What is the sum of all integers between 15 and 415 that are divisible by 7? - explain solution

Question

Question: Using Sigma notation and discrete math rules, calculate the sum of all integers between 15 and 415 that are divisible by 7.

I have a given solution for this but I don't quite understand it and I was wondering if someone could explain.

The first number is 7×⌈(15+1)/7⌉ = 7×3 = 21 and the last number is 7×⌊(415-1)/7⌋ = 7×59 = 413.

The sum of all integers between 15 and 415 that are divisible by 7 is then:

7 × [k = 3 to 59]∑k

= 7 × ([k = 1 to 59]∑k - [k = 1 to 2]∑k)

= 7 × (59×60 - 2×3)/2

= 7 × 1767

= 12369

I don't understand where 7x3 and 7x59 came from in the calculation above. (15+1)/7 is 2.28 so should it not be 7x2.28 instead?

Answer 1

The numbers between $15$ and $415$ are $$ 16, 17, 18, \ldots, 412, 413, 414 $$ Among these, the numbers which are divisible by $7$ are $$ 21, 28, 35, \ldots, 399, 406, 413 $$ Writing these explicitly as multiples of $7$, we get $$ 3\cdot 7, 4\cdot 7, 5\cdot 7, \ldots, 57\cdot 7, 58\cdot 7, 59\cdot 7 $$ We are interested in the sum of all these numebrs. This is where $3\cdot 7$ and $59\cdot 7$ comes from. The formulas that are written there, $$ \left\lceil\frac{15 + 1}{7}\right\rceil\cdot 7 = 21\\ \text{and}\\ \left\lfloor\frac{415-1}{7}\right\rfloor\cdot 7 = 413 $$ is just an algebraic way to express the same idea.

Answer 2

The first multiple of $7$ between $15$ and $415$ is $21$, i.e. $7 \times 3$. The last multiple of $7$ between between $15$ and $415$ is $413$, i.e. $7 \times 59$.

So you have to sum the multiples of $7$ between $7 \times 3$ and $7 \times 59$...

Answer 3

The notation $\lceil x \rceil $ is the "ceiling" function. It means to round UP to the nearest integer. The notation $\lfloor x \rfloor$ is the "floor" function. It means to round DOWN to the nearest integer.

Answer 4

$7\cdot3$ it's the first and $7\cdot59$ it's the last needed number between $15$ and $415$.

Also, $$\sum_{k=1}^nk=\frac{n(n+1)}{2}$$ and from here $$7\sum_{k=3}^{59}k=7\left(\frac{59\cdot60}{2}-1-2\right)=12369.$$

Answer 5

A bit more to add to the conversation, the line you seem to be confused by is the one that begins with this:

"The first number is $7\times \lceil(15+1)/7\rceil$"

You are correct that the expression on the right simplifies to $(15+1)/7$ and so the entire expression simplifies to $7\times \lceil \frac{16}{7}\rceil$ which is approximately $7\times \lceil 2.28\rceil$... (note that $\frac{16}{7}$ is not precisely equal to $2.28$, only approximately equal. Only use equals signs for equality!)

What you seem to be forgetting is that $7\times \lceil 2.28\rceil$ is not the same thing as $7\times 2.28$. The bars you see there represent the ceiling function which "rounds up" whatever is put in between the bars to the next highest integer (if what is put in is an integer already it doesn't change it).

For example $\lceil 4.2\rceil = 5$ and $\lceil 2\rceil = 2$. Negative arguments here, remember that "up" means to the right on the number line, not just larger in terms of absolute value, so $\lceil -13.57\rceil = -13$.

The expression we are interested then continues to be simplified as $7\times \lceil \frac{16}{7}\rceil \color{grey}{(\approx 7\times \lceil 2.28\rceil)} = 7\times 3 = 21$

This was all just a fancy way of finding what integer strictly larger than $15$ was the next multiple of $7$. This could have been done strictly by inspection, remembering our multiples of seven and knowing that $21$ is the next multiple of seven after $15$.

Answer 6

Multiples of $7$ look like $1 \times 7$, $2 \times 7$, $3 \times 7$, $4 \times 7$, and so on. $2 \times 7 = 14 < 15$ and $3 \times 7 = 21 > 15$ so we can be sure that the multiples of $7$ greater than $15$ start at $21$.

The number $x$ such that $x \times 7 = 15$ is, unsurprisingly, $\frac{15}{7}$. The first whole number $>x$ is $\lceil \frac{15+1}{7} \rceil$ since if we didn't add $1$ and $x$ were already a whole number we would get $x$ (which isn't $>x$).

So doing $\lceil \frac{15+1}{7} \rceil$ is just working out that $\frac{15}{7} = 2.28$ so the next whole number is $3$, so that's where we should start our multiples of $7$ from, and the first one will be $$21 = 3 \times 7 = \lceil \frac{15+1}{7} \rceil \times 7.$$

The same idea works for the multiples of $7$ less than $415$.