Question

Find the sum of all 2 digit numbers which are divisible by 2 and 15.
[AP, Ezie]

Answer 1

$\large\underline{\text{Note}}$

Many people get confused because of 'and,' 'or.' The difference between them is the first requires all conditions, the last one requires just at least one condition. This confusion also exists in equation solving.

$\text{ex) What are the roots of }x^{2}=1\text{?}$

$\rightarrow\text{The roots are }x=1\text{ or }x=-1\text{. \green{(correct)}}$

$\rightarrow\text{The roots are }x=1\text{ and }x=-1\text{. \red{(incorrect)}}$

As $x$ cannot be $1$ and $-1$ at the same time.

Here, we need to find the sum of numbers divisible by 2 'and' 15. What if it were divisible by 2 'or' 15? To get rid of confusion, I tried both below.

$\large\underline{\text{Question}}$

The numbers are divisible by 2 and 15. Find the sum of such two-digit numbers.

$\large\underline{\text{Step 1. LCM(Least Common Multiple)}}$

We need to find the sum of integers that are both divisible by 2 and 15.

Such numbers have two factors 2 and 15. Now, we know that the number contains both 2 and 15.

Such numbers are the multiples of $LCM(2,15)=30$. (explained in the attachment)

$\large\underline{\text{Step 2. Arithmetic progression}}$

The algebraic form of the A.P. is $30n$, and we know that $10\leq30n\leq 99$.

Such $n$ are $n=1,2,3$.

The multiples of 30 of two digits are 30, 60, and 90.

$\large\underline{\text{Conclusion}}$

Hence, the sum of such numbers is 180.

$\bigstar\large\underline{\text{Bonus question}}\bigstar$

The numbers are divisible by 2 or 15. Find the sum of such two-digit numbers.

$\large\underline{\text{Step 1. Sets}}$

The numbers divisible by 2 or 15 are their multiples.

The universal set of numbers are numbers between 10 and 99, including two ends. (explained in the attachment)

Now, we know that the possible multiples of 2 are $a_{n}=2n$. For multiples of 15, it is $b_{n}=15n$. For multiples of 30, it is $c_{n}=30n$.

$\large\underline{\text{Step 2. Arithmetic progression}}$

We know that $10\leq a_{n},b_{n},c_{n}\leq 99$.

Such $n$ for $a_{n}$ are in $5\leq n\leq 49$. For $b_{n}$, it requires $1\leq n\leq 6$, and for $c_{n}$ it requires $1\leq n\leq 3$.

$\large\underline{\text{Step 3. Arithmetic series}}$

A series stands for the sum of the sequence.

Here it is denoted as summation notation.

$\red{\bigstar}\displaystyle\sum^{49}_{k=1}a_{k}-\sum^{4}_{k=1}a_{k}\text{ stands for the sum of }a_{n}\text{ from 5th to 19th term.}$

$\red{\bigstar}\displaystyle\sum^{6}_{k=1}b_{k}\text{ stands for the sum of }b_{n}\text{ from 1st to 6th term.}$

$\red{\bigstar}\displaystyle\sum^{3}_{k=1}c_{k}\text{ stands for the sum of }c_{n}\text{ from 1st to 3rd term.}$

However, we know that $\displaystyle\sum^{n}_{k=1}k=\dfrac{n(n+1)}{2}$.

$\red{\bigstar}\displaystyle\sum^{49}_{k=1}a_{k}-\sum^{4}_{k=1}a_{k}\text{ (the sum of possible multiples of 2)}$

$\implies\displaystyle\sum^{49}_{k=1}a_{k}-\sum^{4}_{k=1}a_{k}=2\cdot\dfrac{49\cdot50}{2}-2\cdot\dfrac{4\cdot5}{2}$

$\implies\displaystyle\sum^{49}_{k=1}a_{k}-\sum^{4}_{k=1}a_{k}=2450-20$

$\implies\displaystyle\sum^{49}_{k=1}a_{k}-\sum^{4}_{k=1}a_{k}=2430$

$\red{\bigstar}\displaystyle\sum^{6}_{k=1}b_{k}\text{ (the sum of possible multiples of 15)}$

$\implies\displaystyle\sum^{6}_{k=1}b_{k}=5\cdot\dfrac{6\cdot7}{2}$

$\implies\displaystyle\sum^{6}_{k=1}b_{k}=105$

$\red{\bigstar}\displaystyle\sum^{3}_{k=1}c_{k}\text{ (the sum of possible multiples of 30)}$

$\implies\displaystyle\sum^{3}_{k=1}c_{k}=30\cdot\dfrac{3\cdot4}{2}$

$\implies\displaystyle\sum^{3}_{k=1}c_{k}=180$

$\text{(Sum of numbers divisible by 2 or 15)}$

$=\displaystyle\left(\sum^{49}_{k=1}a_{k}-\sum^{4}_{k=1}a_{k}\right)+\sum^{6}_{k=1}b_{k}-\sum^{3}_{k=1}c_{k}$

$=2430+105-180$

$=2355$

$\large\underline{\text{Conclusion}}$

In conclusion, the sum of such numbers is 2355.

Thank you for reading!

Answer 2

Answer:

To find the sum of all 2-digit numbers that are divisible by both 2 and 15, we need to find the multiples of the least common multiple (LCM) of 2 and 15, which is 30.

The first 2-digit number divisible by 30 is 30 itself, and the last one is 90. So, we need to find the sum of all the multiples of 30 between 30 and 90.

To do this, we can use the formula for the sum of an arithmetic series:

S = (n/2)(a + l)

where S is the sum of the series, n is the number of terms, a is the first term, and l is the last term.

In this case, a = 30, l = 90, and the common difference is 30, so we can find n as follows:

l = a + (n - 1)d

90 = 30 + (n - 1)30

60 = 30(n - 1)

n - 1 = 2

n = 3

So, there are three 2-digit numbers that are divisible by both 2 and 15: 30, 60, and 90.

The sum of these numbers is:

30 + 60 + 90 = 180

Therefore, the sum of all 2-digit numbers that are divisible by both 2 and 15 is 180.