Why $1472222^{5555} + 14145555^{2228}$ is divisible by $7$?

Question

Maybe someone can show why $1472222^{5555} + 14145555^{2228}$ is divisible by $7$? I just have written $$1472222^{5555} + 14145555^{2228}\equiv3^{5555}+4^{2228}\pmod 7$$ but what to do next? I don't see.

Answer 1

Hint: Take a look at $4^1, 4^2, 4^3, \ldots$ modulo $7$. Pretty soon you will find a pattern. Figure out where $5555$ fits in that pattern. Then do the same for $3^1, 3^2, 3^3, \ldots$ and $2228$.

Answer 2

It is not divisible by 7. You can use the modular arithmetic to find this out. There are two main principle from the field that is needed for this. If $a\equiv b \text{ (mod } k)$ then, $a^k \equiv b^k \text{ (mod }k)$ and if $a \equiv b\text{ (mod } k)$ and if $c\equiv d \text{ (mod } k)$ then, $ac \equiv bd \text { (mod } k)$. \begin{equation} \end{equation} Now, \begin{equation} 1472222 \equiv 3 \text { (mod } 7) \end{equation} \begin{equation} 3^{5555} = {(3^{1111})}^5 = {(3^{11})}^{505} = 177147^{505} \end{equation} \begin{equation} 177147 \equiv 5 \text { (mod } 7) \end{equation} Thus, \begin{equation} 5^{505} = (5^5)^{101} =3125^{101} \end{equation} \begin{equation} 3125 \equiv 3 \text { (mod } 7) \end{equation} \begin{equation} 3^{101}=3^{33}\times3^{33}\times3^{33}\times 9 = \underbrace{3^6 \times ...\times 3^6}_\text{15 times} \times 3^{11} \end{equation} By Fermat's little theorem, $a^{p-1}\equiv 1 \text{ (mod }p)$, the above equation can be written as with respect to $\text{ mod 7}$ as \begin{equation} \underbrace{1\times...\times1}_\text{15 times}\times 5 \end{equation} Thus, $1472222^{5555} \equiv 5 \text{ (mod } 7)$ Similarly, \begin{equation} 14145555 \equiv 4 \text{ (mod } 7) \end{equation} \begin{equation} 4^{2228} = 256^{557} \equiv 4^{557} \text{ (mod } 7) = 4^{500}\times 4^{56}\times 4 \end{equation} This can be written as: \begin{equation} 4^{500}\times 4^{56}\times 4\equiv4^{125}\times 2^9 \text{ (mod } 7) \equiv 4^{125} \text{ (mod } 7) \end{equation} \begin{equation} 4^{125} = 4^{120} \times 4^5 \equiv 4^{31} \text{ (mod }7) = 4^{16}\times4^{16}\times \frac{1}{4} \equiv 4 \text{ (mod} 7) \end{equation} Thus, $1472222^{5555}+14145555^{2228} \equiv 2 \text{ (mod } 7)$ and hence not divisible by 7.

I might have done some steps wrong. Please notify me if any. I will be glad to correct them. Thank you.

Answer 3

$1472222=1470000+2100+70+49+3$, so $1472222\equiv3\pmod7$.

$14145555=14140000+4900+630+21+4$, so $14145555\equiv4\pmod7$.

You can now use Fermat’s little theorem: $x^6\equiv1\pmod7$, if $x$ is not divisible by $7$.

Now $5555\equiv5\pmod6$ and $2228\equiv2\pmod6$, so you have to evaluate $3^5+4^2$ modulo $7$.

But $3\cdot5\equiv1$, so you simply have $5+16=21$, which is indeed congruent to $0$ modulo $7$.p