why did you add 4 when a number is divisible, what i mean is: so 7744 is divisible by 32, and then u said we have XXXX47744<-- why add a 4 to each one of them?
either of the below holds, then it is divisible by 2^(n+1): 1. the last n digits is divisible by 2^n but not by 2^(n+1), and the (n+1)th digit is odd 2. the last n digits is divisible by 2^(n+1), and the (n+1)th digit is even
we know that 112 is divisible by 8 because 12 is divisible by 4 but not by 8, and 1 is odd
yeah i know, but the question just wanted a general expression i guess
since k can be any value greater than or equal to 9, then we can't really find the number of k-digit lovely numbers divisible by 512 unless we're given k, correct?
@DHMO I suppose. The question is exactly what I said though, and I interpreted it as saying the diatomic compounds they form are more stable in the first case.
A have another number theory question @DHMO, I think we discussed it a few days ago, but here it is: Fix k >= 2 a positive integer. Find the number of positive integers less than b^3 which are divisible by b+1 and do not contain any even digits in their base b representation..
@Semiclassical Random stuff. It's most been redoing the previous things because exams are coming and I want an equilibrium between math and other school stuff.
One piece of test-taking strategy: If you pick specific values for $p,q$ then the form of the answers is such that they'll all be different. So if you can find a pair of $p,q$ giving a simple answer, then you'll know which answer must be correct.
Okay. My point is that the phrase 'the distance from the midpoint of the base to the side of the triangle' means precisely the length of this segment FC.
Now, you could also do this problem for general p,q. For that, you'd want to use the characterization of F you gave earlier, i.e. F is chosen so that AFD is a right triangle.
This is a strange definition, considering the context. A surface of resolution need not be an embedded surface. Like, take $C$ to be the curve $(x - 1/2)^2 + z^2 = 1$ and rotate it around the $z$-axis. That has two singularities.
@Alex It is the group of cosets of $n\Bbb Z$. That is, it is the group with underlying set $\{a + n\Bbb Z : a \in \Bbb Z\}$, with addition defined as $(a + n\Bbb Z) + (b +n\Bbb Z) = (a + b) + n\Bbb Z$.
@Fargle Thanks. This might be a stupid question, you say that $\mathbb{Z}/ n \mathbb{Z}$ is the group of cosets of $n \mathbb{Z}$. How is $n \mathbb{Z}$ defined?
The problem is: Prove that there is an infinity of $a$'s satisfying this property: $a\vert (x+y)^5-x^5-y^5\Leftrightarrow a\vert(x+y)^7-x^7-y^7$ for all $(x,y)$
I know that the only solutions for $a$ are the one who can be writen as $2^n$...
