Recent content by physics kiddy

Homework Statement It's a question from IE Irodov's General Physics --> An elevator car whose floor to ceiling is equal to 2.7m starts ascending with constant acceleration 1.2m/s2. 2s after the start a bolt begins to fall from the ceiling of the car .Find (a) the time after which bolt...

Well, I have found a solution : Let us assume that 42n-1 is divisible by 15. So, we have 16n-1 is divisible by 16-1 ... Since 16n-1 = 16n - 1n ... it takes the form of an-bn is divisible by a-b. Just prove that it is possible and you have it ...

Here's the problem : Let X = {1,2,3,4 ... 10}. Find the number of pairs {A,B} such that A \subseteq X and B \subseteq X, A \neq B and A \cap B = {5,7,8}. My attempt: Once we know that the remaining numbers are 1,2,3,4,6,9,10 ... a total of 7 numbers, we can use permutation to know that...

Let me see if I can do ... Once we have figured out that 100! has 2^97 * 3^48 in it. Factorise 12^19. It's (2*3*2)^49 = 2^98*3^49. So the number is 2*3 = 6...

Yea, that also works and even better, no need to search other factors. Thanks a lot. I tried it also.

Oh my god! It was awfully easy: Factors of 88 --> 1*88 2*44 4*22 8*11 (n+10+k)(n+10-k) = 88 Assuming n+10+k = 44 and n+10-k = 2; we have 2n+20=46 =>2n=26 => n = 13 And putting it in n2+20n+12, we have 441 whose root is 21 ... 1st number = 13 again taking n+10+k = 22 and n+10-k...

(n+10+k)(n+10-k) = 88 Since difference of factors is 3 ... so, (n+10+k)-(n+10-k)=3 => 2k=3 =>k=2/3

I guess next we need to factorize 88. It's 11*2^3 ..

That's what I can do without much effort ... it's (n+10+k)(n+10-k) = 88 What next ?

Homework Statement Let Sn = n2+20n+12, n is a positive integer. What is the sum of all possible values of n for which Sn is a perfect square ? Homework Equations The Attempt at a Solution Well, I tried to factorise it : n2+20n+12 = n2+20n+100-88 =(n+10)2-88. And I conclude that...

I got it. Thanks a lot for teaching me something new. Such fabulous proofs of number theory make me a fan of it. I want to master it but don't know where to begin. Please suggest me some books. Thanks again ...

I now understand what you people want to say ... but I don't find any way to solve the problem that way ... For example I considered the number 6 = 2 * 3. Simply, I can say that 2(3) when divided by k = 6 is divisible by k as n = 1 here but k = (6)^2 can't divide 2(3). How do I put it in...

I request you to consider it once again. Let me be clear with it ... If m(m+1) = k^n, then m(m+1)/k*k ... n = 1 alright upto here ? Then it's clear that k must divide either m or m+1 but it can't divide both. I don't believe there's anything wrong with it. I would like to seek opinion of...

I have found a solution to the problem. It was a bit difficult to type, so I wrote it and attached two .jpg files. Let me know if it's correct ...

I believe you proved it. Since k divides either of m or m+1, it is not expressible in the form of kn because then it would have to divide both m and m+1. Is that enough or needs a better proof ? Thanks for the hint ...