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This is a short number puzzle:

Find a number $x$, which leaves remainder $1$ when divided by $2,3,4,5$ and $6$.

Here is the some more explanation:
$x$ divided by $2$ gives remainder $1$
$x$ divided by $3$ gives remainder $1$
$x$ divided by $4$ gives remainder $1$
$x$ divided by $5$ gives remainder $1$
$x$ divided by $6$ gives remainder $1$

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closed as off-topic by GentlePurpleRain, DrunkWolf, Deusovi, AJL, 2012rcampion Jan 1 '16 at 17:13

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – GentlePurpleRain, DrunkWolf, Deusovi, AJL, 2012rcampion
If this question can be reworded to fit the rules in the help center, please edit the question.

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Brute force, everyone.

There are 1.667 correct answers just between 1 and 100.000

You may check this fiddle:

https://jsfiddle.net/g4tk05pn/

So while the question is not spesific like smallest number possible, there is not just one answer.

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  • $\begingroup$ Cool.. nice try..!!!! and got it worked..!! cheers..!! $\endgroup$ – Ragu Swaminathan Jan 1 '16 at 14:42
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The answer is simple, it's 61.

LCM of 2,3,4,5,6 (which is 60), plus 1.

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  • $\begingroup$ Who would down-vote this? $\endgroup$ – klm123 Jan 1 '16 at 13:53
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The Chinese remainder theorem renders this type of problem easy. However, in this case the answer is even easier:

The answer is $1$, since:
$$ 1~\text{mod}~2 \equiv 1 \\ 1~\text{mod}~3 \equiv 1 \\ 1~\text{mod}~4 \equiv 1 \\ 1~\text{mod}~5 \equiv 1 \\ 1~\text{mod}~6 \equiv 1 $$

The Chinese remainder theorem tells us that there are an infinite number of numbers satisfying this type of system of equations. There is a bit of difficulty in that our moduli are not pairwise coprime, so we must take the least common multiple of the moduli instead of just multiplying them:

$\text{LCM}(2,3,4,5,6)=60$, so our final answer is $x\equiv 1~\text{mod}~60$

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  • $\begingroup$ Note that you can use WolframAlpha to solve this type of problem. $\endgroup$ – 2012rcampion Jan 1 '16 at 7:10
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It's 721, or in other words, 6! + 1.

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  • $\begingroup$ 1 is the trivial solution. If you add more numbers in the factorial, eg. 7! + 1, 8! + 1, etc, you'll have infinitely many more solutions. $\endgroup$ – cst1992 Jan 1 '16 at 7:07

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