Recent content by mathworker

Hint: if you use 16bit - 1) 0x10 + 0xfe = 0x10e 2) 0x20 + 0xfe = 0x11e And in 8 bit: 1) 0x10 + 0xfe = 0x0e 2) 0x20 + 0xfe = 0x1e So it is more about overflow than reset.

Hello I am reading "The Theory of Numbers, by Robert D. Carmichael" and stuck in an exercise problem, Find numbers x such that the sum of the divisors of x is a perfect square. I know sum of divisors of a x = p_1^{{\alpha}_1}.p_2^{{\alpha}_1}...p_n^{{\alpha}_1} is Sum of divisors...

B is playing a game in which he has to choose a number A between 0 and 1/2 and his probability of winning if he choose A is A^2. I have encountered this at middle of some-other problem and couldn't make a move...

Prove that least two humans named John should have born on same day(means with same D.O.B) if world has only 24 million humans and their alphabet chart contain just 4 letter 'J','H','O' and 'N' and there can't be more than one million humans of same name and world started just 2739 years ago...

I am studying Introduction to linear Algebra by Gilbert Strang while calculating the particular solution P for $Ax=b$,he made the free variables $0$ to calculate the particular solution and said that P along with linear combinations of null space solutions make up the complete set. I understood...

In, \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc} L.H.S to be greater than one (a,b) should be (2,3) substituting them rest is linear equation \frac{1}{2}+\frac{1}{3}+\frac{1}{c}=1+\frac{1}{6c} c=5

yeah I got it max value is not 1 its \frac{13}{12} - - - Updated - - - But,I guess (2,3,5) is the only solution...(Smirk)

Since modulo is "zero" there is no remainder. \frac{(ab-1)(bc-1)(ca-1)}{abc} is not a fraction abc-a-b-c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc} is not a fraction So the little terms must sum it up to zero or 1 so, \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=0(let)...

Re: ALGEBRA Profit problem. look at the below quotes carefully,

as C_0,C_1,C_2,...,C_n are combinatorial coeffs of (1+x)^n. n is certainly an integer which is \geq 1 . the rest is obvious...:)

use (2^n-1)*n\geq\frac{(2^n-1)}{n} and then AM\geq GM

As you have not changed x to u you can't change limts from \int_0^1 to \int_1^2:rolleyes:

Thanks for the link,I have gone through it. But as far as I understood partition function doesn't give the number of partitions of specific cardinality.I mean if we want only the partitions that contains r terms for example or can we define a restricted partition function that can do the job?If...

Find the number of different combinations of r natural.numbers that add upto n I tried this for quite a fair amount.of.time but.couldn't figure it out.(Punch)

Re: find f and t It may not be an ellegent one i just gave it a shot$$ \begin{align*}c^2+2c+1&=4a^4\\c&=2a^4-b^2\end{align*}$$ I gave a try at a=2 so b is 5 and c is 7. And 7 is also the first prime to satisfy both equations even individually too. So r is -52 and t is -210