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There is a relevant identity we can find. The general thing we can prove is \[\sum_{n=0}^N{N\choose n}^2={2N\choose N}\] The \({2N\choose N}\) value is the \(x^N\) term of the generating function \((1+x)^{2N}\). Now use symmetry and the same generating function written as \((1+x)^N(1+x)^N\) \[\sum_{n=0}^N{N\choose n}^2=\sum_{n=0}^N{N\choose n}{N\choose N-n}\] We can find the \(x^N\) term as a sum of what can multiply to \(x^N\). For each \(0\leq n\leq N\), choose the \(x^n\) term and the \(x^{N-n}\) term from each \((1+x)^N\). Then each \({N\choose n}x^n{N\choose N-n}x^{N-n}\) contributes to the sum for the \(x^N\) term and exactly matches the summation after symmetry. Now with \(N=2025\) specifically, \[y=\sum_{n=0}^{2025}{2025\choose n}^2={4050\choose2025}\] This can be used to show \(x=y/y=1\).
Let \(a=5^\circ\) and then there is some symmetry and cancellation that happens. \[\left({\sin(45^\circ-a)+\sin(45^\circ+a)\over\cos(a)}\right)^2\] \[=\left({\sin(45^\circ)\cos(a)-\cos(45^\circ)\sin(a) +\sin(45^\circ)\cos(a)+\cos(45^\circ)\sin(a)\over\cos(a)}\right)^2\] \[=\left(2\sin(45^\circ)\cos(a)\over\cos(a)\right)^2 =\left(2\sin(45^\circ)\right)^2=\left({2\sqrt{2}\over2}\right)^2 =\left(\sqrt{2}\right)^2=2\]
Both triangles share \(\angle B\) so their area can be described in the following ratio since we know \(BA=2BD\) and \(BC=2BE\). \[{\text{Area}(\triangle ABC)\over\text{Area}(\triangle DBE)} ={{1\over2}(BA)(BC)\sin(B)\over{1\over2}(BD)(BE)\sin(B)} ={(2BD)(2BE)\over(BD)(BE)}=2\cdot2=4\]
By adding \(x^2+1\) to each side, we can complete a square. \[x^4+2x^2+1=651^y+x^2 \Rightarrow (x^2+1)^2=651^y+x^2\] Then subtract the \(x^2\) and we can factor a difference of squares. Also factor \(651=3\cdot7\cdot31\). \[(x^2+1)^2-x^2=(3\cdot7\cdot31)^y=(x^2+x+1)(x^2-x+1)\] Now consider greatest common divisor and subtract these 2 factors \[\gcd(x^2+x+1,x^2-x+1)=\gcd(x^2-x+1,2x)=g\] So \(x^2-x+1\) is always odd which means \(g\) is also odd. But also \(g\mid2x\) so \(g\mid x\). This means \[g=\gcd(x^2-x+1,x)=\gcd(x,1)=1\] This means \(A=x^2+x+1\) and \(B=x^2-x+1\) are coprime and each must be a \(y\)th power of an integer. All possible cases can be determined by the factors of \(651\) and noting that \(A>B\) and \(A-B=2x\). \[(A,B)=(651^y,1^y),(217^y,3^y),(93^y,7^y),(31^y,21^y)\] First focus on \(y=1\). We can look at the values of \(x=(A-B)/2\) which are \(325,107,43,5\). Out of these, only \(x=5\) looks reasonable to get \(A,B\) to match the values in a pair here and it turns out that it works, so \(x=5,y=1\) is a solution.
Now we would like to ensure there are no other solutions. Note that for \(y\geq2\), the smallest possible value of \(A\) is \(31^2=961\) which requires \(x>30\). If \(y\) is even, then both \(A\) and \(B\) must be squares, but we can see clearly (knowing \(x>30\)) that they lie between the consecutive squares: \[(x-1)^2<B<x^2<A<(x+1)^2\] So \(y\) must be odd and \(y\geq3\). First consider the case where \(A=651^y,B=1\). Then \(B=x^2-x+1=1\) only has solutions \(x=0,1\) and neither of these will make \(A=x^2+x+1=651^y\).
Next, in other cases, for some integers \(a,b\) from our other cases, \(B=b^y\) and \(A=a^y\geq(b+10)^y\) (since the smallest \(b-a\) is \(10\)). From this, \[a^y-b^y\geq(b+10)^y-b^y>10b^y\] Since \(A-B=a^y-b^y=2x\), \(x>5\cdot b^y\). But then \(x^2>25\cdot b^{2y}\). So now the value of \(B\) satisfies \[B=x^2-x+1>(x-1)^2>25\cdot b^{2y}-50b^y+1=25(b^{2y}-2b^y)+1\] Since \(b\geq3\), \(2b^y<b^{y+1}\). Also since \(y\geq3\), \(2y\geq y+3\), and \(b\geq3\), we find \(b^{2y}\geq9\cdot b^{y+1}\). So we continue the big inequality sequence: \[>25(9\cdot b^{y+1}-b^{y+1})+1>200b^{y+1}>b^y=B\] This is a contradiction, so no other solutions can exist.
In this cryptarithmetic puzzle, we can see that \(A=0\) or \(A=1\). If \(A=0\) then we would need a carry of \(1\). The maximum possible carry anywhere in this puzzle is \(3\) since there are \(4\) numbers added. This means \(B\geq7\). But for \(B=7,8,9\), we cannot get the carry to satisfy \(B+\text{carry}\equiv3\) modulo \(10\). So instead we must have \(A=1\). \[\begin{array}{ccccc} &1&B&C&X\\ &&1&B&C\\ &&&1&B\\ +&&&&1\\ \hline &1&3&9&4\\ \end{array}\] Now we know \(B+1+\text{carry}\) does not carry to the next column so we can only have \(B=1\) or \(B=2\). If \(B=1\), this forces \(C+B+1+\text{carry}=C+2\) to carry \(1\). We would need \(C+2+\text{carry}=19\) which cannot happen because \(C\leq9\) and \(\text{carry}\leq3\). So we must have \(B=2\). \[\begin{array}{ccccc} &1&2&C&X\\ &&1&2&C\\ &&&1&2\\ +&&&&1\\ \hline &1&3&9&4\\ \end{array}\] Now we can see \(C+2+1+\text{carry}=C+3+\text{carry}=9\) does not carry, so \(C\leq6\). In the right column, \(X+C+2+1\leq X+9\leq18\) so the largest carry possible is \(1\). We have either \(C=5\) or \(C=6\). If \(C=6\), then the right column cannot carry. But \(X+C+2+1=X+9>4\) so we would need \(X=5\) with a carry. The only possibility left is \(C=5\) forcing a carry from the right, so \(X+C+2+1=X+8=14\) and we find \(X=6\). The completed puzzle is \[\begin{array}{ccccc} &1&2&5&6\\ &&1&2&5\\ &&&1&2\\ +&&&&1\\ \hline &1&3&9&4\\ \end{array}\]
We can reduce the power of the numerator by subtracting away integer multiples of \(10\). This does create a different number we have to work with. \[\left\lfloor{10^{2025}\over10^{25}+3}\right\rfloor \equiv\left\lfloor{10^{2025}\over10^{25}+3} -10^{2000}{10^{25}+3\over10^{25}+3}\right\rfloor \equiv\left\lfloor{-3\cdot10^{2020}\over10^{25}+3}\right\rfloor\] \[\equiv\left\lfloor{-3\cdot10^{2020}\over10^{25}+3} +3\cdot10^{1975}{10^{25}+3\over10^{25}+3}\right\rfloor \equiv\left\lfloor{3^2\cdot10^{1975}\over10^{25}+3}\right\rfloor\] We have reduced the power of \(10\) by \(50\) and introduced \(3^2\). This pattern continues 39 more times to get \[\equiv\ldots\equiv \left\lfloor{3^{80}\cdot10^{25}\over10^{25}+3}\right\rfloor\] We can't quite repeat the pattern because \(3^{80}\cdot10^0\) is not a multiple of 10, but we can do something a little different. \[\equiv\left\lfloor{3^{80}\cdot(10^{25}+3-3)\over10^{25}+3}\right\rfloor \equiv\left\lfloor3^{80}-{3^{81}\over10^{25}+3}\right\rfloor\] We can find \(3^4\equiv1\) modulo 10 so \(3^{80}=(3^4)^{20}\equiv1\) modulo 10 also. The other fraction is clearly not an integer so let \({3^{81}\over10^{25}+3}=Z+\epsilon\) for an integer \(Z\) and \(0<\epsilon<1\). So now we have \[\equiv\lfloor1-(Z+\epsilon)\rfloor=1-\lceil Z+\epsilon\rceil =1-\left\lceil{3^{81}\over10^{25}+3}\right\rceil\] Now is where things get tricky. There is another little nice thing we can do, but beyond that it actually looks like something you just need large number arithmetic for. Anyway, we can compute \[{3^{81}\over10^{25}}-{3^{81}\over10^{25}+3} ={3^{82}\over10^{50}+3\cdot10^{25}}\approx1.33\cdot10^{-11}\] It's small, and (verifiable with large number arithmetic) does not make our ceiling expression overflow to the next integer so we can find (modulo 10) \[1-\left\lceil{3^{81}\over10^{25}}\right\rceil\] This still requires us to determine the 26th digit from the right of \(3^{81}\) which is not easy to do without a proper calculator. Anyway by calculator, \(3^{81}/10^{25}\approx44342648824303.77\) so taking away the higher digits, \[1-\lceil3.77\rceil=1-4=-3\equiv7\] Using a large number calculator like Python can confirm that the last digit is \(7\).
As \(x\to0\), \(x^2\to0\) and \(\cot^2(x/3)\to\infty\). Keep \(x^2\) on the numerator since taking its derivative makes it simpler and put \(1/\cot^2(x/3)\) on the denominator, so we have something in a \(0/0\) form compatible with LHopital's rule. \[=\lim_{x\to0}{x^2\over{\tan^2(x/3)}} =\lim_{x\to0}{2x\over2\tan(x/3){1\over3}\sec^2(x/3)} =3\lim_{x\to0}{x\over\tan(x/3)\sec^2(x/3)}\] This is still in the \(0/0\) form so apply LHopital's rule again. \[=3\lim_{x\to0}{1\over{1\over3}\sec^2(x/3)\sec^2(x/3) +\tan(x/3)\cdot2\sec(x/3)\cdot{1\over3}\sec(x/3)\tan(x/3)}\] \[=3\lim_{x\to0}{1\over{1\over3}\sec^4(x/3)+{2\over3}\tan^2(x/3)\sec^2(x/3)}\] Now we can directly substitute \(x=0\). \[=3\lim_{x\to0}{1\over{1\over3}\cdot1+{2\over3}\cdot0\cdot1} =3\left({1\over{1\over3}}\right)=3\cdot3=9\]
A faster way to solve this actually involves recognizing the square. \[=\lim_{x\to0}\left(x\cot(x/3)\right)^2\] Focus on the limit inside the square, which as rewritten can be split up to familiar limits. \[\lim_{x\to0}{x\cos(x/3)\over\sin(x/3)} =\lim_{x\to0}{x\over\sin(x/3)}\cdot\lim_{x\to0}\cos(x/3)\] \[=3\lim_{x\to0}{x/3\over\sin(x/3)}\cdot\lim_{x\to0}\cos(x/3) =3\cdot1\cdot\cos(0/3)=3\cdot1\cdot1=3\] So the limit is \(3^2=9\).
Square the equation with \(x\) and then we can solve it easily. \[x^2=t+{1\over t}+2=98+2=100\Rightarrow x=\pm10\] But clearly \(x=10\) since \(t>0\).
Find the largest root of the polynomial \(11x^3-111x^2-111x+11\).
By the rational root theorem, possible rational roots are \[{\pm1,\pm11\over\pm1,\pm11}\Rightarrow\pm1,\pm11,\pm{1\over11}\] We can try \(\pm1\) first since they are simplest and find that \(x=-1\) is a root. Then after division, we have \((x+1)(11x^2-122x+11)=0\). With whatever technique you choose (quadratic formula, guessing, etc), we can finish factoring to \((x+1)(11x-1)(x-11)=0\). Therefore the roots are \(x=-1,1/11,11\) and \(11\) is the largest.
Let \((0,0)\) be the coordinates of the bottom left corner. Then the top left to bottom right diagonal goes through \((0,6),(6,0)\). We can find the equation for it is \(y=6-x\). Next, the other diagonal goes through \((3,0),(6,6)\) which has equation \(y=-6+2x\). Solving for the intersection, we find it at \((4,2)\). It has \(y\) coordinate \(2\) which is \(4\) away from the top edge at \(y=6\). So the height of the triangle is \(4\) and the base (top edge) is \(6\). The area is \({1\over2}(4)(6)=12\).
There does not appear to be anything special about this linear system so we can solve it with general gaussian elimination. Here, we multiply things to common multiples to avoid fractions. \[ \left(\begin{array}{ccc|c} 5&-7&3&46\\ 2&-4&1&8\\ 6&2&-5&42\\ \end{array}\right) \sim \left(\begin{array}{ccc|c} 30&-42&18&276\\ 30&-60&15&120\\ 30&10&-25&210\\ \end{array}\right) \sim \left(\begin{array}{ccc|c} 5&-7&3&46\\ 0&-18&-3&-156\\ 0&52&-43&-66\\ \end{array}\right) \] \[ \sim \left(\begin{array}{ccc|c} 5&-7&3&46\\ 0&6&1&52\\ 0&52&-43&-66\\ \end{array}\right) \sim \left(\begin{array}{ccc|c} 5&-7&3&46\\ 0&156&26&1352\\ 0&156&-129&-198\\ \end{array}\right) \sim \left(\begin{array}{ccc|c} 5&-7&3&46\\ 0&6&1&52\\ 0&0&-155&-1550\\ \end{array}\right) \] \[ \sim \left(\begin{array}{ccc|c} 5&-7&3&46\\ 0&6&1&52\\ 0&0&1&10\\ \end{array}\right) \sim \left(\begin{array}{ccc|c} 5&-7&0&16\\ 0&6&0&42\\ 0&0&1&10\\ \end{array}\right) \sim \left(\begin{array}{ccc|c} 5&-7&0&16\\ 0&1&0&7\\ 0&0&1&10\\ \end{array}\right) \] \[ \sim \left(\begin{array}{ccc|c} 5&0&0&65\\ 0&1&0&7\\ 0&0&1&10\\ \end{array}\right) \sim \left(\begin{array}{ccc|c} 1&0&0&13\\ 0&1&0&7\\ 0&0&1&10\\ \end{array}\right) \] The solution is \(x=13,y=7,z=10\).
Find the largest integer \(x\) for which there exists an integer \(y\) such that \(x^2=y^2+27\).
Subtract \(y^2\) from each side and factor the difference of squares. \[x^2-y^2=27\Rightarrow(x+y)(x-y)=27\] Let \(A=x+y\) and \(B=x-y\). Both of these are integers so they factor \(27\). Notice that \(A+B=2x\) so maximizing \(A+B\) will find us a maximum of \(x\). Ignore the order for now, we just need to pick 2 numbers that multiply to \(27\) from the possibilities: \[(1,27),(3,9),(-1,-27),(-3,-9)\] The case that maximizes \(A+B\) is \((1,27)\) with \(2x=28\Rightarrow x=14\). This corresponds to \(y=\pm13\).
Find the number of factors of 18900 that are square-free.
First factor it \(18900=2^2\cdot5^2\cdot189=2^2\cdot3^3\cdot5^2\cdot7\). Square-free numbers cannot have a prime square as a factor, so all prime exponents must be \(0\) or \(1\). This gives 2 choices for each of 4 prime factors so \(2^4=16\).
Let \(a,b\) be the widths along the horizontal axis (left then right) and let \(c,d\) be the widths along the vertical axis (bottom then top). Then by the pythagorean theorem \[a^2+c^2=10^2,\ b^2+d^2=25^2,\ b^2+c^2=436,\ a^2+d^2=x^2\] Add the first two to form an equation, and same with the last two \[a^2+b^2+c^2+d^2=10^2+25^2,\ a^2+b^2+c^2+d^2=x^2+436\] Then we can solve \[x^2+436=10^2+25^2=100+625=725\Rightarrow x^2=289\Rightarrow x=17\] In general, the thing we have shown here is the British flag theorem.
How many \(z\in\mathbb{Z}\) satisfy \({7\over z}\leq-{4\over11}\)
Since \(z\neq0\), consider both positive and negative cases, since that determines whether inequalities need to be flipped. First \(z>0\). Multiply each side by \(11z\). \[77\leq-4z\Rightarrow z\leq{-77\over4}<0\] So both \(z>0\) and \(z<0\) which means there are no positive solutions. Now for \(z<0\). Also multiply each side by \(11z\). \[77\geq-4z\Rightarrow z\geq{-77\over4}=-19.25\] So the integers that work are \(z=-1,-2,\ldots,-19\), 19 in total.
First evaluate each greatest common divisor \[\gcd(40,92)=\gcd(40,12)=\gcd(4,12)=\gcd(4,0)=4\] \[\gcd(70,30)=\gcd(10,30)=\gcd(10,0)=10\] Then use the formula \(\text{lcm}(a,b)=ab/\gcd(a,b)\) \[\text{lcm}(4,10)={4\cdot10\over\gcd(4,10)}={40\over2}=20\]
This is doable by manually calculating powers of \(\phi\). By expanding \(\phi^n=\phi^{n-1}\cdot\phi\), we can see a general recurrence for \(\phi^n=a_n+b_n\sqrt{5}\) \[\begin{align}&a_1=b_1={1\over2}\\ &a_n={a_{n-1}+5b_{n-1}\over2}={a_{n-1}+b_{n-1}\over2}+2b_{n-1}=b_n+2b_{n-1}\\ &b_n={a_{n-1}+b_{n-1}\over2}\\\end{align}\] But this is still tedious and requires following 2 variables that depend on the previous. This number \(\phi\) is the golden ratio and related to the Fibonacci and Lucas sequences. We can show this.
Use the relationship \(\phi^2=\phi+1\). Now notice that \[\phi^n=\phi^{n-2}\phi^2=\phi^{n-2}(\phi+1)=\phi^{n-1}+\phi^{n-2}\] This relationship suggests the Fibonacci sequence, defined as \(F_0=0,F_1=1,F_n=F_{n-1}+F_{n-2}\). Notice that \(\phi^1=F_1\phi+F_0\). Next by induction, we prove \(\phi^n=F_n\phi+F_{n-1}\) for \(n\geq1\). \[\phi^n=\phi^{n-2}\phi^2=\phi^{n-2}(\phi+1)=\phi^{n-1}+\phi^{n-2} =(F_{n-1}\phi+F_{n-2})+(F_{n-2}\phi+F_{n-3})\] \[=(F_{n-1}+F_{n-2})\phi+(F_{n-2}+F_{n-3})=F_n\phi+F_{n-1}\] Now we can show \[\phi^n=F_n\left({1+\sqrt{5}\over2}\right)+F_{n-1} ={F_n+F_n\sqrt{5}+2F_{n-1}\over2}={F_{n+1}+F_{n-1}+F_n\sqrt{5}\over2}\] We can use the Lucas numbers formula \(L_n=F_{n-1}+F_{n+1}\) here. To prove it, note that the Lucas numbers are defined by \(L_0=2,L_1=1,L_n=L_{n-1}+L_{n-2}\). The base case for \(L_1\) can be shown true easily. Then by induction \[L_n=L_{n-1}+L_{n-2}=F_n+F_{n-2}+F_{n-1}+F_{n-3}\] \[=(F_n+F_{n-1})+(F_{n-2}+F_{n-3})=F_{n+1}+F_{n-1}\] So the final formula is \[\phi^n={L_n+F_n\sqrt{5}\over2}\] The problem asks for \(x=a-b\) where \(a=L_9/2\) and \(b=F_9/2\) come from \(\phi^9\). \[x=(L_9-F_9)/2=(76-34)/2=42/2=21\]
First compute \(y-1\). \[y-1={2025^3+1080^3\over2025^3+945^3}-{2025^3+945^3\over2025^3+945^3} ={1080^3-945^3\over2025^3+945^3}\] Next we can find that each of these numbers is divisible by \(135\) which allows a good amount of factoring to happen. Invert it for \(x=1/(y-1)\) and then simplify and apply \(a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)\). \[x={1\over y-1}={2025^3+945^3\over1080^3-945^3} ={135^3\over135^3}\cdot{15^3+7^3\over8^3-7^3} ={(15+7)(15^2-15\cdot7+7^2)\over(8-7)(8^2+8\cdot7+7^2)}\] \[={22\over1}\cdot{225-105+49\over64+56+49}=22\cdot{169\over169}=22\]
Find the largest prime divisor of the three digit number that is \(23\) times the sum of its digits.
Write an equation for this problem where \(a,b,c\) are the 3 digits. \[100a+10b+c=23(a+b+c)\] Then we can rearrange it apply divisibility rules to reduce the search to almost nothing. \[77a=13b+22c \Rightarrow 13b=77a-22c=11(7a-2c)\] So now we see that \(11\mid13b\) but \(11\nmid13\) so we must have \(11\mid b\). But since \(b\) is a single digit, \(b=0\). This means \(7a-2c=0\) and in single digits the only solution is \(a=2,c=7\). Therefore \(207\) is the number and we can verify \(207=(2+7)\cdot23=9\cdot23=3^2\cdot23\). So the largest prime divisor is \(23\).
If we can find out the arc angle that \(x\) subtends, we can identify its measure. Focus on the small triangle mostly outside the circle. We know one angle is \(60^\circ\). The \(84^\circ\) angle helps us find another angle of this triangle is \(180^\circ-84^\circ=96^\circ\). Then the remainding angle of the triangle (the top angle) is \(180^\circ-96^\circ-60^\circ=24^\circ\). Supplementary to it is a large angle inside the circle of measure \(180^\circ-24^\circ=156^\circ\). This large angle subtends an arc of measure \(2\cdot156^\circ=312^\circ\). Notice how this big arc is exactly the parts of the circle that \(x\) does not subtend so \(x\) subtends an arc of measure \(360^\circ-312^\circ=48^\circ\). An angle at a point on the circle has half the measure of arc it subtends so \(x=24\).
Find the difference between the min and max values of \(0.4x^3-0.6x^2-7.2x+4.5\) in the internal \([-3,4]\).
Since the interval is closed, we consider the endpoints as well as any extrema. Look at the derivative to find critical points. \[1.2x^2-1.2x-7.2=0 \Rightarrow x^2-x-6=0 \Rightarrow (x+2)(x-3)=0\] So the points we have to test are \(-3,-2,3,4\). Since we only care about the difference, ignore the \(4.5\) constant, since all it does is shift the graph. \[\begin{array}{rll} -3: & 0.4(-27)-0.6(9)-7.2(-3) & = -10.8-5.4+21.6 & = 5.4 \\ -2: & 0.4(-8)-0.6(4)-7.2(-2) & = -3.2-2.4+14.4 & = 8.8 \\ 3: & 0.4(27)-0.6(9)-7.2(3) & = 10.8-5.4-21.6 & = -16.2 \\ 4: & 0.4(64)-0.6(16)-7.2(4) & = 25.6-9.6-28.8 & = -12.8 \\ \end{array}\] So we pick the largest and smallest and find the difference \[(8.8)-(-16.2)=8.8+16.2=25\]
In the sequence \(12,y,x,40\), every term after the second is the sum of the previous two.
This gives us equations: \[x=12+y,\quad40=y+x\] Use the first to replace \(y\) in the second. \[40=(x-12)+x=2x-12\Rightarrow2x=52\Rightarrow x=26\] We can also find \(y=14\).
We can see there are \(9\) vertices arranged in a circle and carefully check than each has \(6\) edges. Then \(9\cdot6=54\) counts each edge twice so there are \(54/2=27\) edges in total.
A cyclic group of order \(4004\) is generated by \(a\). Find the order of \(a^{715}\).
We need to find some power \(p\) such that \(a^{715p}\) is the identity. Since \(a\) generates the group, the identity is \(a^{4004}\) so we need \(715p\) to be the smallest \(p\) that makes \(715p\) a multiple of \(4004\). This would be \[{4004\over\gcd(715,4004)}={4004\over\gcd(715,429)}={4004\over\gcd(286,429)}\] \[={4004\over\gcd(286,143)}={4004\over\gcd(0,143)}={4004\over143}=28\] So the order of \(a^{715}\) is \(28\).
We could do the full manual arithmetic, but notice how the digits are rotated, so each digit ends up in each of the 4 positions exactly once. This means if we add them together, we get \((5+7+8+9)(10^3+10^2+10+1)\). Now notice that \(11\times101=1001=10^3+10^2+10+1\). So \(x\) is just \(5+7+8+9=29\).
Find the number of rectangles that can be formed using integer lattice points \(x,y\). \(0\leq x\leq2,0\leq y\leq4\).
A rectangle is uniquely determined by its bounds on each axis. For those bounds, we choose 2 valid \(x\) coordinates and 2 valid \(y\) coordinates. There are \(3\) possible values for \(x\) and \(5\) possible values for \(y\). So the number of rectangles we can form is \[{3\choose2}{5\choose2}=3\cdot{5\cdot4\over2}=3\cdot10=30\]
Find the larger of two primes that sum to \(33\).
Since \(33\) is odd, we must have 1 odd and 1 even prime. The only even prime is \(2\) and it so happens that \(33-2=31\) is prime so the larger of these primes is \(31\).