solve any new mathematical problems. Preparing for IMO (1st stage) , any suggestions? — Aryan Malik, Jun 30 '24 at 15:35
$\displaystyle \left(7^{7}\right)^{7} \not= 7^{\left(7^{7}\right)}$ — Felix Marin, Jun 30 '24 at 15:49
$7^4 \equiv 1 \pmod{100}$. And $7^2\equiv 1 \pmod 4$ and $7^1 \equiv 1 \pmod 2$. So $7^{7^{7^{7}}}\equiv 7^{{{7\mod 4}^{7\mod 2}}^{7\mod 1}}\equiv 7^{3^{1^1}}\pmod {100}$. BTW this is definitely a duplicate. — fleablood, Jun 30 '24 at 15:49
You want a hint for how to solve any IMO problem? @AryanMalik — Thomas Andrews, Jun 30 '24 at 20:12

score 2 · Accepted Answer · edited Jun 30 '24 at 15:20

First:

You can prove that $\forall k \in \mathbb{N} \cup \{0\}$ we have:

$7^{4k+1} \equiv 7 \pmod{100}$.

$7^{4k+2} \equiv 49 \pmod{100}$.

$7^{4k+3} \equiv 43 \pmod{100}$.

$7^{4k+4} \equiv 1 \pmod{100}$.

Second:

$7^{7^{7}} \equiv (-1)^{7^{7}} \equiv -1 \equiv 3 \pmod{4}$.

So,$7^{7^{7^{7}}} \equiv 43 \pmod{100}$.

$7^{7} \equiv 3 \pmod{4}$.

So, $7^{7^{7}} \equiv 43 \pmod{100}$.

Also, $7 \equiv 3 \pmod{4}$.

So, $7^{7} \equiv 43 \pmod{100}$.

So, $7^{7^{7^{7}}} - 7^{7^{7}} - 7^7 - 7 \equiv 43-43-43-7 \equiv 50 \pmod{100} $

score 1 · Answer 2 · answered Jun 30 '24 at 15:19

Other than the methods already presented, I will present another way by taking modulo $25$ and $4$. Note $\newcommand{\Mod}[3]{#1\equiv#2\ (\mathrm{mod}\ #3)}$ $\Mod7{-1}{4}$ so $$\Mod{7^{7^{7^7}}-7^{7^7}-7^7-7}{(-1)^{7^{7^7}}-(-1)^{7^7}-(-1)^7-(-1)\equiv2}{4}$$ Now, $\Mod{7^2}{-1}{25}$ so $\Mod{7^4}{1}{25}$. Now, $\Mod{7^{7^7}\equiv7^7}{-1}{4}$ so $$\Mod{7^{7^{7^7}}-7^{7^7}-7^7-7}{7^{3}-7^{3}-7^3-7\equiv7-7\equiv0}{25}$$ Therefore, the number is $50\ \mathrm{mod}\ 100$.

Hope this helps. :)

score 0 · Answer 3 · answered Jun 30 '24 at 14:38

We start by noticing that

$7^0 = 1$
$7^1 = 7$
$7^2 = 49$
$7^3 = 343$ thus $7^3 = 43$ mod($100$)
$43*7 = 301$ thus $7^4 = 1$ mod($100$)

Now it's easy, since we need to check the exponents mod($4$):

$7 = 3$ mod($4$), thus $7^7 = 7^3$ which is $43$ mod($100$)
$43 = 3$ mod($4$), thus $7^{7^7} = 7^3$ which is $43$ mod($100$)
The same argument goes for the last one: $7^{7^{7^7}} = 7^3$ which is $43$ mod($100$)

At last, we have $43 - 43 - 43 - 7 = -50$ which is $50$ mod($100$)

Calculate the last 2 digits of the following expression

3 Answers3