3816547290
A Special Number: the only one in which
Every digit is used,
Each digit is used only once, and
The first n digits are divisible by n, for n=1..10.
3 is divisible by 1,
38 is divisible by 2,
381 is divisible by 3,
3816 is divisible by 4,
38165 is divisible by 5,
381654 is divisible by 6,
3816547 is divisible by 7,
38165472 is divisible by 8,
381654729 is divisible by 9,
3816547290 is divisible by 10.
Note by
Faizan Khan
2 years, 9 months ago
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[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
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Top NewestOn Brilliant there are question related to this number. This number can found out without hit-n-trial method.
Clearly the 5th and 10th digits are 5 & 0 respectively,
Now 4th digit can only be from 2 & 6 as 3rd digit is odd and for the number to be divisible by 4 its last two digits must be divisible by 4,
Sum of first 3, next 3 & next 3 digits must be divisible by 3,
Hence,
We get the following combinations for 4th,5th &6th digits - 258 or 654,
Now with use of divisibility of 8 we can figure out 6th,7th & 8th digits,
With help of less hard work we can figure out the number.
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I would like to clarify your solution-We know that a number is divisible by \(4\) when last two digits are divisible by \(4\) and there are no two digit numbers having \(4\) and \(8\) as last digits and are divisible by \(4\) therefore for \(4^{th}\) digit we can say that at the \(4^{th}\) place only \(2\) or \(4\) are possible digits.
Similarly , as \(8\) is divisible by \(4\) the \(8^{th}\) digit would be either \(2\) or \(4\). Then, as \(Akshay\) told we can proceed to get the answer.
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Hmmm... Well answered brother.. Thankss
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Good job, Falzan!
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i've always been intersred in relation between numbers .... awesome !!
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That's brilliant
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By the way I already knew that number.
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Wow, beautiful!
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Great Work.. Its Awesome// how did you found it?
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