Iâ€™m confused. If a = 30 and b= 30, then 30 % 60 + 30 % 60 == 60?

It wouldnâ€™t be 60 % 60

nvm I needed to distribute the c

can someone please break this part down for me?

We can %60 on every number in the array and get [30, 20, 30, 50, 20] and now we can see 30 + 30 = 60. So the two integers whose sum is divisible by 60 are 30 and 150.

Where did we get 30 and 30 from?

30 % 60 = 30

150 % 60 = 30

Because last_element = first_element + difference * (number_of_element - 1) , the sum can be expressed as (2 * first_element + difference * (number_of_elements - 1)) * number_of_elements)â€¦

Is not the last part missing the / 2 part

```
the sum can be expressed as (2 * first_element + difference * (number_of_elements - 1)) * number_of_elements) / 2
```

because:

The sum of an arithmetic sequence is:

(first_element + last_element) * number_of_element / 2

last_element = first_element + difference * (number_of_element - 1)

so

(first_element + (first_element + difference * (number_of_element - 1))) * number_of_element / 2 or

(2 * first_element + difference * (number_of_elements - 1)) * number_of_elements) / 2

Hereâ€™s the animated proof form wikiepedia if you are interested.

â€śfromâ€ť

From:

n = 10

for (i = 0; i < 10; i++) {

for (j = 0; j <= i; j++) {

doSomething();

}

}

To:

n = 10

for (i = 0; i < n; i++) {

for (j = 0; j <= i; j++) {

doSomething();

}

}

Below is an example in Java

import java.util.Arrays;

import java.util.List;

public class Mod {

```
public static void main(String[] args) {
List<Integer> list = Arrays.asList(30, 20, 150, 110, 200);
Integer offset = 60;
list.forEach(m -> {
list.forEach(o -> {
if (m!=o && (m % offset + o % offset) % offset == 0)
System.out.println(m);
});
});
}
```

}

test

- (first_element + last_element) * number_of_element / 2
- last_element = first_element + difference * (number_of_element - 1)
- (first_element + first_element + difference * (number_of_element - 1)) * number_of_element / 2 (substituting last_element in eq 1)
- (2 * first_element + difference * (number_of_element - 1)) * number_of_element / 2 (simplifying eq 3)

the divided by 2 part is missing in the end for the above mentioned equation

I would format this statement over a few lines. Itâ€™s a bit hard to grok when the line break is in the middle of a code statement:

Because last_element = first_element + difference * (number_of_element - 1) , the sum can be expressed as (2 * first_element + difference * (number_of_elements - 1)) * number_of_elements). In big O complexity analysis, we drop the constant terms (first_element, diffence and -1), so this really becomes O(n^2).

I still donâ€™t get modular arithmetic. I donâ€™t understand the properties of it in a equation.

dopecoder is right it should be (2 * first_element + difference * (number_of_element - 1)) * number_of_elements / 2

not

(2 * first_element + difference * (number_of_element - 1)) * number_of_elements

This helped me understand the distributive property of modulo.

https://www.splashlearn.com/math-vocabulary/algebra/distributive-property

The key takeaway from the example in this section of the course for me was that I need to run modulo one additional time, unlike with multiplication. Eg, Typically

(a + b) * m == am + bm

By contrast

(a + b ) % m == (a % m + b % m) % m

I still donâ€™t completely understand WHY I need that third modulo, but câ€™est la vie.

cuz a % m + b % m could still be greater than m

function mod(x, y) {

while (x >= y) {

x -= y

}

return x

}

Is only works when x is > 0.

Shouldnâ€™t **(n+1) + (n+1) + â€¦(n+1) = k(n+1)**?

The video shows **(n+1) + (n+1) + â€¦(n+1) = n(n+1)** which doesnâ€™t make sense?

In the video he discusses how n = number of elements, so if you see that every â€śgroupingâ€ť of elements equates to (n + 1), e.g. 2 + (n-1) == (n + 1), 3 + (n - 2) == (n + 1) and so on, then we know that for every element the â€śgroupingâ€ť equates (n + 1).

Therefore the answer can be simplified to n * (n + 1).

Iâ€™m also lost on this one. I can see that 30 and 150 % 60 == 30, but I donâ€™t see the jump. Was a % 60 done to every single element, and then we looked for the difference or something in the other elements? Maybe if this was written out into the equations? Iâ€™m also lost on what a, b, c represents in those equations. Maybe if someone could do a more fleshed out example?