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...
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...
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 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 ...