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Question: A number when divided by $4$ gives a remainder which is $3$ more than the remainder obtained on dividing the number by $34$. Find the least such number.

My approach: $$\begin{align} &n=34b+r,\quad\ \, 0\le r<34\\[.2em] &n=4a+r\!+\!3,\,\ 0\le r+3<4\quad\Longrightarrow\quad r=0\end{align}$$

therefore we need a number which when divided by $34$ gives a remainder $0$, and a remainder of $3$ when divided by $4$.

$$34x=4y+3,$$ where $x$ and $y$ are integers

$$y=(8x)+(2x-3)/4,$$

I tried putting values of $x=1,2,3,4,5,\ldots$ but not getting an integer value for $y$.

Bill Dubuque
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Fin27
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    Note that if a number is divisible by $34$ then it is even, whereas any number of the form $4k+3$ is odd. – lulu Nov 24 '21 at 15:25
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    $34x=4y+3$ has no integer solutions. – markvs Nov 24 '21 at 15:28
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    It is $\rm\color{#c00}{unsolvable}$, since by the standard CRT Solvability Criterion in the first linked dupe

    $$\begin{align} n&\equiv r!+!3!!!\pmod{!4}\ n&\equiv r\quad! \pmod{!34}\end{align}, \ \text{is solvable}!\iff \underbrace{\gcd(4,34)\mid (r!+!3)-r}_{\textstyle{\rm i.e.}\ \ \ \color{#c00}{2\mid 3}\qquad\ }\qquad$$

     – Bill Dubuque Nov 25 '21 at 11:00
  • Said equivalently, it is unsolvable because reducing the system $!\bmod 2,$ yields a parity contradiction: $,r!+!1\equiv n\equiv r\pmod{!2}.,$ Such parity inconsistency is a prototypical example of the CRT Solvability Criterion - as explained at length in the similar example in the 2nd linked dupe. – Bill Dubuque Nov 25 '21 at 11:00
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    There is nothing at all novel here. We already have plenty of examples of applying this criterion - many more illuminating. The question should be re-closed (and probably deleted too). – Bill Dubuque Nov 25 '21 at 11:02
  • @BillDubuque Thanks : I vote to close in agreement with the above comments of Bill Dubuque. I have used the link in his first comment from above, referred to as the "first linked dupe" to mark this question as a duplicate. I invite someone else to close this question using the "second linked dupe" as a duplicate candidate , which can be found in Bill's second comment. – Sarvesh Ravichandran Iyer Nov 25 '21 at 11:36

1 Answers1

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You want to find the least number $n$ such that \begin{eqnarray*} n\%34&\equiv& r,\\ n\%4&\equiv& r+3, \end{eqnarray*} Of course by definition \begin{eqnarray*} 0\leq&n\%34&\leq33,\\ 0\leq&n\%4&\leq3, \end{eqnarray*} from which it follows that $r=0$. Then $n=34x$ and $n=4y+3$ for some integers $x$ and $y$. But the former implies that $n$ is even, whereas the latter implies that $n$ is odd, a contradiction. Hence no such number exists.

Servaes
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  • Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Nov 24 '21 at 17:31
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    @BillDubuque I understand your sentiment, but I do not feel that this is a duplicate. – Servaes Nov 24 '21 at 21:18
  • No, it is a dupe. That $,r=0,$ here is irrelevant since a parity contradiction arises for any $,r,,$ i.e. recalling that congruences persist mod factors of the modulus (here the factor $2$) we infer $$\begin{align} n&\equiv \color{#c00}r\ \ \ :! \pmod{!34}\ n&\equiv \color{#0a0}{r!+!3}!!!\pmod{!4}\end{align},\Rightarrow, \begin{array}{} n!!!! &\equiv \color{#c00}r\ \ \ \ \ \pmod{!2}\ n!!!!&\equiv \color{#0a0}{r!+!1}\pmod{!2}\ \end{array}\qquad$$ – Bill Dubuque Nov 25 '21 at 08:08
  • Thus $,\color{#0a0}{r!+!1}\equiv \color{#c00}r\pmod{!2},,$ contra successive integers have opposite parity (or, said equivalently, we deduce that $, 0\equiv 1\pmod{!2}\Rightarrow 2\mid 1,$ by subtracting $,r,$ from both sides or prior congruence). – Bill Dubuque Nov 25 '21 at 08:08
  • Above is essentially the same prototypical parity example that I gave in the 2nd dupe link, which is a special case of the proof of the general CRT solvability criterion given in the 1st dupe link. The presentation in the above answer obfuscates the general essence of the matter, i.e. that reducing the congruence pair mod their moduli gcd yields a necessary (& sufficient) condition for solvability. – Bill Dubuque Nov 25 '21 at 08:08
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    well I don't know and not wish to interfere in this matter, but just to put up my opinion that the answer provided by @Servaes is easy to understand compared to the explanations given in the links referenced to. As a person with not so much exposure to such high level of mathematics, this answer is fine for me. Rest is up to the wishes of mods. Peace – Fin27 Nov 25 '21 at 23:11