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A number when divided by $2,3,4,5,6$ leaves remainder $1,2,3,4,5$ respectively but when its divided by $11$ the remainder is $0$. FIND THE NUMBER

I tried taking LCM of $2,3,4,5,6$ and subratcing by $1(59)$ then multiply by $11$. What am i doing wrong? Please Guide.

RobPratt
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Bruh
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    Hint: Remainder is $~-1~$ when divided by each of $~2,3,4,5,6.~$ So, $~60k - 1 = 11a.~$ An inelegant approach to completing the problem is that since $~60 \equiv 5 \pmod{11},~$ you have that $~5k -1~$ must be a multiple of $~11.$ – user2661923 Jul 09 '23 at 16:41
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    Welcome to Mathematics Stack Exchange. You need to solve $x\equiv-1\bmod60$ and $x\equiv0\bmod11$. Do you know the Chinese remainder theorem? An indication of what you are doing wrong is that the remainder when $x$ is divided by $3$ is not the same as the remainder when $11x$ is divided by $3$ – J. W. Tanner Jul 09 '23 at 16:43
  • Beginning with $,11,,$ and given divisors $,(2,3,4,5,6),,$ the pattern of remainders $(1,2,3,x,5)$ is found every $,12^{th},$ multiple of $,11.\quad$ Finally, at $,N=11\cdot49=539,,$ the pattern becomes $(1,2,3,4,5).\quad$ – poetasis Jul 09 '23 at 17:20
  • In your text, is this problem before or after the https://en.wikipedia.org/wiki/Chinese_remainder_theorem ? So: is this problem preliminary motivation for the CRT, or are you supposed to solve this using the CRT? – GEdgar Jul 09 '23 at 17:37
  • As a complete beginner to number theory and these types of questions, here is my attempt:

    2m - 11n = 1

    gcd(2, -11) = gcd(2, 11) = 1 -11 = 2 * -5 - 1 ==> 1 = 2(-5) + 11

    m = -5, n = -1

    x = 11n = -11

    Of course there are an infinite number of solutions but I'm too lazy to find the general formula

     – avighnac Jul 09 '23 at 18:15
  • By CCRT (in first dupe) we know the first $5$ congruences are equivalent to $,x\equiv -1\pmod{60}.,$ Further, by the Easy CRT formula we have $$\begin{align} x&\equiv -1\pmod{60}\ x&\equiv\ ,0\ \pmod{11}\end{align}\iff \bmod 660!:\ x\equiv 11(\color{#c00}{-11^{-1}}\bmod 60)\equiv 11(\color{#c00}{-11})\equiv 539$$ Your error is applying an incorrect CRT method to solve the prior congruence system, viz. $\bmod 60!:\ x = 11\color{#c00}n\equiv -1\iff \color{#c00}{n\equiv -11^{-1}}\ \ $ – Bill Dubuque Jul 09 '23 at 18:28
  • thank you everyone figured out what i was doing wrong!! – Bruh Jul 13 '23 at 15:26

1 Answers1

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Let $n$ be that number then $n+1$ can divisible by $2,3,4,5,6$ without remainder.

$\newcommand{\lcm}{\operatorname{lcm}}\lcm(2,3,4,5,6)=60$

$n+1=60k, k\in Z$

and if the number is divisible by $11$

$n=11t,t\in Z$

by solving

$$n+1=60k, k\in Z$$ $$n=11t,t\in Z$$

$n=60k-1=11t$, $ k , t\in Z$

$t=5k+\frac{5k-1}{11}$

for $k=9, t=49$

then

$$n=539$$

In general $$n\equiv 539 \mod {660}$$

Lion Heart
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