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| 21 | 22 | 23 | 24 | 25 | 26 | 27 |
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Since \(23\) is prime, by Wilson's theorem: \[22!\equiv-1\ (\text{mod}\ 23)\] Now multiply each side by the inverse of \(22\) module \(23\) which is \(-1\equiv22\). \[21!\cdot(22\cdot-1)\equiv-1\cdot-1\ (\text{mod}\ 23) \Rightarrow 21!\equiv1\ (\text{mod}\ 23)\]
It might not be clear, but each letter is a base 10 digit forming a 5 digit number. So the problem is asking us to find a number \(xyzwv\) that reverses to \(vwzyx\) when multiplied by \(4\). This is a cryptarithmetic puzzle that requires us to solve for individual digits of the number, and we are also given that the digits must be distinct.
One observation to make is that the digit to digit carry can never exceed 3, since we start with at most \(4\cdot9=36\) which could carry 3, and at any further step, \(4\cdot9+c\leq39\) since \(c\leq3\). Also for simplicity, assume all modular congruences are mod \(10\) unless stated otherwise. Below is what the multiplication looks like, with the carries. We will solve for each variable. \[ \begin{array}{cccccc} & (c_4) & (c_3) & (c_2) & (c_1) & \\ & x & y & z & w & v \\ \times & & & & & 4 \\ \hline & v & w & z & y & x \\ \end{array} \] Now start with the digits \(v\) and \(x\). We cannot have a carry value so we end up with \(v=4x+c_4\) for some carry \(c\). This means \(x=0,1,2\). Next, \(4v\equiv x\) so we can further narrow it down to \(x=0,2\) since \(4v\) is even. Now think about some cases. If \(x=0\), then \(v=5\) to satisfy \(4v\equiv x\) because \(x\neq v\). But then \(v=4x+c_4=c_4\) would mean we have to carry \(5\), bigger than the limit of \(3\) we established, so we must have \(x=2\) instead. Now suppose \(x=2\). Then we could have either \(v=3,8\) to satisfy \(4v\equiv x\). Since \(v=4x+c_4=8+c_4\geq8\), we can narrow this down to \(v=8\). So only one case remains: \(x=2,v=8\). We could stop here since we found \(x\), but similar reasoning can solve this full digit reversal puzzle. Here is what remains. \[ \begin{array}{cccccc} & (0) & (c_3) & (c_2) & (3) & \\ & 2 & y & z & w & 8 \\ \times & & & & & 4 \\ \hline & 8 & w & z & y & 2 \\ \end{array} \] Next we focus on solving for \(w\) and \(y\). The equations for them are \[4w+3\equiv y,\quad 4y+c_3\equiv w\] Rearrange the second equation to \(w\equiv4y+c_3\) and then add the equations. \[5w+3\equiv 5y+c_3 \Rightarrow 5(w-y)\equiv c_3-3\] From this, we know \(c_3-3\equiv0,5\). But given that \(c_3\in\{0,1,2,3\}\), it must be \(c_3-3\equiv0\Rightarrow c_3\equiv3\). We can also determine that \(w-y\) is even in order to make \(5(w-y)\equiv0\). So our multiply and carry equations become \[4w+3\equiv y,\ 4y+3\equiv w \Rightarrow 3(w+y)\equiv-6 \Rightarrow w+y\equiv-2\] Now combine this with our observation that \(w-y\) is even, and respect the constraint that each digit is unique and we have four cases: \((y,w)=(1,7),(3,5),(5,3),(7,1)\). Next focus on \(z\) in the middle. We have \(4z+c_2\equiv z\Rightarrow 3z+c_2\equiv0\). Since \(c_2\in\{0,1,2,3\}\), we can find possible values \(z=0,3,6,9\). But remember that we found \(c_3=3\), so \(4z+c_2\geq30\). The only value which can possibly give us a carry \(3\) is \(z=9\). That also allows us to show \(4\cdot9+c_2\equiv9\) implies \(c_2=3\). Here is the puzzle with our new information. \[ \begin{array}{cccccc} & (0) & (3) & (3) & (3) & \\ & 2 & y & 9 & w & 8 \\ \times & & & & & 4 \\ \hline & 8 & w & 9 & y & 2 \\ \end{array} \] Next, we use our knowledge of the almost complete puzzle with some steps of the multiplication to reduce the cases on \((w,y)\). Since \(c_4=0\), \(4y+3=w\) must give us a single digit \(w\), so the only possibility is \(y=1\). Therefore, the case that works is \(y=1\) and \(w=7\). We can substitute these into our puzzle and check that it is a valid multiplication. \[ \begin{array}{cccccc} & (0) & (3) & (3) & (3) & \\ & 2 & 1 & 9 & 7 & 8 \\ \times & & & & & 4 \\ \hline & 8 & 7 & 9 & 1 & 2 \\ \end{array} \]
This can be solved with some properties of right triangles. First label some vertices and two angles.
By the Pythagorean theorem, \(BC=20\). Then \(BE=EC=10\) since \(\triangle BCD\) is isosceles. Now notice that the angles labeled \(\theta\) are equal, so \(\triangle ABC\sim\triangle EDB\). Both triangles are 3-4-5 right triangles and \(\theta\) is attached to the longer leg so the side lengths of \(\triangle EBD\) are \(7.5,10,12.5\). We actually only need the shorter leg, \(7.5\) which is the height of the blue triangle. We now can use the base height formula for its area and solve the problem. \[25x={1\over2}(20)(7.5)=75\Rightarrow x=3\]
A rectangle is tiled by squares with distinct side lengths. How many neighbors does the smallest square have?
If the smallest square were in a corner, then we would not be able to put 2 larger squares next to it and fill in all gaps. After the first square, it would look something like this.
If the smallest square were on an edge, we can try to fill in gaps by putting 2 larger squares next to it, but we will run into the same problem where we cannot fit a larger square on the 3rd side.
So we could only have the smallest square somewhere in the middle in an arrangement that looks a bit like this. Therefore it must have 4 neighbors. Each edge can only be bordered by 1 other square. If 2 squares bordered an edge of the smallest square, then we would run into the same gap problem with the case that the smallest square is on an edge.
The last more optional part is showing existence. It turns out to be very challenging to actually find a rectangle tiled by unique sized squares. Below is the smallest example possible with 9 squares tiling a 33 by 32 rectangle. This is part of a large dataset available at here on a website dedicated to "squaring the square" and related problems.
To make this a little easier to read, let \(a=\sqrt{y}\) and \(b=\sqrt{z}\). The first equation becomes \(a+b=3\) and we can use this to simplify the 2nd equation to 1 variable. \[\sqrt{a^2+5}+\sqrt{b^2+3}=5 \Rightarrow \sqrt{a^2+5}+\sqrt{(3-a)^2+3}=5\] Then square twice with some rearranging to eliminate the square roots. \[a^2+5+(3-a)^2+3+2\sqrt{(a^2+5)((3-a)^2+3)}=25\] \[2a^2-6a+17+2\sqrt{(a^2+5)((3-a)^2+3)}=25\] \[(a^2+5)((3-a)^2+3)=(4+3a-a^2)^2\] \[a^4-6a^3+17a^2-30a+60=a^4-6a^3+a^2+24a+16\] \[8a^2-27a+22=0 \Rightarrow a={27\pm\sqrt{27^2-4\cdot8\cdot22}\over16} ={27\pm5\over16}\] For \(a=11/8\) we find \(b=13/8\). For \(a=2\) we find \(b=1\). For the first, we get \((y,z)=(121/64,169/64)\). For the second, we get \((y,z)=(4,1)\). Only the latter satisfies \(y>z\). So \(x=y+z=4+1=5\).
If \(x<0\) or \(y<0\) then the sum is below \(1\) so either \(x\geq0\) or \(y\geq0\). If \(x\geq0\) and \(y<0\) then we would have a non integer result. If \(y\leq0\) and \(x<0\) it would also be a non integer. So both \(x\geq0\) and \(y\geq0\).
We will step through values of \(x\) and compute \(4224-4^x\) until we recognize a power of \(2\).
| \(x\) | \(4224-4^x\) | Power of 2? |
|---|---|---|
| \(0\) | \(4223\) | no |
| \(1\) | \(4220\) | no |
| \(2\) | \(4208\) | no |
| \(3\) | \(4160\) | no |
| \(4\) | \(3968\) | no |
| \(5\) | \(3200\) | no |
| \(6\) | \(128\) | yes |
So we find the solution \(x=6\) and \(y=7\).
Our outer hexagon has side length \(s={8\over\sqrt[4]{27}}\). An equaliteral triangle with side length \(s\) has area \({\sqrt{3}\over4}s^2\) which can be found by drawing an altitude from one vertex to the midpoint of the opposite side. Using the many tick marks for line length equality, we can see that the middle size blue triangle has side length \(s/2\) an the small one has side length \(s/4\). So adding the areas of these 3 equilateral triangles: \[{\sqrt{3}\over4}\left({8\over\sqrt[4]{27}}\right)^2 \left(1+\left({1\over2}\right)^2+\left({1\over4}\right)^2\right)\] \[={\sqrt{3}\over4}\cdot{64\over\sqrt{27}}\left(1+{1\over4}+{1\over16}\right) ={16\over3}\cdot{21\over16}=7\]
Find the largest zero of \(x^3-7x^2-10x+16\).
By the rational root theorem, rational zeroes might be \(x=\pm1,\pm2,\pm4,\pm8\). By trying, we find \(x=1\) is a root. Then divide \(x-1\) to factor the polynomial to \((x-1)(x^2-6x-16)\). Now we can factor the quadratic one more easily and the fully factored polynomial is \((x-1)(x+2)(x-8)\). The zeroes are \(x=1,-2,8\) and the largest is \(x=8\).
What is the max area of a rectangle with perimeter 12?
If a rectangle has sides \(x\) and \(y\) then its perimeter is \(2x+2y=12\) and area is \(xy\). We try to maximize \(xy\) subject to \(x+y=6\). Then \(xy=x(6-x)=6x-x^2\). This is a quadratic polynomial with a negative leading coefficient so it has a maximum at \(x={-(6)\over2(-1)}=3\). This also gives us \(y=3\) so the area is maximized at \(xy=9\).
Start with \(7!\) and see what numbers following that which we can factor out of \(6!\). The factorization of \(6!\) is \[6!=2\cdot3\cdot4\cdot5\cdot6=2^4\cdot3^2\cdot5 =8\cdot(2\cdot3^2\cdot5)=8\cdot9\cdot(2\cdot5)=8\cdot9\cdot10\] So then \[7!\cdot6!=7!\cdot8\cdot9\cdot10=10!=x!\Rightarrow x=10\]
Notice that 43 is half of 86 so double equation 1. \[4x+6y=x^2-y^2+2y\] Separate \(x\) and \(y\) \[y^2+4y=x^2-4x\] Notice that if we add 4, we can factor them as perfect squares \[(y+2)^2=(x-2)^2\] Now if we take square roots, we have 2 cases. First \[y+2=x-2\Rightarrow y=x-4\] Then use the first given equation: \[2x+3(x-4)=43\Rightarrow5x-12=43\Rightarrow x=11\Rightarrow y=7\] For the other case, \[y+2=-(x-2)=2-x\Rightarrow y=-x\] Then in the first given equation: \[2x+3(-x)=43\Rightarrow x=-43\Rightarrow y=43\] This satisfies the first 2 given equations, but not \(x+y>0\). Therefore, the solution is \(x=11\) and \(y=7\).
Start with the left side and do some algebraic manipulation. \[\left(1-{1\over x}\right)^{1-x}=\left({x-1\over x}\right)^{1-x} =\left({x\over x-1}\right)^{x-1}=\left(1+{1\over x-1}\right)^{x-1}\] This clearly matches the structure to give us \(x-1=12\Rightarrow x=13\). In order to show that this is the only solution, we need to analyze the function more rigorously. To do this, let \(f(x)=\left(1+{1\over x}\right)^x\). We will show that \(f(x)\) is increasing on \((0,\infty)\) with range \((1,e)\) which shows our \(x-1=12\) must be unique in this interval. Then we will show that \(f(x)\) is increasing on \((-\infty,-1)\) with range \((e,\infty)\) which shows that there cannot be any more solutions in that interval. Note that \(f(x)\) is undefined at \(x=-1\) and \(x=0\) and that we have a negative exponent base in \((-1,0)\) which makes \(f(x)\) generally complex valued in that interval and not well behaved for real numbers.
We will also state a few other pieces of useful things that will help us in this rigorous analysis. We already defined our function to analyze: \[f(x)=\left(1+{1\over x}\right)^x\] Now let's define the log of it, and compute its 2 derivatives. \[g(x)=x\ln\left(1+{1\over x}\right),\quad g'(x)=\ln\left(1+{1\over x}\right)-{1\over x+1},\quad g''(x)={-1\over x(x+1)^2}\] Since \(f(x)=e^{g(x)}\) and \(e^x\) is monotonic, the monotonicity of \(g(x)\) and \(f(x)\) will be the same. To show monotonicity, we will show \(g'(x)\) has a fixed sign in an interval, which implies \(g(x)\) is monotonic, and finally \(f(x)\) is monotonic.
First, consider the interval \((0,\infty)\). We will first determine the limits. On the right, we have the well known limit \[\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\left(1+{1\over x}\right)^x=e\] On the left, we can do this by starting with taking the log, then change the form a bit so it matches the \(\infty/\infty\) case of LHopital's rule. \[\ln(L)=\lim_{x\to0^+}x\ln\left(1+{1\over x}\right) =\lim_{x\to0^+}{\ln\left(1+{1\over x}\right)\over{1\over x}}\] Now use LHopital's rule and simplify. \[=\lim_{x\to0^+}{{-x^{-2}\over1+{1\over x}}\over-x^{-2}} \cdot{-x^2\over-x^2}=\lim_{x\to0^+}{1\over1+{1\over x}}\cdot{x\over x} =\lim_{x\to0^+}{x\over x+1}=0\] So we find that \(L=e^0=1\). This establishes the endpoints of the range on the interval. Now we will show that \(f(x)\) is increasing on this interval. Focus on \(g'(x)\). We can substitute directly to find \[\lim_{x\to\infty}g'(x)=\ln(1)-0=0\] Then notice how \(g''(x)<0\) on this interval since \(x\) and \(x+1\) are positive. This tells us that \(g'(x)\) decreases on \((0,\infty)\) approaching zero, which means \(g'(x)>0\) and \(g(x)\) is increasing on the interval. Because \(g(x)\) preserves monotonicity, \(f(x)\) increases on \((0,\infty)\) with the range \((1,e)\).
Now we look at the behavior or \((-\infty,-1)\). First the infinite limit. Let \(t=-x\). \[\lim_{x\to-\infty}\left(1+{1\over x}\right)^x =\lim_{t\to\infty}\left(1-{1\over t}\right)^{-t} =\lim_{t\to\infty}\left(t\over t-1\right)^t=L\] Now take the log and evaluate it using LHopital's rule with the \(0/0\) form. \[\ln(L)=\lim_{t\to\infty}t\ln\left({t\over t-1}\right) =\lim_{t\to\infty}{\ln(t)-\ln(t-1)\over{1\over t}}\] \[=\lim_{t\to\infty}{{1\over t}-{1\over t-1}\over{-1\over t^2}} =\lim_{t\to\infty}{{-1\over t(t-1)}\over{-1\over t^2}} =\lim_{t\to\infty}{t^2\over t(t-1)}=1\] So then \(\lim_{x\to-\infty}f(x)=L=e^1=e\). Now we find the other limit. \[L=\lim_{x\to-1^-}\left(1+{1\over x}\right)^x\Rightarrow \ln(L)=\lim_{x\to-1^-}x\ln\left(1+{1\over x}\right)\] Notice that \(x\) is negative and \(|x|>1\). Also notice that \(0<1+{1\over x}<1\) but gets close to zero as \(x\to-1^-\). We can formalize it more rigorously, but there is always an \(x\) closer to \(-1\) we could pick to make \(\ln(1+1/x)\) as close to \(-\infty\) as we want. This is multiplied by the negative and bounded \(x\) so the limit is \(\infty\), implying \[\lim_{x\to-1^-}f(x)=L=e^\infty=\infty\] For the monotonicity behavior, we can direct substitute to find that \[\lim_{x\to-\infty}g'(x)=\ln(1)+0=0\] Then we find \(g''(x)>0\) on our interval because \(x<0\) and \((x+1)^2>0\). This means \(g'(x)\) "starts at" \(0\) toward the beginning of the interval and increases. That tells us that \(g'(x)>0\) so \(g(x)\) is increasing on \((-\infty,-1)\). Finally, the monotonicity behavior is preserved by the log operation so \(f(x)\) is increasing on \((-\infty,-1)\) with range \((e,\infty)\).
Bringing this back to the original problem, \(f(x)\) in well defined only on \((-\infty,-1)\cup(0,\infty)\). On those intervals, it has distinct monotonic ranges \((e,\infty)\) and \((1,e)\). Therefore, if we find a solution on either interval, it must be unique.
How many factors are there fore 320?
\[320=32\cdot10=2^5\cdot2\cdot5=2^6\cdot5\] Then we can pick a power for each prime (including 0) so counting factors: \[(6+1)(1+1)=7\cdot2=14\]
We are not give this, but we have to assume that \(\angle ACB\) is a right angle angle on a square. The angles are the same regardless of scaling so let \(AC=CB=1\) and \(AB=CD=\sqrt{2}\). If we extend the horizontal line through \(B\) to the right more, we can form a right angle and an angle equal to \(\angle CAB\). This will tell us that \(m\angle CBD=135^\circ\). Now we can use the law of sines. \[{\sin(135^\circ)\over\sqrt{2}}={\sin(D)\over1}\Rightarrow \sin(D)={\sqrt{2}/2\over\sqrt{2}}={1\over2}\] From this, we find \(D\) is at an angle of \(30^\circ\). Now we know 2 of the angles in the triangle with \(x^\circ\) and can solve: \[x=180-135-30=15\] One little thing about extending the problem is that if instead we were given the ratio \(r\) of \(BC\) to \(AC\), we could instead find the angle \(x\), and come up with a few different formulas for it (equivalent on \(0,\infty\)). Stewart's theorem is relevant as well as various annoying algebra with right triangle trigonometry. \[\arctan\left({1\over r}\right)-\arcsin\left({r\over1+r^2}\right),\quad \arccos\left({r\over1+r^2}\right)-\arctan(r)\] \[\arccos\left({r+r\sqrt{r^4+r^2+1}\over\left(r^2+1\right)^{3/2}}\right),\quad \arctan\left({r^2+1-\sqrt{r^4+r^2+1}\over r^3}\right)\]
A coin is flipped until there are either 4 heads in a row or a tail followed by 3 heads. What is the reciprocal of the chance that the last flip is heads?
This problem seems to be stated incorrectly. As stated, it gives ending conditions of \(HHHH\) or \(THHH\) which always ends in heads. There are some possible alternatives it might have meant.
One possibility is that it meant to ask for the probability that \(HHHH\) occurs before \(THHH\). In this case, it can only happen in the first 4 flips (probability \(1/16\)). If a \(T\) shows up, \(THHH\) would be completed first.
Another possibility is it meant "a tail" instead of "a tail followed by 3 heads" which would still require \(HHHH\) in the first 4 flips. If a tails showed up, then it ends right there. The probability of this is \(1/16\).
Find the smallest positive integer that is twice the sum of its digits.
Single digits have a sum equal to themselves so it cannot be one of those. Try solving an equation for 10-19, using \(x\) as the ones digit. \[2(1+x)=10+x \Rightarrow 2+2x=10+x \Rightarrow x=8\] Therefore, the smallest such number is 18.
Find the smallest prime that is the sum of three different primes.
If \(2\) was included, then the other primes would be odd and the sum would be even, so we can only have odd primes. The first 3 primes are \(3+5+7=15\) which is not prime. Now consider if \(11\) is allowed. \[3+5+11=19,\quad3+7+11=21,\quad5+7+11=23\] We get \(19\). We have already explored all possibilities with primes up to \(11\). The next smallest sum of 3 primes is \(3+5+13=21>19\) so there are no more that need to be checked.
Take logs, the base is actually not important, but use the same one. \[u\log(2)=\log(81)=4\log(3),\quad v\log(3)=\log(32)=5\log(2)\] Then multiply the equations. \[uv\log(2)\log(3)=20\log(2)\log(3)\Rightarrow uv=20\]
One way to solve this could be to recognize one side is a square. But it turns out that we still have to eliminate square roots so that does not really help us because we will be squaring each side anyway. \[x^2=(x+28)\left(2-\sqrt{4+x}\right)^2=(x+28)\left(8+x-4\sqrt{4+x}\right)\] \[=x^2+36x+224-4(x+28)\sqrt{4+x} \Rightarrow \sqrt{4+x}={9x+56\over x+28}\] Now we square both sides again and deal with large numbers. \[(4+x)(x+28)^2=(9x+56)^2\] \[x^3+60x^2+1008x+3136=81x^2+1008x+3136\] \[x^3+60x^2=81x^2\Rightarrow x^3-21x^2=0\Rightarrow x^2(x-21)=0\] From here, we find \(x=0\) and \(x=21\) and it turns out \(x=0\) is extraneous and \(x=21\) is a solution.
Of the 64 ways to flip a coin 6 times, how many have more heads than tails?
The number of ways to flip \(6\) coins with \(n\) tails is \({6\choose n}\). So we just need \(n\) (number of tails) to be greater than \(6-n\) (number of heads). \[{6\choose4}+{6\choose5}+{6\choose6}={6\cdot5\over2}+{6\over1}+1=15+6+1=22\]
Put both sides in base 2 exponents first, then we can take logs to solve them. \[2^{x^2+1}2^{528+4x}=2^{50x}\Rightarrow x^2+4x+529=50x\Rightarrow x^2-46x+529=0\Rightarrow(x-23)^2=0\] So the solution is \(x=23\).
Given that the inscribed shapes are a regular triangle, and regular pentagon, we already know many of the angle measures. Note that they should have marked the top angle of the pentagon as the same measure as well.
If we draw a radius from the shared vertex to the circle center, it bisects the angles of measure \(60^\circ\) and \(108^\circ\). If we focus on the side with \(x^\circ\), then the angles are \(30^\circ\) and \(54^\circ\). So we can find the difference: \(x=54-30=24\).
We are seeing some symmetry here. Consider the terms at index \(t\) and \(-t\) \[{1\over1+24^t}+{1\over1+24^{-t}}={1\over1+24^t}+{24^t\over24^t+1} ={1+24^t\over1+24^t}=1\] So we can simplify opposite terms to \(1\). The sum becomes \[{1\over2}+{1\over1+24^0}+\sum_{t=1}^{24}1={1\over2}+{1\over2}+24=25\]
Noticing the symmetry, add all 3 equations and then it becomes clear how to solve for \(x\). \[21+25+19=5(a+b+c)=65\Rightarrow a+b+c=13\] Then \(x=2(a+b+c)=2\cdot13=26\).
Find the number of non-prime factors of \(1200\).
First factor it. \[1200=2^4\cdot3\cdot5^2\] The number of factors is \((4+1)(1+1)(2+1)=30\). Then subtract away the primes \(2,3,5\) and there are 27 factors which are not prime.
To solve this, break it up into coprime factors \(100=4\cdot25\) to use the Chinese remainder theorem. Since \(2^2\) is a factor, \[2^{18}\cdot3^{27}\equiv0\ \text{mod}\ 4\] Finding things modulo \(25\), start with \(2^{18}\). By going up in some powers of \(2\), we find \(2^{7}=128\equiv3\) is convenient. Then \[2^{18}\equiv2^7\cdot2^7\cdot2^6\equiv3\cdot3\cdot16\equiv9\cdot-9\equiv-81 \equiv-6\ \text{mod}\ 25\] Then for \(3^{27}\), we can use \(3^3=27\) as a convenient starting point. \[3^{27}\equiv(3^3)^9\equiv2^9\equiv512\equiv12\ \text{mod}\ 25\] Then finally, \(-6\cdot12\equiv-72\equiv3\ \text{mod}\ 25\). Now since the numbers are small, we can cycle through multiples of \(25\) to find that the solution is \(2^{18}\cdot3^{27}\equiv28\ \text{mod}\ 100\).
We can square things to eliminate square roots. \[x+7+x-25+2\sqrt{(x+7)(x-25)}=64\] \[\sqrt{(x+7)(x-25)}=41-x\] \[(x+7)(x-25)=x^2-18x-175=1681-82x+x^2\] \[64x=1856\Rightarrow x=29\]
Find the largest number \(x\) that is a factor of \(5250\) and \(1260\) and satisfies \(\gcd(x,1001)=1\).
If \(x\) is a factor of those 2 numbers, then it is a factor of their gcd. \[\gcd(5250,1260)=\gcd(1260,210)=\gcd(210,0)=210\] Now factor \(210=2\cdot3\cdot5\cdot7\) and \(1001=7\cdot11\cdot13\). A factor of \(210\) that is coprime to \(1001\) must not contain the factor \(7\) so we take as many of the other factors we can: \(2\cdot3\cdot5=30\).
Find the largest \(y\) that satisfies \(z+y\leq64,\:z\geq y+2\).
Write the 2nd one as \(y+2\leq z\). Add the inequalities since \(a\leq b\) and \(c\leq d\) implies \(a+b\leq c+d\). \[2y+z+2\leq64+z\Rightarrow 2y\leq62\Rightarrow y\leq31\] So the largest is \(y=31\). If \(y>31\), then we could show \(z<33\) from the first inequality, but also \(z>33\) from the 2nd inequality.