Show that $2222^{5555} + 5555^{2222}$ is divisible by 7 (without modular arithmetic)

Question

I tried using the following approach:

$$x=2222^{5555}+5555^{2222} = (2222^5)^{1111}+(5555^2)^{1111}$$

Now we know $(x^n+y^n)$ is divisible by $(x+y)$ for odd natural number $n$. So,

$$x=(2222^5+5555^2)k,\ k\in N$$

$$x=(1111^2)(32\cdot1111^3+25)k$$

The term in parentheses was found to be $54165190296027657$, which is divisible by $7$. Thus $7 | x$.

Another method (edit):

We can also write $$2222^{5555} = (7\times 317+3)^{5555} = 7p+3^{5555}$$ Similarly $$5555^{2222}=(793\times7+4)^{2222} = 7q+4^{2222}$$

So we have $$x=7(p+q)+(3^5+4^2)r$$ Here we have $3^5+4^2$ is divisible by $7$. So $x$ is also divisible by $7$.

But is there a simple way without multiplication of large numbers? Thank you!

The thing you need to work out immediately in this kind of problem is value of $2222$ and $5555$ modulo $7$, because that would simplify things a lot — user160738, Jun 17 '17 at 05:46
Well, modulo $7$ the term in parentheses is $25\cdot(1111^3 + 1)$. Just do $1111 \mod(7)$ for an easy sol'n then. — Chris, Jun 17 '17 at 05:46
https://math.stackexchange.com/questions/279333/what-will-be-the-ones-digit-of-the-remainder-in-left55552222-2222555 — lab bhattacharjee, Jun 17 '17 at 05:47

score 4 · Accepted Answer · edited Jan 08 '20 at 15:47

4

We have

\begin{align} 2222^5 + 5555^2&=2222^5 +5555^5-5555^5+ 5555^2\\ &=2222^5 +5555^5-5555^2( 5555^3-1)\\ \end{align}

$2222^5 +5555^5$ is divisible by $2222+5555=7(1111)$ as $5$ is odd.

$$5555^3-1=(7\times793+4)^3-1=7p+4^3-1=7p+63$$

is divisible by $7$.

edited Jan 08 '20 at 15:47

J. W. Tanner

60,406

answered Jun 17 '17 at 06:09

CY Aries

23,393

So there are several ways to reduce calculations. Thanks! – jonsno Jun 17 '17 at 06:11

Michael Rozenberg · Answer 2 · 2017-06-17T06:10:05.820

1

Because $2226$ divided by $7$, $5551$ divided by $7$,

$a^n-b^n=(a-b)(a^{n-1}+...+b^{n-1})$, for odd $n$ we have $a^n+b^n=(a+b)(a^{n-1}-...+b^{n-1})$ and $$2222^{5555}+5555^{2222}=$$ $$=(2226-4)^{5555}+4^{5555}+$$ $$+(5551+4)^{2222}-4^{2222}+$$ $$-\left(4^{5555}-4^{2222}\right)$$ and $$4^{5555}-4^{2222}=4^{2222}\left(4^{3333}-1\right),$$ where the last expression divided by $4^3-1$, which divided by $7$.

edited Jun 17 '17 at 06:10

answered Jun 17 '17 at 05:52

Michael Rozenberg

194,933

Nice answer! I tried something similar but calculations became complicated. – jonsno Jun 17 '17 at 06:11

score 1 · Answer 3 · answered Jun 17 '17 at 05:54

1

Apparently, you know that $x^n+y^n$ is divisible by $x+y$ for $n$ odd. So $7\mid 1113\mid 1111^3+2^3$.

answered Jun 17 '17 at 05:54

Hagen von Eitzen

374,180

Sorry but didnt understand how we got $1111^3 + 2^3$ as a factor? – jonsno Jun 17 '17 at 06:06

Show that $2222^{5555} + 5555^{2222}$ is divisible by 7 (without modular arithmetic)

3 Answers3