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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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Observations to give lower and upper bounds: So we know for sure Now the whole thing becomes Contradiction ... and now I realise my implicit assumption that Going back to those two observations at the beginning, So we seek a number which, So we try just a few nearby values of the integer: And we have the solution,

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I think it's a Explanation: According to my calculator (this answer was posted before the no-computers tag was added), If we're not allowed to use computers, I would The same trick is used to solve similar questions on our sister site Mathematics: Finding the first digit of $2015^{2015}$ and What's the first digit of 2410^2410?

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let me try:

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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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This is one way to do it with the help of binary numbers. It is called exponentiation by squaring. EDIT Thanks to the few comments that pointed out the mistakes in my calculations. I was very lucky to get the right answer. While revising, I also noticed a mathematical mistake that explains why I need 4 digits to get the right answer. I'll leave the old ...

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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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It's different:

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Professor Halfbrain's theorem is Proof

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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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Evaluation:

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My solution: Just normal Math

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Of course, $x=0$ is an answer, so let's look for non-zero ones from now on. If the given expression is a perfect square, so is Now we try to estimate it by "nearby" perfect squares. One could rightly object against the "obviously"s above since we are dealing with possibly negative numbers here. Fortunately, this is easily settled: Thus ...

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

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I think the answer is Proof Computer check

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Step 1: Step 2:

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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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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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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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