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Why isn't this code working to find the sum of multiples of 3 and 5 below the given number?

I've started with some problems on HackerRank, and am stuck with one of the Project Euler problems available there.
The problem statement says: Find the sum of all the multiples of 3 or 5 below N
I've calculated the sum by finding sum of multiple of 3 + sum of multiples of 5 - sum of multiples of 15 below the number n
function something(n) {
n = n-1;
let a = Math.trunc(n / 3);
let b = Math.trunc(n / 5);
let c = Math.trunc(n / 15);
return (3 * a * (a + 1) + 5 * b * (b + 1) - 15 * c * (c + 1)) / 2;
}
console.log(something(1000)); //change 1000 to any number
With the values of num I've tried, it seems to work perfectly, but with two out of five test cases there, it returns a wrong answer (I can't access the test cases).
My question is what is the problem with my code? as the logic seems to be correct to me at least.
Edit: Link to problem page

Some of the numbers in the input are probably larger than what javascript can handle by default. As stated in the discussion on the hackkerrank-site, you will need an extra library (like: bignumber.js) for that.
The following info and code was posted by a user named john_manuel_men1 on the discussion, where several other people had the same or similar problems like yours
This is how I figured it out in javascript. BigNumber.js seems to store the results as strings. Using the .toNumber() method shifted the result for some reason, so I used .toString() instead.
function main() {
var BigNumber = require('bignumber.js');
var t = new BigNumber(readLine()).toNumber();
var n;
for(var a0 = 0; a0 < t; a0++){
n = new BigNumber(readLine());
answer();
}
function answer() {
const a = n.minus(1).dividedBy(3).floor();
const b = n.minus(1).dividedBy(5).floor();
const c = n.minus(1).dividedBy(15).floor();
const sumThree = a.times(3).times(a.plus(1)).dividedBy(2);
const sumFive = b.times(5).times(b.plus(1)).dividedBy(2);
const sumFifteen = c.times(15).times(c.plus(1)).dividedBy(2);
const sumOfAll = sumThree.plus(sumFive).minus(sumFifteen);
console.log(sumOfAll.toString());
}
}

Coin Change Algorithm JS

I have been trying to come up with a solution for this algorithm for 3-4 days but nothing seems to work and the available solutions are a bit more advanced for me. It has to be solved with conditionals only so no recursion or dynamic programming.
I need to determine the least amount of coins necessary to give change given the following denominations: 1, 0.5, 0.2, 0.1, 0.05, 0.02 and 0.01.
Input is the following:
Price of an item
Sum paid by customer
Current ideas:
let price = +gets();
let paidSum = +gets();
//gets is used to accept number input
let change = paidSum - price;
I figured I could use Math.floor to isolate the integer part and subtract it but then I have no idea what to do with the remaining sum.
Would modulo work to test whether the remaining sum contains any of the remaining values for change and then subtract again until I reach zero?
I do realize this isn't the best formulated question but I am at a loss here and I've done every other task apart from this. Thanks.

Simpler, reverse and map the denominations in cents and return a new array with the number of coins you need for each denomination.
const coinsCents = [1, 2, 5, 10, 20, 50, 100]
const getChange = (amountInCents) => {
return coinsCents.reverse().map(coin => {
let amountCoin = Math.floor(amountInCents/coin)
amountInCents -= amountCoin * coin
return amountCoin
}).reverse()
}

With the denominations you have specified, the problem is simpler than the general change making problem. In this actual case we can be sure that using the largest denomination, that is not greater than the amount to pay, always leads to an optimal solution.
So then there is no need for recursion or dynamic programming. Just a simple loop will do.
I will here ignore the additional "layer" of getting the price of the bill and the amount that the customer pays. In the end the only thing that counts is the change amount to pay back to the customer. So this snippet asks for that change amount and returns the coins that need to be given as change.
function getChange(amount) {
amount *= 100; // Convert to number of cents
var denominations = [1, 2, 5, 10, 20, 50, 100]; // cents
var result = [];
while (amount > 0) {
var coin = denominations.pop(); // Get next greatest coin
var count = Math.floor(amount/coin); // See how many times I need that coin
amount -= count * coin; // Reduce the amount with that number of coins
if (count) result.push([coin/100, count]); // Store count & coin
}
return result;
}
// I/O management
change.oninput = function () {
var coins = getChange(this.value);
result.textContent = coins.map(([coin, count]) => `${count} x $${coin}`).join(" + ");
};
To be paid to customer: <input id="change">
<div>Coins to pay: <span id="result"></span></div>

var coins;
var coinArray = {};
var output = {};
/* Method to get coin value without decimal point - it is required because
* javascript will consider 5.6 as 6 if we do Math.round()
*/
function getRoundFigureCoinValue(x) {
return (x * 10 - ((x * 10) % 10)) / 10;
}
// Method to calculate possible combination of coins
function calculateCoins(input) {
let largestPossibleCoin = 1;
if (input) {
coins.forEach((x) => {
if (input >= x) {
largestPossibleCoin = x;
}
});
let remainingCents = input % largestPossibleCoin;
output[largestPossibleCoin] = getRoundFigureCoinValue(
(input / largestPossibleCoin).toFixed(1)
);
if (remainingCents && input > 1) {
calculateCoins(remainingCents);
}
return largestPossibleCoin;
}
}
// Method to be called to get output.
function calculatePossibleCoinCombinations(value) {
if (isNaN(value) || +value <= 0) {
console.log('Invalid input');
return;
} else {
console.log('Possible combinations are:')
value = +value;
}
coins = [1, 5, 10, 25];
while (coins.length) {
let largestPossibleCoin = calculateCoins(value) || 0;
let outputString = '';
coins = coins.filter((x) => x < largestPossibleCoin);
Object.keys(output).forEach((key) => {
outputString += `${output[key]} - ${key} cents; `;
})
console.log(outputString);
output = {};
}
}
/*
Sample inputs:
calculatePossibleCoinCombinations('89');
calculatePossibleCoinCombinations(10);
calculatePossibleCoinCombinations(0);
calculatePossibleCoinCombinations('someString');
calculatePossibleCoinCombinations(-10)
*/

Find n logarithmic intervals between two end points

I am trying to find n logarithmic intervals between two numbers.
eg: for a function
logDiv (10, 10000, 3) where 10 is the starting point, 10000 the ending point, and 3 the number of intervals, I would like to get the following output:
(* {10, 100, 1000, 10000} *)
What I have tried:
function logInterval(total_intervals, start, end) {
var index, interval, result = [];
for (index = 0; index < total_intervals; index++) {
interval = (index/total_intervals * Math.log((end - start) + 1) - 1 + start);
result.push(interval);
}
return result;
}
var intervals = logInterval(5, 1, 500);
https://jsfiddle.net/qxqxwo3z/
This was based on my (poor) understanding of the following solution I found in stack exchange mathematica:
logspace [increments_, start_, end_] := Module[{a}, (
a = Range[0, increments];
Exp[a/increments*Log[(end - start) + 1]] - 1 + start
)]
https://mathematica.stackexchange.com/questions/13226/how-can-i-get-exactly-5-logarithmic-divisions-of-an-interval
Please can someone help me with this? Its not necessary to follow any of my above attempts, just explaining what I tried.

The best way is to start with the logarithmic difference of the end value and the start value, divided by the intervals.
x = (Math.log(end) - Math.log(start)) / total_intervals;
For the factor, you need to do the reverse operation
factor = Math.exp(x);
For getting an array you can multiply the start value with the factor and insert it into the array. The next value is the last value multiplied by the factor, until all items are generated.
function logInterval(total_intervals, start, end) {
var x = (Math.log(end) - Math.log(start)) / total_intervals,
factor = Math.exp(x),
result = [start],
i;
for (i = 1; i < total_intervals; i++) {
result.push(result[result.length - 1] * factor);
}
result.push(end);
return result;
}
console.log(logInterval(3, 10, 10000));
console.log(logInterval(5, 1, 500));
console.log(logInterval(12, 220, 440)); // some frequencies

Thanks to the above answer by nina and a lot more googling and stack overflow answers, I found that this works for me:
function logInterval(total_intervals, start, end) {
var startInterVal = 1, endInterval = total_intervals,
minLog = Math.log(start), maxLog = Math.log(end),
scale = (maxLog-minLog) / (endInterval-startInterVal),
result = [];
for (i = 1; i < total_intervals; i++) {
result.push(Math.exp(minLog + scale*(i - startInterVal)));
}
result.push(end);
return result;
}
https://jsfiddle.net/qxqxwo3z/1/

Find out how many thousands and hundreds and tens are there in a amount

I am having a asp application and in that amount column is there. I need to find out how many thousands and hundreds and tens are there in that amount
For example
if i am having amount as 3660 means
1000's - 3
100's - 6
10's - 6
like this i need
Can any body help me

The simple answer is to divide the number by 1000 whatever is the quotient that is the number of 1000's in the amount. Then divide the remainder with the 100's the quotient will be the number of 100's. And then again divide the remainder with 10, the quotient will be the number of 10's
Something like this:
quotient = 3660 / 1000; //This will give you 3
remainder = 3660 % 1000; //This will give you 660
Then,
quotient1 = remainder/ 100; //This will give you 6
remainder1 = remainder % 100; //This will give you 60
And finally
quotient2 = remainder1 / 10; //This will give you 6

Is it not easier to use type coercion and change the data type to string?
Then you can easily check the value by checking the value at selected index position,
var number = 1234;
number2 = new String(number);
var thousands = number2[0];
var hundreds = number2[1];
and so on....
It may not be usable in what you're doing, it was for me :)

If the "javascript" tag is the correct one, then you've already gotten some answers. If the "asp-classic" tag is actually the correct one, then chances are your scripting language is VBScript, not Javascript.
Did you just pick multiples of 10 as an example, or is that the actual multiple you're looking for? Because if it's the latter, then all you need to do is split the number into digits — that's what the base 10 number system means, after all.
Function SplitNum(theNum)
dim L, i, s, n
n = CStr(theNum)
L = Len(n)
s = ""
for i = 1 to 3
if s <> "" then s = "," & s
if i >= L then
s = "0" & s
else
s = Left(Right(n,i+1),1) & s
end if
next
if L > 4 then s = left(n,L-4) & s
SplitNum = s
End Function
If your actual divisors are something other than multiples of 10, you'll need to do arithmetic. The integer division operator in VBScript is \. (Integer division is basically the "opposite" of the modulus function.)
Function GetMultiples(theNum)
dim q, r
q = theNum \ 1000 & ","
r = theNum Mod 1000
q = q & r \ 100 & ","
r = r Mod 100
q = q & r \ 10
GetMultiples = q
End Function

Try this out...
Here is a fiddle that demonstrates how to use the output..
http://jsfiddle.net/Villarrealized/L3AxZ/1/
function getMultiplesOfTen(number){
number = parseInt(number);
if(typeof(number)!=="number") return number;
var result = {};
(function breakDown(num){
if(isNaN(num))return num;//if it's invalid return
if(num<=0)return false;
num = num.toFixed(0);//get rid of decimals
var divisor = Math.pow(10,num.length-1),//ex. when num = 300, divisor = 100
quotient = Math.floor(num/divisor);
result[divisor]=quotient;//add it to our object
breakDown(num % divisor);//break down the remainder
})(number);
//return result as an object
return result;
}
This function will return an object with the multiple of ten as the key and the number as the value
ex. getMultiplesOfTen(150)=={100:1,10:5} == 1 multiple of 100 and 5 multiples of 10.

Let's say we have the number 7354. To find the thousands:
variable a = 7354 / 1000
variable b = a % 10
The number which is stored in variable b now is the number if the thousands.
To find the hundreds:
variable c = 7354 / 100
variable d = a % 10
The number which is stored in variable d now is the number if the hundreds.
To find the tens:
variable e = 7354 / 10
variable f = a % 10
The number which is stored in variable f now is the number if the tens.
To find the ones:
7354 % 10
This works for every number in the place of 7354, even for bigger numbers than 7354.

The first digit in the quotient of 1,592÷64
1
is in the Choose... ones tens hundreds thousands place.

Decide where the first digit of the quotient should be placed. Do not complete the division.
The first digit of the quotient for 2,370÷24
2,370÷24
should be in the Choose... ones tens hundreds thousands place

2021 version here:
Cast your number as a string, spread it and reverse it like so:
x = 1234
x = [...x.toString()].reverse() // [4, 3, 2, 1]
thous = x[3] // 1
hunds = x[2] // 2
tens = x[1] // 3
units = x[0] // 4
y = [...x.toString()].reverse()[3] // 1 because 1000 has 3 zeros
I suppose you can use some fancy system of powers of 10 to get those indexes. So let's do exactly that and get #Villarrealized 15 lines of 2013 code condensed into just a few lines of 2021 code:
function placeValues(someNumber = 1234) {
x = [...someNumber.toString()].reverse().reduce((p, c, i) => {
p[10**i] = c;
return p;
}, {});
return x; // {1000:1, 100:2, 10:3, 1:4}
}

Finding the nth item in a repeating list of fixed items

I have to determine the mathematical formula to calculate a particular repeating position in a series of numbers. The list of numbers repeats ad infinitum and I need to find the number every n numbers in this list. So I want to find the *n*th item in a list of repeating y numbers.
For example, if my list has 7 digits (y=7) and I need every 5th item (n=5), how do I find that item?
The list would be like this (which I've grouped in fives for ease of viewing):
12345 67123 45671 23456 71234 56712 34567
I need to find in the first grouping number 5, then in the second grouping number 3, then 1 from the third group, then 6, then 4, then 2, then 7.
This needs to work for any number for y and n. I usually use a modulus for finding *n*th items, but only when the list keeps increasing in number and not resetting.
I'm trying to do this in Javascript or JQuery as it's a browser based problem, but I'm not very mathematical so I'm struggling to solve it.
Thanks!
Edit: I'm looking for a mathematical solution to this ideally but I'll explain a little more about the problem, but it may just add confusion. I have a list of items in a carousel arrangement. In my example there are 7 unique items (it could be any number), but the list in real terms is actually five times that size (nothing to do with the groups of 5 above) with four sets of duplicates that I create.
To give the illusion of scrolling to infinity, the list position is reset on the 'last' page (there are two pages in this example as items 1-7 span across the 5 item wide viewport). Those groups above represent pages as there are 5 items per page in my example. The duplicates provide the padding necessary to fill in any blank spaces that may occur when moving to the next page of items (page 2 for instance starts with 6 and 7 but then would be empty if it weren't for the duplicated 1,2 and 3). When the page goes past the last page (so if we try to go to page 3) then I reposition them further back in the list to page one, but offset so it looks like they are still going forwards forever.
This is why I can't use an array index and why it would be useful to have a mathematical solution. I realise there are carousels out there that do similar tasks to what I'm trying to achieve, but I have to use the one I've got!

Just loop every 5 characters, like so:
var data = "12345671234567123456712345671234567";
var results = [];
for(var i = 4; i < data.length; i += 5){
results.push(data[i]);
}
//results = [5, 3, 1, 6, 4, 2, 7]
If you want to use a variable x = 5; then your for loop would look like this:
for(var i = x - 1; i < data.length; i += x){...
There is no need to know y

If your input sequence doesn't terminate, then outputting every nth item will eventually produce its own repeating sequence. The period (length) of this repetition will be the lowest common multiple of the period of the input sequence (y) and the step size used for outputting its items (x).
If you want to output only the first repetition, then something like this should do the trick (untested):
var sequence = "1234567";
var x = 5;
var y = sequence.length;
var count = lcm(x, y);
var offset = 4;
var output = [];
for (var i = 0; i < count; i += x)
{
j = (offset + i) % y;
output.push(sequence[j]);
}
You should be able to find an algorithm for computing the LCM of two integers fairly easily.

A purely mathematical definition? Err..
T(n) = T(n-1) + K For all n > 0.
T(1) = K // If user wants the first element in the series, you return the Kth element.
T(0) = 0 // If the user want's a non-existent element, they get 0.
Where K denotes the interval.
n denotes the desired term.
T() denotes the function that generates the list.
Lets assume we want every Kth element.
T(1) = T(0) + K = K
T(2) = T(1) + K = 2K
T(3) = T(2) + K = 3K
T(n) = nk. // This looks like a promising equation. Let's prove it:
So n is any n > 1. The next step in the equation is n+1, so we need to prove that
T(n + 1) = k(n + 1).
So let's have a go.
T(n+1) = T(N+1-1) + K.
T(n+1) = T(n) + K
Assume that T(n) = nk.
T(n+1) = nk + k
T(n+1) = k(n + 1).
And there is your proof, by induction, that T(n) = nk.
That is about as mathematical as you're gonna get on SO.
Nice simple recurrence relation that describes it quite well there.

After your edit I make another solution;)
var n = 5, y = 7;
for (var i = 1; i<=y; i++) {
var offset = ( i*y - (i-1)*n ) % y;
var result = 0;
if (offset === n) {
result = y;
} else {
result = (n - offset) > 0 ? n - offset : offset;
}
console.log(result);
}
[5, 3, 1, 6, 4, 2, 7] in output.
JSFIDDLE: http://jsfiddle.net/mcrLQ/4/

function get(x, A, B) {
var r = (x * A) % B;
return r ? r : B;
}
var A = 5;
var B = 7;
var C = [];
for (var x = 1; x <= B; ++x) {
C.push(get(x, A, B));
}
console.log(C);
Result: [5, 3, 1, 6, 4, 2, 7]
http://jsfiddle.net/xRFTD/

var data = "12345 67123 45671 23456 71234 56712 34567";
var x = 5;
var y = 7;
var results = [];
var i = x - 1; // enumeration in string starts from zero
while ( i <= data.length){
results.push(data[i]);
i = i + x + 1;// +1 for spaces ignoring
}