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This could be evaluated by using the sum of integers and sum of cubes formulas. But it is even simpler if we are aware of the following well known fact \[\left(\sum_{i=1}^n{i}\right)^2=\sum_{i=1}^n{i^3}\] In this problem, \(n=2025\) but it does not matter which \(n\) we use. Both parts of the fraction are equal so the answer is 1.
Here, \(x^x\) grows faster than \(x!+x\). If we use the continuous factorial (gamma function) this problem will be more complicated. But for integers, we can see that \(x=2\) works and from there the right grows faster than the left. When \(x\geq3\) \[x^x=x\cdot x^{x-1}=x^{x-1}(x-1)+x^{x-1}>x!+x\] Here we can see \(x^{x-1}>x\) since \(x-1\geq2\). Also \[x^{x-1}(x-1)=x(x-1)\cdot x^{x-2}>x(x-1)\cdot(x-2)^{x-2}\geq x(x-1)(x-2)! =x!\] So in integers the only solution is \(x=2\). A plot of the graphs suggests that when using the gamma function this is still true but that would require other techniques to prove.
First use the base change formula to write it with natural logarithms. \[\begin{align}&={\ln(100)\over\ln(30)}+{\ln(10)\over\ln(30)} +{\ln(27)\over\ln(30)}={\ln(100)+\ln(10)+\ln(27)\over\ln(30)}\\ &={\ln(27000)\over\ln(30)}={\ln(30^3)\over\ln(30)}={3\ln(30)\over\ln(30)}=3\\ \end{align}\]
To evaluate this, we can factor the expression and look at the pattern \[\begin{align} \prod_{n=2}^x{n^2\over(n-1)(n+1)}&={2\cdot2\over1\cdot3}{3\cdot3\over2\cdot4} {4\cdot4\over3\cdot5}\ldots{(x-1)(x-1)\over(x-2)x}{x\cdot x\over(x-1)(x+1)}\\ &={2\over1}{2\cdot3\over3\cdot2}{3\cdot4\over4\cdot3} \ldots{(x-1)x\over x(x-1)}{x\over x+1}={2x\over x+1}\\ \end{align}\] And we can see this is \(2\) as \(x\to\infty\). To prove this more formally, we can use induction. When \(x=2\) it is true trivially. Then assume \[\prod_{n=2}^x{n^2\over(n-1)(n+1)}={2x\over x+1}\] Now we can find \[\begin{align}&\prod_{n=2}^{x+1}{n^2\over(n-1)(n+1)} ={2x\over x+1}{(x+1)^2\over x(x+2)}={2(x+1)\over x+2}\end{align}\] Which shows that the formula is correct. So the product inside the parenthesis is \(2\) and the answer is \(2^2=4\)
Using the digit squares modulo 10, the only possibilities are \(x=0,1,5\) since the ones digit comes from \(X^2\). If \(X=0\), then we would need \(B=C=0\) so the digits are not distinct. If \(X=1\), then both 3 digit numbers must be equal so we would have \(A=C\), not distinct. By elimination it must be \(X=5\).
To solve for what the original letters are, start with \(X=5\). Then \[(100A+10B+5)\times5=500+50B+25\] This must be a 3 digit number so \(A=1\). Now try again with \(X=5,A=1\). \[(100+10B+5)\times5=500+50B+25=100C+10B+5\] We can derive the following \[520+40B=100C\] Which shows \(520+40B\) must be a multiple of \(100\). So the only possibilities are \(B=2,B=7\). With \(B=2\) we can see \(125\times5=625\) which has \(C=6\). With \(B=7\) we can see \(175\times5=875\) which has \(C=8\). So there are 2 solutions both with \(X=5\).
By observing a pattern, we can see \[y={1\over1\cdot2}+{1\over2\cdot3}+{1\over3\cdot4}+{1\over4\cdot5} +{1\over5\cdot6}+{1\over6\cdot7}=\sum_{n=1}^6{1\over n(n+1)}\] By using partial fractions this becomes \[\sum_{n=1}^6\left({1\over n}-{1\over n+1}\right)\] This is a telescoping series and evaluates to \(1-1/7=6/7\). The problem asks for \(7\) times this which is \(6\).
This is a linear system. First simplify the 2nd equation by dividing 5 \[x+y=13,\quad2x+y=20\] The \(y\) coefficientns match so we can cancel it. Subtract the first equation from the second \[(2x+y)-(x+y)=20-13\Rightarrow x=7\] This one turns out to be simple to solve with the right choices.
The smaller circle has \(1/2\) of the radius of the larger circle so its area is \((1/2)^2=1/4\) of the larger circle. Then we divide by 2 because the shaded area is half of the smaller circle. So the ratio is 8 to 1.
The fractions follow a pattern \[y={1\over2}{2\over3}\ldots{9\over10}={1\over\cancel{2}} {\cancel{2}\over\cancel{3}}\ldots{\cancel{9}\over10}={1\over10}\] Then \(x=1/y=1/(1/10)=10\)
Please see the January 20, 2025 problem for the work associated with the following result: \[\cos\left({\pi\over5}\right)={1+\sqrt{5}\over4}\] So then we just need the double angle formula \[\cos\left({2\pi\over5}\right)=2\cos^2\left({\pi\over5}\right)-1 =2\left({1+\sqrt{5}\over4}\right)^2-1={6+2\sqrt{5}\over8}-1 ={\sqrt{5}-1\over4}\] Then \[y={1+\sqrt{5}\over4}-{\sqrt{5}-1\over4}={1\over2}\] And finally \[x=\left({1\over2}\right)^{-3}+\left({1\over2}\right)^{-1}+1=8+2+1=11\]
Find the maximum area of a rectangle whose sides are natural numbers and perimeter is 14.
Let \(a,b\) be the side lengths. Then the constraint is \(2a+2b=14\Rightarrow a+b=7\) and we maximize \(ab=a(7-a)=7a-a^2\). Its derivative is \(7-2a\) so the maximum occurs at \(a=3.5\). When \(a<3.5\) the derivative is positive and when \(a>3.5\) the derivative is negative so the function is strictly decreasing as we move away from \(a=3.5\). So we only have to test the integer points closest to this whicha re \(a=3\) and \(a=4\). Both give a rectangle with area \(3\times4=12\).
Using the lengths shown, for the upper diagram we have \[x+\triangle=13+\square\] and for the lower diagram we have \[x+\square=17+\triangle\] Adding them we get \[2x+\triangle+\square=30+\triangle+\square\] So \(2x=30\Rightarrow x=15\).
Find the sum of the roots of \((x-7)(x-5)+(x-7)(x-15)\)
Simplify to \[(x-7)((x-5)+(x-15))=(x-7)(2x-20)=0\Rightarrow x=7,10\] So the sum of roots is \(7+10=17\)
Find the count of base 3 three digit numbers.
The smallest is \(3^2\) and the largest is \(3^3-1\) so the total is \[1+(3^3-1-3^2)=3^3-3^2=27-9=18\]
\(x\) is the largest integer such that \(1\) through \(37\) can be partitioned into \(x\) subsets, all with the same sum.
The sum is \(37(38)/2=37\cdot19\) so \(x\) must be a factor of this, meaning \(x=1,19,37\) are possible, since \(x\leq37\). Clearly \(x=37\) is wrong because each number is different. Partitioning it into \(x=1\) subsets would work. But there is also a way to partition it into \(x=19\) such subsets: \[\{1,36\},\{2,35\},\{3,34\},\ldots,\{17,20\},\{18,19\},\{37\}\] Therefore the answer is \(x=19\).
How many natural numbers leave a remainder of \(4\) when divided into \(364\)?
This is counting how many \(m\) satisfy \[364\equiv4\pmod{m}\] Which means \[m\mid(364-4)\Rightarrow m\mid360\] Factoring we have \(360=2^3\cdot3^2\cdot5\) so it has \(4\cdot3\cdot2=24\) divisors. But we also need \(m>4\) to have a remainder of \(4\) so we exclude the divisiors \(1,2,3,4\) leaving 20 divisors.
Find the coefficient of the \(x^2\) in the Taylor expansion of \(f(x)=e^{3x^2+6x}\) at \(x=0\)
The \(x^2\) term has coefficient \(f''(0)/2!\) so we need some derivatives \[f'(x)=e^{3x^2+6x}(6x+6)\] \[f''(x)=e^{3x^2+6x}(6x+6)^2+e^{3x^2+6x}\cdot6\] Then \(f''(0)/2=(6^2+6)/2=(36+6)/2=42/2=21\)
The expression for \(y\) is the largest number in the "factorial base" system with 21 digits. In general, we can show by induction that \[\sum_{n=1}^x n\cdot n!=(x+1)!-1\] Clearly it is true for \(x=1\) so then \[\sum_{n=1}^{x+1}n\cdot n!=(x+1)!-1+(x+1)(x+1)!=(x+2)(x+1)!-1=(x+2)!-1\] completes the induction. Therefore, \(y=22!-1\) so \(x=22\).
One thing we can do to start is adding the 2 equations and then adding 1 which allows us to do a nice factorization \[yz+y+z+1=288+41+1=330\Rightarrow(y+1)(z+1)=330\] Next substitute \(y=41-z\) so we have a quadratic equation to solve \[(42-z)(z+1)=330\Rightarrow-z^2+41z+42=330\Rightarrow z^2-41z+288=0 \Rightarrow(z-9)(z-32)=0\] So \(z=9\Rightarrow y=32\) and \(z=32\Rightarrow y=9\) which makes sense since the 2 equations are symmetric. In both cases, \(x=|y-z|=23\).
Find the number of ways to make 99 cents using some of the coins from 15 dimes, 15 nickels, 15 pennies.
Subtracting pennies from 99 must leave a remaining total divisible by 5. So we can only use 4, 9, or 14 pennies. If 4 pennies are used, then we can make the remaining 90 cents with 2-9 dimes and nickels for the rest (8 ways) since with 0 or 1 dimes there would not be enough nickels. If 9 pennies are used, then we can make 85 cents with 1-8 dimes and nickels for the rest (8 ways). And if 14 pennies are used, we can also make 80 cents with 1-8 dimes and nickels for the rest (8 ways). In total, there are 24 ways.
Let \[f(x)=1+x+x^2+x^3+\ldots={1\over1-x}\] then \[f'(x)=1+2x+3x^2+\ldots=\sum_{n=1}^\infty nx^{n-1}={1\over(1-x)^2}\] Now notice that \[36y=\sum_{n=1}^\infty{n\over6^{n-1}}=\sum_{n=1}^\infty n(1/6)^{n-1} =f'(1/6)={1\over(1-1/6)^2}={36\over25}\] So we have \(y=1/25\) and \(x=25\)
Let \[x=\sqrt{171-\sqrt{1352}}+\sqrt{171+\sqrt{1352}}\] Then \[x^2=2\cdot171+2\sqrt{171^2-1352}=342+2\cdot167=676\] Finally we need the positive root so \(x=26\)
How many multiples of 4 are between 55 and 166?
The first multiple of 4 in the interval is 56 and the last is 164. To count how many are multiples of 4, we have \[1+(164-56)/4=1+108/4=1+27=28\]
We can rearrange the left a bit \[\left({29\over30}\right)^{-29}=\left({30\over29}\right)^{29} =\left(1+{1\over29}\right)^{29}\] So this matches with \(x=29\)