What will be the ones digit of the remainder in: $$\frac{\left|5555^{2222} + 2222^{5555}\right|} {7}$$
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3What have you tried? What do you know? Does Modulo Arithmetic sound familiar? How about Euler's Theorem? – Calvin Lin Jan 15 '13 at 16:01
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I tried it like 5555...will occur 2222 times, and 5555/7 gives remainder = 4. So 4 will occur 555 times. That's what i thought and similar way for 2222. – Ravi Jan 15 '13 at 16:04
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It's somewhat confusing - the remainder when dividing by $7$ is already always a one-digit number, so do you mean the ones-digit of the quotient, or do you really mean the ones-digit of the remainder? Specifically, if someone asked "what is the one's digit of the remainder in $$\frac{99}{7}$$ what do you think the answer is supposed to be? – Thomas Andrews Jan 15 '13 at 16:23
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Its the one's digit of remainder only. I ain't getting how to procede in such a question.. – Ravi Jan 15 '13 at 16:24
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Well the answer is 0. – Ravi Jan 15 '13 at 16:25
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The number is completely divisible by 7 – Ravi Jan 15 '13 at 16:26
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What number, Ravi? $99$ is not divisible by $7$, and neither is the numerator in this problem. – Thomas Andrews Jan 15 '13 at 16:26
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I am saying about my question..Its answer is 0 – Ravi Jan 15 '13 at 16:30
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Sir, can you please tell me stepwise how to solve it? – Ravi Jan 15 '13 at 16:32
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Sorry, I was wrong, the remainder is zero. My problem is that I think you mean a different question, but that there might have been a translation or transcription error. Essentially, the "one's digit" part of the question is redundant - an unnecessary condition of the problem, unless the original question was asking for something else. – Thomas Andrews Jan 15 '13 at 16:36
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How did you solve it ? Did you use binomial theorem or something else? – Ravi Jan 15 '13 at 16:39
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Related: http://math.stackexchange.com/questions/247411/congruence-and-division – lab bhattacharjee Feb 02 '13 at 19:38
4 Answers
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Hint: $ 5555 \equiv 4 \pmod{7}$, $2222 \equiv 3 \pmod{7}$.
Edit: Calculate that $4^3 \equiv 1, 3^6 \equiv 1 \pmod{7}$. Hence, this implies that $4^{3k} \equiv (4^3)^k \equiv 1^k \equiv 1 \pmod{7}, 3^{6j} \equiv (3^6)^j \equiv 1^j \equiv \pmod{7}$.
Now, $2222 \equiv 2 \pmod{3}$, and $5555 \equiv 5 \pmod{6}$. Hence,
$$ 5555^{2222} + 2222^{5555} \equiv 4^{2222} + 3^{5555} \equiv 4^ 2 + 3^ 5 \pmod{7}.$$
Calvin Lin
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Actually when you divide 5555 by 7 then 4 occurs 557 times, and when you divide 2222 by 7 then 3 occurs 1853 times. But that doesn't help either. – Ravi Jan 15 '13 at 16:17
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And i don't think we have to apply Euler 's theorem here. By the way , if we ignore the mod function then can you tell how to procede? – Ravi Jan 15 '13 at 16:18
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1@Ravi When I said to check your arithmetic, I meant $4^2 + 3^5 \neq 269$. You'd have to apply something to reduce the powers from 2222/5555 to some much more manageable. It need not be Euler's theorem, but that's a fast guaranteed approach. You may look at $2^3, 5^3 \pmod{7}$ and decide that's sufficient. – Calvin Lin Jan 15 '13 at 16:26
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@Ravi Taking modulus $n$ is equivalent to finding the remainder when a number is divided by $n$. That is why I asked if you knew what modulo arithmetic is. Considering that you're talking about "don't have to apply Euler's Theorem", it should be safe to assume you understand it (perhaps not very well). – Calvin Lin Jan 15 '13 at 16:27
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Well, yea at my level I haven't been taught much about Euler' s theorem except when dealing with complex numbers. – Ravi Jan 15 '13 at 16:29
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@Ravi That's not the Euler's Theorem I'm referring to. Alright, I'd add an edit. – Calvin Lin Jan 15 '13 at 16:30
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This is not a solution. I was trying to find out the possible values of $n$ such that $${\underbrace{22\cdots22}_{n\text{ digits }}}^{\underbrace{55\cdots55}_{n\text{ digits }}}+{\underbrace{55\cdots55}_{n\text{ digits }}}^{\underbrace{22\cdots22}_{n\text{ digits }}}$$ is divisible by $7$.
Let $$\underbrace{11\cdots11}_{n\text{ digits }} =\frac{10^n-1}9=M$$
Clearly $M$ is odd.
Now, $$(2M)^{5M}+(5M)^{2M}\equiv (2M)^{5M}+(-2M)^{2M}\pmod 7\equiv (2M)^{5M}+(2M)^{2M}=(2M)^{2M}\{(2M)^{3M}+1\}$$ which will be divisible by $7$
if (i) $7\mid \{(2M)^{3M}+1\}\iff 7\mid \{M^{3M}+1\}$ as $2^3\equiv1$
$\iff 7\mid \{\left(\frac{10^n-1}9\right)^{3M}+1\}\iff 7\mid \{(10^n-1)^{3M}+1\}$ as $9^3=3^6\equiv1\pmod 7$
So, we need $(10^n-1)^{3M}\equiv-1\pmod 7\implies (10^n-1)^3\equiv-1$ as $3M\equiv3\pmod {\phi(7)}$ as $M$ is odd.
Taking Discrete Logarithm wrt a primitive root $3\mod 7,$
$3\cdot ind_3(10^n-1)\equiv 3\pmod 6=6c+3$ for some integer $c$
So, $$ ind_3(10^n-1)=2c+1$$
So, $10^n-1\equiv 3^{2c+1}\pmod 7\equiv 3,6,5\iff 10^n\equiv4,6,7\pmod7$
$10^n\not\equiv7\pmod 7$ as $(10,7)=1$
$10^1\equiv3,10^2\equiv3^3\equiv2,10^3\equiv2\cdot10\equiv6,10^4\equiv2^2=4,10^5\equiv2\cdot6\equiv5,10^6\equiv 6^2\equiv1\pmod 7$
So, $n\equiv3,4\pmod 6$
or if (ii)$7\mid (2M)^{2M}\iff 7\mid M^{2M}\implies 7\mid M\implies 10^n\equiv1\pmod 7\iff 6\mid n$
So, if $n\equiv0,3,4\pmod 6,$ $${\underbrace{22\cdots22}_{n\text{ digits }}}^{\underbrace{55\cdots55}_{n\text{ digits }}}+{\underbrace{55\cdots55}_{n\text{ digits }}}^{\underbrace{22\cdots22}_{n\text{ digits }}}$$ is divisible by $7$.
lab bhattacharjee
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${\rm mod}\ 7\!:\ 4^{2222}\!+(-4)^{5555} \equiv 4^{2222}(1-4^{3333})\equiv 4^{2222}(1-(\color{#c00}{2^6})^{1111})\equiv 0\,$ by $\,\color{#c00}{2^6\equiv 1}$
${\rm mod}\ 7\!:\ \ \ a^{k}+\ (-a)^{k+3n} \ \equiv\ \,a^k\, (1\,-\,a^{3n})\ \ \equiv\ \ a^k\, (1\,-\,(\color{#c00}{b^6})^{n})\equiv 0\,$ by $\ \color{#c00}{b^6\equiv 1}\,$ if $\,0\not\equiv a\equiv b^2$
Bill Dubuque
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in this question i would use remainder theorem since 5555 and 2222 are positive integers the expression is positive so we have find remainder of expression (5555^2222+22222^5555) when divided by 7 5555 divided by 7 leaves remainder 4 2222 diided by 7 leaves remiander 3 remainder will be of for 4^2222+ 3^5555 so expression can be written as 16^ 1111+ 243^ 1111 is always divisible by 7
