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question:John is building a tree house to serve as his hideout whenever he wanted to spend some time alone with himself. The first thing he did is to gather some materials. For the pillars, he needs two sets of wood with different lengths. If the first set of wood is 4 feet long and the second set is 5 times longer than the first set, how long is the second set of wood?
answer:If the first set of wood is 4 feet long and the second set is 5 times longer than the first set, then the second set of wood is 4 feet * 5 = boxed{20} feet long.
question:Write down a necessary but not sufficient condition for the inequality |x2|<1 to hold.
answer:The inequality |x-2|<1 can be rewritten as -1<x-2<1, which simplifies to 1<x<3. Therefore, the statement "1<x<4" cannot imply "1<x<3" directly, but "1<x<3" can imply "1<x<4". Thus, "1<x<4" is a necessary but not sufficient condition for "1<x<3". Hence, the answer is boxed{1<x<4}.
question:Let n>1 be any integer. Define f,g as functions from {0,1,2,cdots,n1 } to {0,1,2,cdots,n1} defined as begin{align*} &f(i)=2i pmod{n} &g(i)=2i+1 pmod{n} end{align*} Show that for any integers ell,m in {0,1,2,cdots,n1 } , there are infinitely many compositions of f,g that map ell to m
answer:1. **Claim 1:** For any ( m ), we can find a combination of compositions ( h ) such that ( h(0) = m ). **Proof:** Let ( m = 2^k a_k + cdots + 2a_1 + a_0 ) with ( a_j in {0, 1} forall j ) be the binary expansion of ( m ). Define ( h_i = a_i g + (1 - a_i)f ) for ( i = 0, 1, cdots, k ). We will show that the function ( h := h_0 circ h_1 circ cdots circ h_{k - 1} circ h_k ) is such that ( h(0) = m ). Note that if ( m = 2^k + 2^{k_1} + 2^{k_2} + cdots + 2^{k_j} ) with ( k > k_1 > cdots > k_j geq 0 ), then ( a_k = a_{k_1} = cdots = a_{k_j} = 1 ) and ( a_i = 0 ) otherwise. So ( h_k = h_{k_i} = g ) and ( h_i = f ) otherwise. So [ h_{k_1 - 1} circ h_{k_1 - 2} circ cdots circ h_k (0) = 2^{k - k_1 - 1}. ] Then, begin{align*} h_{k_2 - 1} circ h_{k_2 - 2} circ cdots circ h_{k_1} (2^{k - k_1 - 1}) &= 2^{k - k_2 - 1} + 2^{k_1 - k_2 - 1}, h_{k_3 - 1} circ h_{k_3 - 2} circ cdots circ h_{k_2} (2^{k - k_2 - 1} + 2^{k_1 - k_2 - 1}) &= 2^{k - k_3 - 1} + 2^{k_1 - k_3 - 1} + 2^{k_2 - k_3 - 1}, &vdots h_{k_j - 1} circ h_{k_j - 2} circ cdots circ h_{k_{j - 1}} (2^{k - k_{j-1} - 1} + cdots + 2^{k_{j - 2} - k_{j-1} - 1}) &= 2^{k - k_j - 1} + 2^{k_1 - k_j - 1} + cdots + 2^{k_{j - 1} - k_j - 1}. end{align*} Finally, begin{align*} h_0 circ h_1 circ cdots circ h_{k_j} (2^{k - k_j - 1} + 2^{k_1 - k_j - 1} + cdots + 2^{k_{j - 1} - k_j - 1}) &= 2^k + 2^{k_1} + cdots + 2^{k_j} &= m. end{align*} Therefore, ( h(0) = m ). (blacksquare) 2. **Claim 2:** For any ( ell ), we can find a combination of compositions ( h' ) such that ( h'(ell) = 0 ). **Proof:** Let ( n = 2^{alpha} cdot c ) where ( c ) is odd. Suppose ( ell + ell_c equiv 0 pmod{c} ) for some ( ell_c ). Let ( p(x) = g(f^{phi(c) - 1}(x)) = 2^{phi(c)} cdot x + 1 ). (Here ( phi ) is Euler's totient function.) Then note that [ p^{ell_c}(ell) = 2^{ell_c cdot phi(c)} cdot ell + 2^{(ell_c - 1) cdot phi(c)} + cdots + 2^{phi(c)} + 1 equiv ell + ell_c equiv 0 pmod{c}. ] Then we have [ f^{alpha}(p^{ell_c}(ell)) = 2^{alpha} cdot p^{ell_c}(ell) equiv 0 pmod{n} = 0. ] Hence ( h' = f^{alpha} circ p^{ell_c} ) is our desired composition. (blacksquare) From the above claims, we see that ( h circ h' ) will map ( ell ) to ( m ). Now for the infinitely many part, note that we can simply go from ( ell to 0 underbrace{to ell to 0 to dots to 0 to ell}_{text{k compositions}} to 0 to m ), and for infinitely many ( k ), we get infinitely many compositions. (blacksquare)
question:a simultaneous throw of pair of dice . find the probability of getting the total more than 7 .
answer:To find the probability of getting a total of more than 7 with a pair of dice, we first need to determine the total number of possible outcomes when two dice are thrown. Since each die has 6 faces, and each face is equally likely to come up, there are 6 * 6 = 36 possible outcomes when throwing two dice. Next, we need to count the number of outcomes that result in a total sum of more than 7. We can list these outcomes as follows: - Rolling a 3 on the first die and a 5, 6 on the second die (2 outcomes) - Rolling a 4 on the first die and a 4, 5, 6 on the second die (3 outcomes) - Rolling a 5 on the first die and a 3, 4, 5, 6 on the second die (4 outcomes) - Rolling a 6 on the first die and a 2, 3, 4, 5, 6 on the second die (5 outcomes) Now, let's count the outcomes for each case: - For a 3 on the first die: 2 outcomes - For a 4 on the first die: 3 outcomes - For a 5 on the first die: 4 outcomes - For a 6 on the first die: 5 outcomes Adding these up, we get 2 + 3 + 4 + 5 = 14 outcomes that result in a total sum of more than 7. Now, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes: Probability = Number of favorable outcomes / Total number of possible outcomes Probability = 14 / 36 To simplify the fraction, we can divide both the numerator and the denominator by 2: Probability = 7 / 18 Therefore, the probability of getting a total of more than 7 with a pair of dice is boxed{7/18} .