All Questions
22
questions
2
votes
5
answers
848
views
$33^{33}$ is the sum of $33$ consecutive odd numbers. Which one is the largest? (Q25 from AMC 2012)
The number $33^{33}$ can be expressed as the sum of $33$ consecutive odd numbers. The largest of these odd numbers is
$\mathrm{A.}\ 33^{32} +32$
$\mathrm{B.}\ 33^{31} +32$
$\mathrm{C.}\ 33^{32} -32$
$\...
- 749
6
votes
1
answer
212
views
Find the remainder when the sum is divided by $1000$
Find $S \pmod{1000}$ given: $$S = \sum_{n=0}^{2015} n! + n^3 - n^2 + n - 1$$
$$S_0 = 0! + 0 - 0 + 0 -1 = 0$$
$$S_1 = 1! + 1 - 1 + 1 - 1 = 1$$
$$S_2 = 2! + 8 - 4 + 2 - 1 = 7$$
This isn't helping, so:...
- 11.2k
2
votes
0
answers
150
views
Evaluate this product $n \times \frac{n-1}{2} \times \dots \times \frac{n-(2^k-1)}{2^k}$
For $k = \lfloor \log_{2}(n+1) \rfloor - 1$ evaluate
$n \times \frac{n-1}{2} \times\frac{n-3}{4} \times \frac{n-7}{8} \times \dots \times \frac{n-(2^{k}-1)}{2^k}$
So the product goes up to $k$ and I ...
- 69
1
vote
1
answer
127
views
On $\lfloor\sqrt n \rfloor+ \sum_{j=1}^n \lfloor n/j\rfloor$ [duplicate]
How do we prove that $\Big[\sqrt n \Big]+ \sum_{j=1}^n \bigg[ \dfrac nj\bigg]$ is an even integer for all $ n \in \mathbb N$ ? (where $\Big[ \space \Big]$ denotes the "greatest integer" function)
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0
votes
1
answer
417
views
amortized analysis
a) define f(k) as the largest power of 2 that divides k.
For example, f(25) = 1, f(42) = 2, f(144) = 16.
What is ${1 \over k}\sum_1^k f(k)$?
b) define f(k) as the square of largest power of 2 that ...
- 147
9
votes
3
answers
2k
views
Which number was removed from the first $n$ naturals?
A number is removed from the set of integers from $1$ to $n$. Now, the average of remaining numbers turns out to be $40.75$. Which integer was removed?
By some brute force, I got $61$. I want to know ...
- 1,949
1
vote
2
answers
133
views
Intuition for the following change of index of summation
I am working through Concrete Mathematics. I came across the following change in index of summation while going through the number theory chapter.
$$\sum_{m|n}^{ } \sum_{k|m}{ } a_{k,m} = \sum_{k|n}^...
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