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Look at the minimum of sizes of each datum. Let $m_i$ be the minimum of sizes of bacteria at $i$th observation. If $m_i+m_{i+2}=2m_{i+1}$ for some $i$, it means that at least one bacterium had size $m_i$ at time $i$ and has growth rate $m_{i+1}-m_i$ since if the $(i+1)$th minimum $m_{i+1}$ was not made by the bacterium it means there was another bacterium ...


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Continue the method mentioned in Christoph's comment. If we get two disjoint pair we have done. If we only get 3 number $a<b<c$, let $d_1=b-a$. Now we have to exclude $(a,b)$ and $(b,c)$. So there are $45-2=43$ pairs of numbers. Apply pigeon hole principle again($36-1=35$ possible differences). If we get two disjoint pair we have done. Otherwise, ...


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Yes, $f(A)=A$ in this case. First let us show that the filtration property you describe implies a seemingly stronger property. Claim: If $\{k_1,\dots, k_n\}\in A$ and $j_i$ are so that $j_i\leq k_i$ for all $i\leq n$ then $\{j_1,\dots, j_n\}\in A$. Note that we do not assume that the $k_i$ or $j_i$ are increasing, though me may assume wlog that at least ...


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only works for symmetric cuts. First lets note that any cut we make works if we rotate it by 90 degrees. So lets solve the problem for cuts we make from bottom row to top row. Now lets number the points as follows\ 1 2 3 4 5 6 7 8 8 7 6 5 4 3 2 1 Now we can pick only one number in the top row. Since picking a second number is same as starting the cut at ...


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Of course, it's a condition that has to be checked for all triples $x,y,z$ with $x \le z$, but it's a simple "algebraic" condition between a few elements. No distinction is needed (or made) for the case where $L$ itself is infinite or not. The condition is what makes a lattic modular, not its size. Maybe your text has some special theorems that only hold ...


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The sum is $$(p-0)+(p-1)+\ldots+(p-(p-1))=1+2+\ldots+p=\frac{p(p+1)}2$$


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These, of course are Fibonacci numbers. The recurrence means that $$\pmatrix{0&1\\1&1}\pmatrix{a_n\\a_{n+1}}=\pmatrix{a_{n+1}\\a_{n+2}}.$$ Write $M=\pmatrix{0&1\\1&1}$ Then, by induction, $$\pmatrix{a_n\\a_{n+1}}=M^{n-1}\pmatrix{a_1\\a_2}=M^{n-1}\pmatrix{1\\2} =M^n\pmatrix{1\\1}=M^{n+1}\pmatrix{0\\1}=M^{n+2}\pmatrix{1\\0}.$$ In the matrix $M^...


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Straightforward use of binomial theorem. $n\binom{n-1}{4}\binom{n-5}{5}(n-10)(n-11)$ I am assuming $n\ge 12$ and the order of the the jars doesn't matter.


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We can determine the number of solutions under these constraints using generating functions (indeed, using one of the main results of the study of generating functions): If we interpret the equation using generating functions as: $$m_1x_1+\cdots+m_rx_r=n \quad\rightarrow\quad \sum_{x_1\ge 0}...\sum_{x_r\ge 0} x^{m_1x_1+...+m_rx_r}$$ Then the sum of all ...


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Let $A = \{ a_1, \ldots, a_{10} \}$ be a subset with ten elements, where $i < j \rightarrow a_i < a_j$, of $B = \{1, \ldots, 37\}$. Obviously $a_{10} - a_1 \leq 36$. Define $d_i := a_{i + 1} - a_{i}$. We have $1 \leq d_i \leq 27$, and $\sum_{i = 1}^9 d_i = a_{10} - a_{1} \leq 36$. If $d_i = d_j$, then $a_i + a_{j + 1} = a_j + a_{i + 1}$. So in order ...


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