For a project, I'm trying to find the sum of the multiples of both 3 and 5 under 10,000 using Swift. Insert NoobJokes.
Printing the multiples of both 3 and 5 was fairly easy using a ForLoop, but I'm wondering how I can..."sum" all of the items that I printed.
for i in 0...10000 {
if i % 3 == 0 || i % 5 == 0 {
print(i)
}
}
(468 individual numbers printed; how can they be summed?)
Just a little walk through about the process. First you will need a variable which can hold the value of your sum, whenever loop will get execute. You can define an optional variable of type Int or initialize it with a default value same as I have done in the first line. Every time the loop will execute, i which is either multiple of 3 or 5 will be added to the totalSum and after last iteration you ll get your result.
var totalSum = 0
for i in 0...10000 {
if i % 3 == 0 || i % 5 == 0
{
print(i)
totalSum = totalSum + i
}
}
print (totalSum)
In Swift you can do it without a repeat loop:
let numberOfDivisiblesBy3And5 = (0...10000).filter{ $0 % 3 == 0 || $0 % 5 == 0 }.count
Or to get the sum of the items:
let sumOfDivisiblesBy3And5 = (0...10000).filter{ $0 % 3 == 0 || $0 % 5 == 0 }.reduce(0, {$0 + $1})
range : to specify the range of numbers for operation to act on.
here we are using filter method to filter out numbers that are multiple of 3 and 5 and then sum the filtered values.
(reduce(0,+) does the job)
let sum = (3...n).filter({($0 % 3) * ($0 % 5) == 0}).reduce(0,+)
You just need to sum the resulting i like below
var sum = 0
for i in 0...10000 {
if i % 3 == 0 || i % 5 == 0 {
sum = sum + i
print(i)
}
}
Now sum contains the Sum of the values
Try this:
var sum = 0
for i in 0...10000 {
if i % 3 == 0 || i % 5 == 0 {
sum = sum + i
print(i)
}
}
print(sum)
In the Bottom line, this should to be working.
var sum = 0
for i in 0...10000 {
if i % 3 == 0 || i % 5 == 0 {
sum += i
print(i)
}
}
print(sum)
Related
My question is as on the title. I'm trying to print even numbers from 1 to 500 suing a while loop and break keyword. Below is my best possible answer I can think of, but this only print number 2. I've been spending hours but I wasn't able to solve it.
var number = 0
while true{
number += 2
print(number)
if number % 2 == 0 && number <= 500 {
break
}
}
You can use Stride
for evenNumber in stride(from: 0, through: 500, by: 2) {
print(evenNumber)
}
To specifically do this with while and break:
var i = 0
while true {
print(i)
i += 2
if i > 500 {
break
}
}
for i in 0...500 {
if i % 2 == 0 {
print(i)
}
}
Use below code
var numbers = 0...500
for number in numbers {
if number % 2 == 0 {
print(number)
}
}
I think it's easier to use build-in stride
let arr = Array(stride(from: 0, to: 502, by: 2))
print(arr)
//
For manually
var counter = 0
var arr = [Int]()
while counter <= 500 {
if counter % 2 == 0 {
print(counter)
arr.append(counter)
}
counter += 1
}
var number = 0
while true {
number += 2
print(number)
// ↓ Your code goes here ↓
if number > 499 {
break
}
}
For some reason, it always gave me the wrong result. It's always isItPrime = true no matter what number was assigned to the "number" variable.
This is my code:
let number = 6
var i = 1
var isItPrime: Bool?
while i < number {
if number % i == 0 {
isItPrime = false
} else {
isItPrime = true
}
i += 1
}
print(isItPrime)
Can somebody explain to me what's wrong with my code and why the isItPrime bool outputs always true ?
Problem 1
The last iteration of your while loop
while i < number {
if number % i == 0 {
isItPrime = false
} else {
isItPrime = true
}
i += 1
}
does overwrite the result.
So you always end up with the following result
if number % (number-1) == 0 {
isItPrime = false
} else {
isItPrime = true
}
Problem 2
Finally every number can be divided by 1, so you should start i from 2.
So
let number = 6
var i = 2
var isItPrime = true
while i < number {
if number % i == 0 {
isItPrime = false
break
}
}
print(isItPrime)
Refactoring
You can write a similar logic using Functional Programming
let number = 5
let upperLimit = Int(Double(number).squareRoot())
let isPrime = !(2...upperLimit).contains { number % $0 == 0 }
Because isItPrime is overwritten in subsequent iterations, the last number which is checked, which is number - 1 will always set isItPrime to true, because number and number - 1 are coprime.
Instead of saving the value to a boolean, just end the loop when you found out that the number is not a prime:
let number = 6
var isItPrime: Bool = true
for i in 2 ..< number {
if number % i == 0 {
isItPrime = false
break // end the loop, as we know that the number is not a prime.
}
}
print(isItPrime)
When dealing with problems like this, don't be afraid to take out a piece of paper and manually see what is going on in your loop.
Your loop will go from i = 1 to number = 5 (because of the < operator.
With that in mind, we perform each iteration manually.
for i = 1, number = 6
6 mod 1 = 0, isItPrime = false
for i = 2, number = 6
6 mod 2 = 0, isItPrime = false
for i = 3, number = 6
6 mod 3 = 0, isItPrime = false
for i = 4, number = 6
6 mod 4 = 2, isItPrime = true
Last iteration of the loop, for i = 5, number = 6
6 mod 5 = 1, isItPrime = true
There we can see that the problem is that the last iteration will always have a module of 1, therefore resulting in in your else clause getting executed.
It is always returning true because your while loop isn't working the way you want. Currently, it loops until i is 1 less than number. During that final run through the loop, number % i == 0 is false, so your code sets isItPrime to true.
To fix this problem, try this code:
let number = 6
var i = 2
var isItPrime: Bool?
while (i < number || isItPrime == false) {
if number % i == 0 {
isItPrime = false
} else {
isItPrime = true
}
i += 1
}
print(isItPrime)
You may have noticed I set i to 2, because any number modulo (%) 1 is 0
I think it's worth pointing out, however, that:
You should probably make this a method
If you initially set isItPrime to true, you can dispense with the else part of your if-else statement
Hope this helps!
// MARK: - Function
func primeNo(_ num:Int,_ divisor :Int = 2){
if divisor == num{
print("Num is prime")
}else{
if num%divisor != 0{
primeNo(num, divisor + 1)
}else{
print("num is not prime")
}
}
}
// MARK: - Use
primeNo(6)
out Put
num is not prime
How do I find all the numbers divisible by another number in swift that have a remainder of 0? This is a Fizzbuzz related question.
Lets say that...
let number = 150
And I want to do something like...
print("Fizz") // for all the numbers where the remainder of number % 3 == 0.
So if number was 15, it would print "Fizz" 5 times.
This will work
let number = 150
for num in 1...number {
if num % 3 == 0 {
print("Fizz :\(num)")
}
}
you can just loop through the number and check with your desired divisible number if the remainder is 0 then print fizz
let number = 15
for i in 0..<number {
if i % 3 == 0 {
print("\(i) Fizz")
}
}
It will print Fizz 5 times with the i value, that which number is Fizz.
Simply try this code: (You can simply replace num with any Int number and divider that is also an Int value which is used to divide all numbers till num. )
override func viewDidLoad() {
let num:Int = 15
let divider:Int = 3
var counter:Int = divider
while counter <= num {
print("Fizz")
counter += divider
}
}
func fizzbuzz(number: Int) -> String {
if number % 3 == 0 && number % 5 == 0 {
return "Fizz Buzz"
} else if number % 3 == 0 {
return "Fizz"
} else if number % 5 == 0 {
return "Buzz"
} else {
return String(number)
}
}
https://www.hackingwithswift.com/guide/ios-classic/1/3/challenge
I am trying to find the odd numbers and a multiple of 7 between a 1 to 100 and append them into an array. I have got this far:
var results: [Int] = []
for n in 1...100 {
if n / 2 != 0 && 7 / 100 == 0 {
results.append(n)
}
}
Your conditions are incorrect. You want to use "modular arithmetic"
Odd numbers are not divisible by 2. To check this use:
if n % 2 != 0
The % is the mod function and it returns the remainder of the division (e.g. 5 / 2 is 2.5 but integers don't have decimals, so the integer result is 2 with a remainder of 1 and 5 / 2 => 2 and 5 % 2 => 1)
To check if it's divisible by 7, use the same principle:
if n % 7 == 0
The remainder is 0 if the dividend is divisible by the divisor. The complete if condition is:
if n % 2 != 0 && n % 7 == 0
You can also use n % 2 == 1 because the remainder is always 1. The result of any mod function, a % b, is always between 0 and b - 1.
Or, using the new function isMultiple(of:, that final condition would be:
if !n.isMultiple(of: 2) && n.isMultiple(of: 7)
Swift 5:
Since Swift 5 has been released, you could use isMultiple(of:) method.
In your case, you should check if it is not multiple of ... :
if !n.isMultiple(of: 2)
Swift 5 is coming with isMultiple(of:) method for integers , so you can try
let res = Array(1...100).filter { !$0.isMultiple(of:2) && $0.isMultiple(of:7) }
Here is an efficient and concise way of getting the odd multiples of 7 less than or equal to 100 :
let results: [Int] = Array(stride(from: 7, through: 100, by: 14))
You can also use the built-in filter to do an operation on only qualified members of an array. Here is how that'd go in your case for example
var result = Array(1...100).filter { (number) -> Bool in
return (number % 2 != 0 && number % 7 == 0)
}
print(result) // will print [7, 21, 35, 49, 63, 77, 91]
You can read more about filter in the doc but here is the basics: it goes through each element and collects elements that return true on the condition. So it filters the array and returns what you want
any of you knows how can I check if the division remainder is integer or zero?
if ( integer ( 3/2))
You should use the modulo operator like this
// a,b are ints
if ( a % b == 0) {
// remainder 0
} else
{
// b does not divide a evenly
}
It sounds like what you are looking for is the modulo operator %, which will give you the remainder of an operation.
3 % 2 // yields 1
3 % 1 // yields 0
3 % 4 // yields 1
However, if you want to actually perform the division first, you may need something a bit more complex, such as the following:
//Perform the division, then take the remainder modulo 1, which will
//yield any decimal values, which then you can compare to 0 to determine if it is
//an integer
if((a / b) % 1 > 0))
{
//All non-integer values go here
}
else
{
//All integer values go here
}
Walkthrough
(3 / 2) // yields 1.5
1.5 % 1 // yields 0.5
0.5 > 0 // true
swift 3:
if a.truncatingRemainder(dividingBy: b) == 0 {
//All integer values go here
}else{
//All non-integer values go here
}
You can use the below code to know which type of instance it is.
var val = 3/2
var integerType = Mirror(reflecting: val)
if integerType.subjectType == Int.self {
print("Yes, the value is an integer")
}else{
print("No, the value is not an integer")
}
let me know if the above was useful.
Swift 5
if numberOne.isMultiple(of: numberTwo) { ... }
Swift 4 or less
if numberOne % numberTwo == 0 { ... }
Swift 2.0
print(Int(Float(9) % Float(4))) // result 1