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How many times can you subtract 10 from 100?
This one is an april fools joke. The string "10" appears once in "100" so it can be removed once.
Find the area of the right triangle with hypotenuse 3 and whose legs differ in length by 1.
Let the legs be \(x,y\) with \(x<y\) and \(y-x=1\). Then \[x^2+y^2=3^2\Rightarrow x^2+(x+1)^2=9\Rightarrow x^2+x-4=0 \Rightarrow x={-1\pm\sqrt{17}\over2}\] Then the area is \[{1\over2}xy={1\over2}\left({\sqrt{17}-1\over2}\right) \left(\sqrt{17}+1\over2\right)={1\over8}(17-1)=2\]
The first equation has a difference of squares and can be factored \[(2x+y)(2x-y)=35\] Then divide the other equation to get \[2x-y=7\] Now add this with the 2nd given equation \[(2x+y)+(2x-y)=12\Rightarrow4x=12\Rightarrow x=3\]
Start by multiplying \(2x\) \[x^2+16=8x\Rightarrow x^2-8x+16=0\Rightarrow(x-4)^2=0\] So the solution is \(x=4\).
First move the 2 to the other side then square \[\sqrt{x^2+24}=x+2\Rightarrow x^2+24=x^2+4x+4\Rightarrow 4x=20\Rightarrow x=5\]
We can also write the equation as \[x!=20x^2\] By trying a few small values, we can find \(x=6\) is the solution. From this point, we would expect \(x!>20x^2\) for every \(x>6\) which can be shown by induction. \[(x+1)!=(x+1)x!>20x^2(x+1)>20(x+1)^2\] Since \(x^2>x+1\) when \(x>6\).
\[ABCDX1=3\times1ABCDX\] \(A,B,C,D\), and \(X\) are distinct digits.
When multiplying the right side, \(3\times X\) must have remainder \(1\), so the only possibility is \(X=7\). We do not need to evaluate this further, but if we do then we find \(ABCD=4285\)
From \(3\times X=21\), we carry 2 so by remainder, \(3D+2=X=7\Rightarrow D=5\). Then \(3D+2=17\) so 1 is carried. Next is \(3C+1=D=5\Rightarrow C=8\). Next we carry 2 and \(3B+2=C=8\Rightarrow B=2\). Next 0 is carried because \(3B+2=8\) so we get \(3A=B=2\Rightarrow A=4\) (remainder mod 10). From here, we carry 1 so \(3+1=A=4\) which works out.
For the geometric sequence \(4,12,36,108,\ldots\) which term of the sequence is \(8748\)?
The sequence starts at \(4\) and terms after that are multiplied by 3 so the \((n+1)\)th term is \(4\cdot3^n\) so we solve \[4\cdot3^n=8748\Rightarrow3^n=2187\Rightarrow n=7\] So it is the 8th term of this sequence that is 8748.
How many roots of \(x^{12}-4096\) are non real complex numbers?
One way we can solve this is by derivative tests. The derivative is \(12x^{11}\) so it has a minumum at 0 and we can tell the function is strictly decreasing when \(x<0\) and strictly increasing when \(x>0\). So it decreases from \(\infty\) to \(-4096\) crossing the \(x\) axis once, then increases from \(-4096\) to \(\infty\) crossing the \(x\) axis again. We find that 2 of the zeroes are real and the other 10 must be complex.
Another method is starting by noticing the nice factorization \[x^{12}-2^{12}=(x^6+2^6)(x^6-2^6)=(x^6+2^6)(x^3+2^3)(x^3-2^3)\] \[=(x^2+4)(x^4-4x^2+16)(x+2)(x^2-2x+4)(x-2)(x^2+2x-4)\] We can see \(x=\pm2\) are real solutions and the quadratic formula shows that the quadratic factors have no real solutions. For the quartic factor, substitute \(y=x^2\) then observe that \(y^2-4y+16\) has no real solution for \(y\), so \(x=\pm\sqrt{y}\) cannot be real.
Start by factoring the middle equation and using trigonometric identities. \[{60\over61}=\cos^4(\theta)-\sin^4(\theta)=(\cos^2(\theta)-\sin^2(\theta)) (\cos^2(\theta)+\sin^2(\theta))=\cos(2\theta)\] Now we can draw a triangle with angle \(2\theta\) which has adjacent 60, hypotenuse 61, and opposite 11. We can use the double angle formula for tangent to find \(y=\tan(\theta)\) with a quadratic equation. \[\tan(2\theta)={11\over60}={2y\over1-y^2}\Rightarrow{11\over60}(1-y^2)=2y \Rightarrow11-11y^2=120y\Rightarrow(11y-1)(y+11)=0\Rightarrow y={1\over11},-11\] So because of the restriction on \(\theta\), we know \(y>0\) so \(y={1\over11}\). Then \[x=\cot(\theta)={1\over\tan(\theta)}={1\over y}=11\]
The mean of
The equations from this are \[x+9=2w,\quad2x+15=2y,\quad3x+18=2z,\quad w+y+z=60\] The simplest way to solve this is by adding the first 3 equations \[6x+9+15+18=2(w+y+z)=2\cdot60=120\Rightarrow6x=120-42=78\Rightarrow x=13\]
How many triangles are there with integer side lengths and a perimeter of 23?
Let \(a,b,c\) be the side lengths and order them \(0<a\leq b\leq c\). We need \(a+b+c=23\) and \(a+b>c\). Consider possible values of \(c\). If \(c\leq7\), then the maximum possible perimeter is \(21\) so \(c>7\).
If \(c=8\), then the only possibility is \(a=7,b=8\), so this is 1 triangle.
If \(c=9\), then \((a,b)\) can be \((7,7),(6,8),(5,9)\) so 3 triangles.
If \(c=10\), then we count \((6,7),(5,8),(4,9),(3,10)\) so 4 triangles.
If \(c=11\), there are 6 triangles: \((6,6),(5,7),(4,8),(3,9),(2,10),(1,11)\).
When \(c\geq12\), we cannot have \(a+b>c\), so there are no more to count. In total there are 14 triangles.
Zia was asked to subtract 3 from a number then divide by 9. Instead he subtracted 9 then divided by 3 to get 43. What is the correct answer?
The starting number can be solved for: \[{x-9\over3}=43\Rightarrow x-9=129\Rightarrow x=138\] Then doing the operations that were supposed to happen: \[{x-3\over9}={138-3\over9}={135\over9}=15\]
Find the number of squares with perimeter 12 that are needded to cover a square with side length 12.
If the perimeter is 12, the side length is 3 so the area is 9. If the side length is 12, the area is 144. So then \(144/9=16\) so we need 16 of these smaller squares. These align to a \(4\times4\) grid in the bigger square.
Find the largest area that can be formed by a rectangle with side length that are integers and with perimeter 18.
If the side lengths are \(a,b\) then we need \(2a+2b=18\Rightarrow a+b=9\). The function to maximize is \(ab=a(9-a)=9a-a^2\). Its derivative is \(9-2a\) which means the maximum occurs at \(a=4.5\). Moving away from this point causes the area to decrease so we consider the closest integer solutions of \(a=4,5\). In both cases, we rectangle with side lengths \(4,5\) and area \(20\).
We are given the fibonacci sequence and a summation with it. The \(f_9\) denominator is constant so focus on the sum of fibonacci squares. There is a formula which can be proven by induction \[\sum_{i=1}^nf_i^2=f_nf_{n+1}\] So \[\sum_{i=1}^8{f_i^2\over f_9}={1\over f_9}f_8f_9=f_8\] By evaluating further terms, we can find \(f_8\) \[f_3=2,f_4=3,f_5=5,f_6=8,f_7=13,f_8=21\]
Square the first equation and we find \[16=(a+b)^2=a^2+b^2+2ab=9+2ab\Rightarrow2ab=7\] Factor the third equation and we find \[x=(a+b)(a^2-ab+b^2)=4(9-ab)=36-4ab=36-14=22\]
Solving the first equation \[1=23t+23^2\Rightarrow23t=1-23^2=-22\cdot24\] Then we can substitute it to find \(x\) \[x={23\cdot24\over23t+23\cdot24}={23\cdot24\over-22\cdot24+23\cdot24} ={23\over-22+23}=23\]
Four numbers including \(x\) average to \(120\). The sum of the other three numbers is four times the value of \(x\).
This problem has a mistake. Instead of "average" it should say "sum".
Let \(y\) be the sum of the other 3 numbers. Then the 2 equations are \[{x+y}=120,\quad y=4x\] Then \(5x=120\Rightarrow x=24\).
Notice the pattern for each consecutive pair of numbers \[-1-1-\ldots-1+x=0\Rightarrow-25+x=0\Rightarrow x=25\]
Two years ago, Mei's age was a multiple of 6. Last year, it was a multiple of 5. If Mei is under 30, what is her age now?
If \(x\) is her age, then \(x-2\) is a multiple of 6 and must be under 30 so the possibilities are \(0,6,12,18,24\). We need to find one such that adding 1 (to get \(x-1\)) results in a number divisible by 5. We find 1 possibility so \(x-2=24\Rightarrow x=26\).