/sci/ - Science & Math

[math]
\text{For each positive integer } n \text{, let}
\\
\qquad \qquad \qquad \qquad \displaystyle S_n = 1 + \frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n},
\\
\qquad \qquad \qquad \qquad \displaystyle T_n = S_1+S_2+S_3+\cdots+S_n ,
\\
\qquad \qquad \qquad \qquad \displaystyle U_n = \frac{T_1}{2}+\frac{T_2}{3}+\frac{T_3}{4}+\cdots + \frac{T_n}{n+1}.
\\
\text{Find, with proof, integers }0<a,b,c,d<1000000\text{ such that }T_{1988}=a S_{1989} - b
\\
\text{and }U_{1988}=c S_{1989} - d \text{.}
[/math]

Here's a more readable version.

>>10615013
[math]\text{Stop writing everything in a math environment}[/math]

Previous Thread >>10613291

>>10615013
T(n) = (n+1)S(n+1) - (n+1)
U(n) = (n+2)S(n+1) - 2(n+1)

S(1989) = (1989-b)/(1989-a)
S(1989) = (3978-d)/(1990-c)

>>10615077
It seems like the denominator of S(n) gets really big really fast.
I think a=b=1989 and c=1990 and d=3978 are the only solutions.

>>10615013
T_n=n/1+(n-1)/2+....+1/n
T_n-aS_{n+1}=(n-a)/1+(n-a-1)/2+....+(1-a)/n-a/(n+1)
Note that the numerator and denominator on the expression on the right always sum to n-a+1, so
T_n-aS_{n+1}=integer+(n-a+1)S_{n+1}.
So you are looking for a integer a such that
(n+1-a)S_{n+1)=integer. (For n=1988).
According to sage denominator of S_{1989}=
6331348150057119915235069099821082987306389226809354560594476756233276880155185572330417230848009667991466133946387880111736660853103021342901126468968070775518495674150647976428809702030951146780353007378945704798504589934680591955907606859720142280821774211819099517686684490454089736078513579802948840358382447020211717828453405828887095705346777754996832835162709207837126305565117558044044048726037363239121613820619792934620634305784565257270555355854818660046975512876668307463841507923795565128454525579231717448096331185154012640369356485023653819141799267510005649961799667051190501548252100568474231877268378179585811790858089862962089169743569114097711613850230322370746045213519206237047980724317373598619616609725041073645736334817783807749137173193617740186371978895068643016900495823239799921883679357025476973010709318056516029067072640000

So the first problem has no solution.

>>10615120
>(n+1-a)S_{n+1)=integer.
>the first problem has no solution.
Don't forget the case a=n+1, integer=0

>>10615013
Denominator of cS_{1989}-d is a multiple of denominator of S_{1989}.
On the other hand, sage says again that denominator of U_{1988} is larger that denominator of S_{1989}, so the second problem has no solution.

>>10615134
You are right, I missed that one.

>>10615136
You are forgetting the 0=0 case.

>>10615111 is the answer
I don't know how to prove it by hand though

>>10615144
Ok, I see the solution is here
>>10615077
I probably mistyped something in the calculation here
>>10615136

>>10615013

thinly veiled homework thread