Number theory – divisibility check in different modules

So I am working on modular arithmetic and have encountered the following problem:

Accept that $ a cdot b mod m = c $ and I know that $ c mod n = d $, Can I still test for divisibility? $ c $ With $ d $? A quick check tells me if that works $ d mod n = 0 $ does not work – but is this review not possible or is there a general way to find out $ c | a $? Maybe, if $ m $ and $ n $ are primes?

Many Thanks!

Number theory – check of divisibility with minimal bits

Suppose we get an infinite stream of integers. $ x_1, x_2, … $,

a) Show that we can calculate whether the sum of all integers seen so far is divisible by a fixed integer $ N $ With $ O (log N) $ Bits of memory.

b) Leave $ N $ an arbitrary number, and let's assume we are given $ N $Hauptfaktorisierung: $ N = p_1 ^ {k_1} p_2 ^ {k_2} … p_r ^ {k_r} $, How would you check if $ N $ divides the product of all integers $ x_i $ So far, use as little memory as possible? Make note of the number of bits used in $ k_1, …, k_r $,

For part a) we know this for every prime number $ p ne 2, 5 $There is an integer
$ r $ so that to see if $ p $ divides a decimal number $ n $we interrupt $ n $ in $ r $-Tuple of decimal places, add these $ r $tuple and check if the sum is divisible by $ p $, But $ N $ is a fixed integer and not necessarily a prime. Is there a way to join the above sentence with any number?

For part b) for the product of $ x_i $ (call $ y $) is divisible by $ N $, then $ y $ must be divisible by everyone $ p_i ^ {k_i} $ (call $ a_i $, Since we get the prime factorization, we can only check if $ y $ is divisible by $ N $ by sharing $ y $ of each $ a_i $ and stop, if it fails, right? Would this lead to the use $ k_1 * … * k_r $ Bits?

Am I at least on the right track or am I completely wrong? Any help that understands this problem would be enormously helpful. Many Thanks.

nt.number theory – integer partitions under divisibility constraint

Consider integer partitions of $ x in mathbb {N} $ of the size $ k $ on the condition that the partition elements are different and the ratio of an element to each smaller element is a natural number.

example $ f (x, k) $:

  • $ f (17.3) = {1, 4, 12 } $
  • $ f (14,3) = {2, 4, 8 } $
  • $ f (101,4) = {1, 2, 14, 84 } $
  • $ f (4,3) = emptyset $

As @Henrik has pointed out, one limitation can be expressed as follows: $ x = a_1 + a_1 a_2 + ldots + a_1 cdots a_k $ and then factor $ x $, $ x-a_1 $, $ x-a_1 a_2 $etc. to find candidates for successive ones $ a_i $but because there are usually several options for each successive one $ a_i $Maybe you have to use a sophisticated search (with backtracking) or a variant of linear programming.

Were these partitions (or series) examined? Is there an efficient algorithm or method to find it? $ x $ and $ k $?