# Tag Info

113

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I think that This is because This works and is valid because

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because

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18 pieces: In 9 parts: In 8 parts: In 7 parts: We need to be able to split it into 9 parts of 56, so it can't hurt to make 9 pieces of 56 and then split those further. Since we need to do better than 19 pieces, we can have at most 18 pieces. This means that most of our 56s are split into exactly two parts (we can have an extra piece for every 56 we don't ...

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Lateral thinking!

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Here's one way I found: Or, using just the characters explicitly allowed in the question:

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Notice 9867312 is a Monday number. The largest Monday number may not contain 5 because in this case it would end in 5, and thus not be divisible by 2, 4 and 8, so it would have at most 6 digits. On the other hand, a Monday number may not have 8 digits. Indeed, if that were the case, the preceding paragrph would imply such a number has each digit but 0 and ...

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If the double factorial is allowed, then I propose WolframAlpha agrees that the result is 19.

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I thought a bit too much but I finally got it:

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That happens because when you square a number(let's say $x$), you will get $x^2$ as the result. Then you subtract 1 from it and you get $x^2 - 1$, which can be rewritten as $x^2 - 1^2$ which is then equal to $(x-1)(x+1)$. Prime numbers are only divisible by $1$ and itself($x$). Also, for any number $x$ the following is true: $x$, $x+1$ or $x-1$ is ...

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let me try:

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The first wrinkle is easy enough — you get one coin for each (nonempty) subset of $n$ travellers ($2^n - 1$ total), minus one coin for the ferry, for a total of $2^n - 2$. If that's divisible by $n$, you go. By Fermat's Little Theorem, any prime number of partygoers will work. But the second wrinkle means that you need to be able to divide your party ...

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I am quite sure it is not the expected answer but it is the immediate answer comes into my mind.

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Professor Halfbrain's theorem is Proof

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It's different:

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My solution: Just normal Math

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Evaluation:

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For the 5s For the 1s (previous edit)

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OEIS doesn't list this sequence. After analyzing the pattern, I come to the conclusion that (one) answer is Explanation: EDIT (05/06/18): I submitted this sequence to OEIS and it has (finally) been approved now.

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Step 1: Step 2:

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Obviously a is one of {1,3,7,9} and a,b are coprime. Also, b can't be a multiple of 3 regardless (else our number is a multiple of 3). That leaves 23 possibilities (4 for a, 6 for b, but we can't have a=b=7), or 22 if you don't count 11111 as "limerick". The only other trick I see is that 1001 = 7x11x13; so mod 1001, aabba = 11b-a. Clearly that isn't going ...

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22 pieces Suppose the weight of the bar is 504. I chose 504 since 7*8*9=504, so, the numbers would be easy to work with. 7 pieces of weight 56 (A) 8 pieces of weight 7 (B) 6 pieces of weight 9 (C) 1 pieces of weight 2 (D) Scenario 1: The two gangsters decide to settle in a peaceful manner Each gets a share of weight 504/9=56. Give 7 A pieces to 7 ...

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We are looking for The only way I know of to find one is I think So

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... ... ... ... ... ... ...

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I have a solution using 448 consecutive integers. The integers are centered on a number which I will call $c$, which is: 342807324437386669890245640930601418907623281362205720311372555412334319102271973072975564592463524267335118648596000454654370063914007920553665901460149824033735950242152959820 $c$ is approximately $3.4*10^{176}$. The integers are the ...

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(Kind of) analytical solution that only requires small amount of calculations, (potentially) doable by hand. First step: we can safely drop 2, 3 and 7 from the equation as those digits are used in 23 and 17. Now, we need to build a prime from: 5, 11, 13, 17, 19, 23, 29 and 31. Second step: let's try to build the shortest number possible from these numbers. ...

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Explanation: Recap to see that all of them fit:

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The answer is I wrote a Java program to find it:

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It has been shown that fewer than 16 pieces is not possible (due to the 8-way split then requiring someone to have a single piece larger than a single portion of the 9-way split). I'll show by contradiction than a split into exactly 16 pieces is not possible: for that purpose let's assume that there exists a solution for 16 pieces. We'll use units of 1/...

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