## How to find the majority integer that is divisible by the integer 10?

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c program to find the majority element in an array

I'm writing a function "most_of" that takes a list of numbers as a argument. The objective of the function is to take the list, iterate over it and find out if the majority of the list integers are divisible by 10.

So for example, if I had passed the argument:

[1,10,10,50,5]

The output would be:

True

Because 3/5 of the integers are divisible by 10. However, if I had passed:

[1,2,55,77,6]

The output would be:

False

Because 4/5 of the list integers are not divisible by 10.

Here is what I have tried:

def most_of(lst): for i in lst: if lst[i] % 10 == 0: lst == True else: lst == False

I'm basically stuck at this point because this doesn't check if the majority of the numbers are divisible by ten, it just divides.

Thanks for the help!

Count how many integers are divisible by ten, and test whether that number is "the majority" - that is, if it's greater than or equal to half the lists' length. Like this:

def most_of(lst): num = sum(1 for n in lst if n % 10 == 0) return num >= len(lst) / 2.0

For example:

>>> most_of([1,10,10,50,5]) True >>> most_of([1,2,55,77,6]) False

**Majority Element,** Function to find Majority element. // in an array. void findMajority( int arr[], int n). {. int maxCount = 0;. int index = -1; // sentinels. for ( int i = 0; i < n; i++). {. int count c) Use the divisibility lemma to prove that an integer is divisible by 10 if and only if its last digit is 0. d) Use the divisibility lemma to prove that an integer is divisible by 4 if and only if its last two digits form a two-digit integer that is divisible by 4. For example, 4332 is divisible by 4 because 32 is divisible by 4.

The objective of the function is to take the list, iterate over it and find out if the majority of the list integers are divisible by 10.

Your list will contain two kind of integers: those that are divisible by 10 and those that aren't. You need to find the number of integers in each of the two categories, compare those numbers and return True or False accordingly. So, your function would look like this:

def most_of(lst): divisible_counter = 0 non_divisible_counter = 0 for element in lst: if element % 10 == 0: divisible_counter +=1 else: non_divisible_counter += 1 if divisible_counter > non_divisible_counter: return True else: return False

Of course, all the above code could be reduced a lot. But I wanted to show an algorithm that would be easier to understand for Python beginners.

**Nature of Mathematics,** Regression analysis The analysis used to determine the relationship between two variables. Remainder When an integer m is divided by a positive integer n, and a quotient 10 1 10 1 10 5 30 Replication On a spreadsheet, the operation of copying a formula from one place to another. Locate the rounding place digit. 1005! / 10^n: A Number Puzzle Date: 9/11/96 at 17:38:17 From: Anonymous Subject: 1005! / 10^n: A Number Puzzle Find the largest positive integer, n, such that 1005! is divisible by 10 to the power of n. I would like to know: a) how one would go about solving such a problem and b) your opinion of the grade level of the problem

A slight modification of the answer by Oscar:

def most_of(lst): return sum(1 if n % 10 == 0 else -1 for n in lst) >= 0

with the same results of course

lst1 = [1,10,10,50,5] lst2 = [1,2,55,77,6] print(most_of(lst1)) # True print(most_of(lst2)) # False

**FINDING MAJORITY FOR INTEGER ELEMENTS *,** For n input elements and a positive integer k, an element in the input is regarded as a with the input integer elements and our goal is to find the majority elements. In the previous [1][2][3][4][7][8][9][10]. In our previous study Initially n elements are divided into n1-1/2^i ordered sets by ordered partitioning [5] which takes Count how many integers are divisible by ten, and test whether that number is "the majority" - that is, if it's greater than or equal to half the lists' length. Like this: def most_of(lst): num = sum(1 for n in lst if n % 10 == 0) return num >= len(lst) / 2.0

you assign your list a bool after you test your first number, but you have to count all numbers which can divide by ten without rest and all other numbers and then compare this counters:

def most_of(lst): divideByTen = 0 otherNumbers = 0 for i in lst: if i % 10 == 0: divideByTen+=1 else: otherNumbers+=1 if(divideByTen > otherNumbers): return True else: return False a = [1,10,10,50,5] b = [1,2,55,77,6] print(most_of(a)) print(most_of(b))

**Python Challenges - 1: Exercises, Practice, Solution,** Write a Python program to check if a given positive integer is a power of two. Note: The majority element is the element that appears more than n/2 be divided by each of the numbers from 1 to 10 without any remainder. Write a C++ program that prompts the user to enter a positive integer and then uses this criteron to determine whether the number is divisible by 11. Hint: you need the following functions to solve the program.

**A Dictionary of Political Economy: Biographical, Bibliographical, ,** If and all other coins are multiples of that unit . 10 . So we may divide an integer by 2 , 3 , 4 , 5 , 3 . Now , in the first system it is Thus , from time immemorial and get And all other coins are | unit exactly by the immense majority of numbers The number p is an integer. We need to determine. Whether p is divisible by 5 or not. As we are not given any additional information about p, let us analyse the individual statements. Step 3: Analyse Statement 1. As per the information given in statement 1, p is divisible by 10. 5 is a factor of 10 – hence, if a number is divisible by 10

**GMAT Club Forum • p and q are integers. If p is divisible by 10^q ,** If p is divisible by 10^q and cannot be divisible by 10^(q + 1), what is the value of q? banksy wrote: p and q are integers. If p is These two are the majority. We have to find the total number of 5 digit integers which are divisible by 5. A number is said to be divisible by 5 if the ones digit is either 0 or 5 . Case I : Consider that the ones digit is 0 _ , and the number of ways in which the other four places can be filled is 9 × 10 × 10 × 10 = 9000 , since all the places can take all the ten

**A Complete Epitome of Practical Navigation and Nautical Astronomy, ,** to millions or more , consisting of any whole number of units , are called Integers , but aid we get better results with fewer figures than if we used vulgar fractions . for 5 divided by 10 , because 5 parts of anything that is divided into 10 parts is but in the majority of instances consist of whole numbers ( or integers ) and Interactive math practice for 4000+ skills. Proven success. Sign up today!

##### Comments

`==`

is for checking if things are equal, not for assigning values. But then you don't want to reassign the`lst`

variable anyway. You need to keep a count of how many items are and are not divisible by 10, then return`True`

or`False`

depending on which is greater.- Possibly better:
`sum(not n % 10 for n in lst)`

, since`0`

is Falsy and`bool`

is a subclass of`int`

. - As influenced by this answer I would replace the middle line by
`num = len(list(filter(lambda x: x%10 ==0, lst)))`

. Its somewhat clumsy to sum something if you are just interested at the number of elements. For huge lists memory efficiency might become an issue, but we should be factor ten below input list size. - @guidot but with your proposal we would be creating a temporary list, which is even worse from the memory-consumption perspective.
- @ÓscarLópez I admitted inferior memory-efficiency already (which is unlikely to cause serious problems for reasonably-sized lists), but increasingly welcome plain code and the question does not mention
*sum*anywhere. - @guidot it's not clumsy, it's the idiomatic way to count items that meet a condition in Python. In an old-fashioned loop, we would have a statement like
`count++;`

somewhere. Here we're doing the same, and efficiently - we process one item at a time, and keep adding`1`

for each item that meets the condition. - This is inferior in respect to elegance and does not really look like idiomatic Python (if expresssion requires no parentheses, strange white space omissions...).