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The average age group of students and teachers is \(25\). The students' average is \(20\). The teachers' average is \(35\). Find the ratio of students to teachers.
Suppose there are \(s\) students and \(t\) teachers. Then we know the sum of the students' ages is \(20s\) and the sum of the teachers' ages is \(35t\). Then when added together, we have the average as \(\frac{20s+35t}{s+t}=25\). We can take this equation and find the ratio \(s/t\).
\[\frac{20s+35t}{s+t}=25 \Rightarrow 20s+35t=25(s+t)=25s+25t \Rightarrow 10t=5s \Rightarrow \frac{s}{t}=\frac{10}{5}=2\]Find the ratio of the \(x^5\) term and the \(x^6\) term in the expansion of \((x+1)^7\).
These are just the binomial coefficients.
\[{7\choose5}/{7\choose6}=\frac{7!}{5!2!}\frac{6!1!}{7!}=\frac{6}{2}=3\]We need to find the area of \(x\). Let \(r\) be the side length of one of the squares, so we need to find \(x=r^2\). We will draw 2 lines and label some points.
Using the Pythagorean theorem, we can find \(\overline{DF}=\overline{EF}=r\sqrt{5}\). Also, \(\angle DFE\) is a right angle so \(\triangle DEF\) is isosceles. We also have that \(\triangle DEF \cong\triangle EFB\). This means \(\overline{EB}=r\sqrt{10}\). Next, we find \(\overline{CE}\) which when added to \(\overline{EB}\) will give us an expression with \(r\) that is equal to \(13/\sqrt{2}\). Consider the following triangle taken out of the diagram. We need to find its height \(h\), the red segment drawn. This will be the same as \(\overline{CE}\) because \(\overline{AC}\parallel\overline{DE}\)
Here \(h\) is an altitude so it splits this right triangle into 2 similar right triangles. Using the following ratio, we can find \(h\).
\[{h\over3r}={r\over r\sqrt{10}} \Rightarrow h={3r\over\sqrt{10}}\]So now \(\overline{CE}+\overline{EB}={3r\over\sqrt{10}}+r\sqrt{10}\). This is a leg of the large triangle, having length \(13/\sqrt{2}\) so we solve for \(r\).
\[{3r\over\sqrt{10}}+r\sqrt{10}={13\over\sqrt{2}} \Rightarrow {3r+10r\over\sqrt{10}}={13\over\sqrt{2}} \Rightarrow 13r\sqrt{2}=13\sqrt{10} \Rightarrow r=\sqrt{5}\]So now we have \(x=r^2=5\).
First, take the numbers \(36,35,\ldots,27\) and reduce them modulo \(13\). This gives us the following equivalent expression.
\[10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1\mod13\]From here, multiplying and reducing modulo 13 gets us the answer of 6.
Now use the half angle identity \(\sin^2(x)={1\over2}(1-\cos(2x))\).
\[\begin{align}&{1\over{1\over2}(1-\cos(\pi/4))\cdot{1\over2}(1-\cos(3\pi/4))} ={4\over(1-\sqrt{2}/2)\cdot(1+\sqrt{2}/2)}\\& ={4\over1^2-(\sqrt{2}/2)^2}={4\over1-1/2}={4\over1/2}=8\end{align}\]Find the area bounded by the \(x\)-axis and \(f(x)=4+2x-2x^2\).
This is a quadratic function with a negative leading coefficient and a maximum of 4 so the area we are looking for can be found by integrating between its zeroes. To find them, factor.
\[4+2x-2x^2=-2(x^2-x-2)=-2(x+1)(x-2)\]The zeroes are \(-1\) and \(2\). Next, the integral.
\[\int_{-1}^2(4+2x-2x^2)dx=\left[4x+x^2-{2\over3}x^3\right]_{-1}^2 =4(2-(-1))+(2^2-(-1)^2)-{2\over3}(2^3-(-1)^3)=9\]\(32\) times the probability that Zack, who flips two dimes, and Xander, who flips three dimes, get the same number of heads.
The number of heads possible are 0, 1, or 2. Consider the probability for each outcome of Zack's flips and multiply that by the probability that Xander flips the same number of heads.
\[{{2\choose0}\over2^2}{{3\choose0}\over2^3} +{{2\choose1}\over2^2}{{3\choose1}\over2^3} +{{2\choose2}\over2^2}{{3\choose2}\over2^3} ={1\over4}{1\over8}+{2\over4}{3\over8}+{1\over4}{3\over8}={10\over32}\]The problem asks for 32 times this value which is 10.
This turns out to be the intersection of a hyperbola and circle. The "big brain" reasoning with observing that \(122=11^2+1^2\) leads to noticing that the numbers work out nicely. But in general, it's not too easy to find solutions for intersections of general quadratic curves. We normally have to do some form of substituting one variable in place of the other. First, rewrite one of the equations.
\[x+y+xy=23\Rightarrow y(1+x)=23-y\Rightarrow y={23-x\over1+x}\]Now substitute this into the circle equation. Prepare for some algebra mess.
\[\begin{align}& x^2+\left({23-x\over1+x}\right)^2=122\\& x^2(1+x)^2+(23-x)^2=122(1+x)^2\\& x^4+2x^3+x^2+529-46x+x^2=122=244x+122x^2\\& x^4+2x^3-120x^2-290x+407=0 \end{align}\]This is difficult to factor. The rational root theorem helps us find \(x=1\) and \(x=11\), so we factor it to this.
\[(x-1)(x-11)(x^2+14x+37)=0\]The quadratic part is positive when \(x\) is positive so it will not give any solutions we care about. The only solutions in positive \(x\) are \(x=1\) and \(x=11\) which correspond to \(y=11\) and \(y=1\) respectively. Finally, we take the solution where \(x>y\) so \(x=11\).
How many non-congruent triangles?
This problem is confusing as worded and solving it is still a mystery. The least weird explanation I have seen is that it should actually be how many triangles are congruent to at least one other, so the big triangle is excluded since there is only 1 and the others are counted so \(3+9=12\).
Find the smallest postage value that cannot be placed on an envelope using at most three stamps worth 1, 2, or 5 cents.
We can solve this by brute force. With one stamp, we have \(1,2,5\) as possibilities. Then for 2 stamps, add 1, 2, or 5 to any of these values. The list of possibilities for 2 stamps is 2,3,4,6,7,10. Then we repeat for the list of possibilities for 3 stamps which is 3,4,5,6,7,8,9,11,12,15. Now we combine these numbers into a new list so the possible postage values are 1,2,3,4,5,6,7,8,9,10,11,12,15. The smallest that does not appear in this list is 13.
Unfortunately the \(\pi\) day problem does not involve \(\pi\). We can solve this problem by looking at the prime factorizations of these 3 numbers.
\[\begin{align}&882=2\cdot3^2\cdot7^2\\&336=2^4\cdot3\cdot7\\& 980=2^2\cdot5\cdot7^2\end{align}\]Then by taking the highest prime powers common to all of them, the greatest common divisor is \(2\cdot7=14\).
First, the area of the rectangular part at the bottom is \(3\cdot4=12\). Then the top part is a triangle with a base of 4 and height of 1.5 so its area is \({1\over2}\cdot4\cdot1.5=3\). The area is \(12+3=15\).
\(S=\{1,2,3,4,5\}\). Find the number of \(U\subset S\) for which \(\exists\:a<b<c\) where \(a,c\in U\) and \(b\notin U\).
Consider the opposite of the condition. For every \(a<b<c\) with \(a,c\in U\), we have \(b\in U\). This says there are no gaps in the ordering of the elements. So we need to subtract this amount from the number of nonempty subsets which is \(2^5-1=31\). When 1 is the first, there are 5 subsets with no gaps, adding the next integer each time. Starting with 2, there are 4 subsets. Continuing this reasoning, we have 5+4+3+2+1=15 contiguous subsets. The answer is \(31-15=16\). In general, if \(S=\{1,2,\ldots,n\}\), then the number of such subsets is \(2^n-1-n(n+1)/2\).
\(f(x)=\tan(x)\) and \(\theta=\arctan(4)\). Find \(f'(\theta)\).
The derivative is \(f'(x)=\sec^2(x)=1+\tan^2(x)\). Then \(f'(\theta)=1+(\tan(\theta))^2=1+4^2=17\)
Find the number of positive integer pairs \((a,b),a\leq b\) for which \({1\over a}+{1\over b}={1\over72}\).
Since, \(0<a\leq b\), \({1\over a}\geq{1\over b}\). We can use this to find bounds for \(a\). Since \({1\over a}+{1\over b}\leq{2\over a}\), we have \({2\over a}\geq{1\over72}\Rightarrow a\leq144\). Since \({1\over a}={1\over72}-{1\over b}<{1\over72}\), \(a>72\).
Now, let \(a=72+k\) for some \(0<k\leq72\). Then
\[\begin{align}& {1\over72+k}+{1\over b}={1\over72} \Rightarrow 72+k+b={(72+k)b\over72} \\& \Rightarrow 72^2+72k+\cancel{72b}=\cancel{72b}+kb \Rightarrow b={72^2\over k}+72 \end{align}\]Therefore, we need \(k\) to be a factor of \(72^2\) which is below \(72\) in order for \(b\) to be an integer. This includes the \(12\) factors of \(72\). Since \(72=2^3\cdot3^2\), we can use the prime factorization to find factors of \(72^2\) which are below \(72\) and not factors of \(72\): \(2^4,2^4\cdot3,2^5,2^6,3^3,2\cdot3^3\). This is \(6\) more factors for a total of \(18\) possible values of \(k\). Each choice of \(k\) corresponds to the integer pair \(72+k,{72^2\over k}+72\) which satisfies the equation.
How many counting numbers under \(400\) have an odd number of factors?
The numbers that have an odd number of factors are all perfect squares. To see why, consider the prime factorization \(n=p_1^{a_1}p_2^{a_2}\ldots\). The number of factors is \((a_1+1)(a_2+1)\ldots\) so we would need each of these numbers multiplied to be odd, which is equivalent to each power \(a_i\) being even. The perfect squares under \(400\) are \(1^2,2^2,\ldots,19^2\) so the answer is \(19\).
Find the number of ways that the vertices of a decagon can be labeled from 1 to 10 such that the labels on the vertices of any edge differ by no more than 2.
Consider where the 1 is put. There are 10 possible vertices to choose. Now, the 1 must be next to the 2 and 3. We have 2 ways to put these numbers next to the 1. From here, we have a chain of forcing the next pair of numbers. The 4 and 5 must go next to the 2 and 3, but there is only 1 way to put them so the labels on the same edge differ by no mare than 2. This continues with the pair 6,7 and finally the pair 8,9. The last number is 10, which is acceptable to put in its final location. Going around the decagon, the order is \(8-6-4-2-1-3-5-7-9-10\) where the 10 and 8 connect around. Since we can choose 10 places for the 1 and choose 2 arrangements for the 2 and 3 to be next to the 1, there are \(10\times2=20\) possible such labelings.
The sum of the squares of five consecutive positive integers starting with \(x\) is \(2655\).
We can solve this with the sum of squares formula which is:
\[\sum_{k=1}^nk^2={n(n+1)(2n+1)\over6}\]Let one summation have \(n=x+4\) and the other have \(n=x-1\). By subtracting them, we get the sum \(x^2+(x+1)^2+(x+2)^2+(x+3)^2+(x+4)^2\).
\[\begin{align}& \sum_{k=1}^{x+4}k^2-\sum_{k=1}^{x-1}k^2 ={(x+4)(x+5)(2x+9)\over6}-{(x-1)x(2x-1)\over6}\\& ={1\over6}(30x^2+120x+180)=5x^2+20x+30=2655 \end{align}\]This gives us a quadratic equation: \(5x^2+20x-2625=5(x+25)(x-21)=0\). So we must have \(x=21\), the positive solution. The other solution \(x=-25\) works if we allow \(x\) to be negative.
Since \(143=11\cdot13\), all terms for \(i\geq13\) are \(\equiv0\mod143\). This reduces the work to computing
\[\sum_{i=1}^{12}i!\mod143\]It requires a little bit of tedious multiplication and division, but work can be reused since if \((n-1)!\equiv a\), then \(n!\equiv an\), reducing modulo \(143\) at each step. The answer is \(22\).
First square both sides twice with a little adjustment.
\[\begin{align}& 506+(506+x)^{1\over2}=x^2 \\& (506+x)^{1\over2}=x^2-506 \\& 506+x=x^4-1012x^2+506^2 \\& x^4-1012x^2-x+505\cdot506=0 \end{align}\]Then we could tediously factor this polynomial to get \((x-23)(x+22)(x^2+x-505)=0\) which has the solution \(x=23\). We would also have to check for extraneous solutions and find that none of the others are solutions.
Alternatively, we can observe that the left side is the function \(f(x)=\sqrt{506+x}\) iterated to \(f(f(x))=x\). Since \(f(x)\) is monotonic, if we choose \(x\) to be a non fixed point, it will not return to \(x\) under iterations of \(f\). To see why,
\[f(x)<x \Rightarrow f(f(x))<f(x)<x \]Since \(f(x)\) is monotonically increasing and \(y<z\) implies \(f(y)<f(z)\). A similar result can be shown for \(f(x)>x\). Thus, we must have the fixed point \(f(x)=x\) which is easier to solve for than the quartic equation above resulting from an algebra mess.
\[\sqrt{506+x}=x \Rightarrow 506+x=x^2 \Rightarrow (x+22)(x-23)=0\]We find \(x=-22\) is an extraneous solution since a square root cannot be negative so the answer is \(x=23\).
Find the number of \(S\subset\{1,2,3,4,5,6\}\) for which the sum of \(S\)'s elements is divisible by 3.
First, split \(S\) into 3 sets by modulo 3: \(S_0=\{3,6\},S_1=\{1,4\},S_2=\{2,5\}\). Now we can choose any of the 4 subsets of \(S_0\) and add it to a subset of \(S_1\cup S_2\) and not change its sum modulo 3, so we consider subsets of \(S_1\cup S_2\) which sum to a number divisible by 3 and multiply that result by 4. Now consider 3 cases for how many elements we pick from \(S_1\). If 0, then we cannot use any of \(S_2\) because the sum would be 1 or 2 (mod 3), which is 1 possibility. If 1, then we must use exactly 1 element from \(S_2\), so this is 4 possibilities since there are 2 ways to choose 1 element from a set of 2. If 2, then we have to choose both elements in \(S_2\) as well, which is 1 possibility. This gives us an answer of \(4(1+4+1)=24\). This total includes the empty set \(\varnothing\) and \(S\) itself.
There does not seem to be the nicest solution that does not require pulling some numbers out of nowhere. Let \(a=\sqrt[3]{\sqrt{5}+2}\) and \(b=\sqrt[3]{\sqrt{5}-2}\). Then the quantity we are looking for is \((a+b)^4\). We can start by writing \((a+b)^3\). Note that \(ab=1\) which can be shown easily.
\[(a+b)^3=a^3+3a^2b+3ab^2+b^3=a^3+b^3+3ab(a+b) =(\sqrt{5}+2)+(\sqrt{5}-2)+3(a+b)\]Then if we let \(u=a+b\), this can be written as the cubic equation \(u^3=2\sqrt{5}+3u\). By some lucky guessing, we can find \(u=\sqrt{5}\) is a solution and the other 2 are not real. So the quantity we are looking for is \((a+b)^4=u^4=25\).
Another thing we can do is relating this problem to another simpler one.
\[y=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\]First, by writing \(y^3\), we will find
\[y^3=(2+\sqrt{5})+(2-\sqrt{5}) +3\left(\sqrt[3]{2+\sqrt{5}}\sqrt[3]{2-\sqrt{5}}\right)y\]That part in parenthesis simplifies to \(-1\) so simplifying the equation some more gives \(y^3=4-3y\). This is much easier to factor than the other polynomial, and we find \(y=1\) with the other 2 solutions not real.
Now how does this relate to the original problem? Notice that \(y+2\sqrt[3]{\sqrt{5}-2}=u\) using \(u\) as defined in the previous work. Let \(z=2\sqrt[3]{\sqrt{5}-2}\). Then we find \(z^3=8\sqrt{5}-16\). The next part uses a little pulling stuff out of nowhere. What if we could write \(z=c\sqrt{5}-d\) for some numbers \(c,d\)? Well then, \(z^3\) would be (if you work out a bit of algebra)
\[z^3=\sqrt{5}(5c^3+3cd^2)-d^3-15cd^2\]Notice that the coefficients sum nicely, \(5+3=8\) and \(-1-15=-16\) which matches the other way we computed \(z^3\). So we have \(c=d=1\) and find that \(z=\sqrt{5}-1\). From this, we get the same \(u=y+z=1+(\sqrt{5}-1)=\sqrt{5}\) as before.
Find the number of ways to make \(\$12.85\) with only dimes and quarters.
The number of quarters used must be odd, otherwise the amount remaining could not be created with only dimes. For every odd amount of quarters used, the remaining amount can be created with only dimes. So the smallest number of quarters that can be used is 1, and the largest is \(12\cdot4+3=51\). Counting these gives us a total of 26 possible amounts of quarters to select while creating the rest of the amount in dimes.
First, thinking about powers of 3, we can find \(3^4=81\equiv-1\mod41\). Then \(3^{123}=(3^4)^{30}\cdot3^3\equiv(-1)^{30}\cdot27\equiv27\mod41\).
Find the number of positive integers less than 100 that are not divisible by 2, 3, or 7.
This can be solved with the inclusion-exclusion principle. Start with all positive integers from 1 to 100, then subtract those which are divisible by one of the numbers. Some are taken away 2 or 3 times because they are divisible by more than 1 of those numbers. So next we add back the numbers which are divisible by 2 of them, and finally subtract the ones which are divisible by all 3. This works because the 3 numbers given are all prime, or more generally pairwise coprime.
\[\begin{align}&100-\left\lfloor{100\over2}\right\rfloor-\left\lfloor{100\over3}\right\rfloor -\left\lfloor{100\over7}\right\rfloor+\left\lfloor{100\over2\cdot3}\right\rfloor +\left\lfloor{100\over2\cdot7}\right\rfloor+\left\lfloor{100\over3\cdot7}\right\rfloor -\left\lfloor{100\over2\cdot3\cdot7}\right\rfloor\\& 100-50-33-14+16+7+4-2=100-97+27-2=28 \end{align}\]Use partial fractions to rewrite \({1\over n(n+1)}={1\over n}-{1\over n+1}\) and we see a telescoping series.
\[62\sum_{n=2}^{30}{1\over n(n+1)} =62\sum_{n=2}^{30}\left({1\over n}-{1\over n+1}\right) =62\left({1\over2}-{1\over3}+{1\over3}-{1\over4} +\ldots+{1\over30}-{1\over31}\right) =62\left({1\over2}-{1\over31}\right)=31-2=29\]Since each of \(i=1,2,\ldots,30\) are coprime to 31, \(i^{30}\equiv1\mod31\) by Fermat's little theorem which lets us simplify it to this.
\[\sum_{i=1}^{30}(i^{30})^{10}\equiv\sum_{i=1}^{30}1^{10} \equiv\sum_{i=1}^{30}1\equiv30\mod31\]Find the number of ways to partition 6 named objects into two proper subsets.
Select a subset for the first partition (\(2^6=64\) possible subsets) and put the remaining objects in the other partition. Since they must be proper subsets, neither partition can contain all of the objects so that eliminates 2 choices of subsets for the first partition (all or none). Finally, divide by \(2!\) because the order of the partitions does not matter so \(62/2!=31\).