How many integers between 100 and 200 are divisible by 6? (are there any shortcut method to get this?)

first such number -- 102

last such number -- 198

(198-102)/6 = 16

but !! .....

let's look at the first 4 of our numbers:
102 108 114 120

if I follow my above method
I would get (120-102)/6 = 3
notice I would not be counting the very first number

so our answer of 16 should be increased by 1

number of integers between 100 and 200 which divide evenly by 6 are 17

Can you generalize my method?

17

To find the number of integers between 100 and 200 that are divisible by 6, you can use a shortcut method called the "divisibility rule."

The divisibility rule for 6 states that a number is divisible by 6 if it is divisible by both 2 and 3.

First, let's consider the divisibility by 2. Every other number is divisible by 2, so half of the numbers between 100 and 200 (inclusive) will be divisible by 2.

Next, we need to consider the divisibility by 3. To determine if a number is divisible by 3, we add up the digits of the number. If the sum is divisible by 3, then the number itself is divisible by 3.

In this case, we want to find the number of integers divisible by 6, which means they must be divisible by both 2 and 3.

To determine the number of integers divisible by 3, we need to find the number of multiples of 3 between 100 and 200.
The first multiple of 3 greater than or equal to 100 is 102, and the last multiple of 3 less than or equal to 200 is 198. We can find the number of multiples of 3 by counting in increments of 3 from 102 to 198.

102, 105, 108, 111, ..., 195, 198

To calculate the number of terms in this sequence, we can use the formula for the nth term of an arithmetic sequence:

nth term = first term + (n - 1) * common difference.

Here, the first term is 102, the common difference is 3, and we want to find the number of terms. We can set up the equation as follows:

198 = 102 + (n - 1) * 3

Simplifying the equation, we get:

198 - 102 = 3n - 3

96 = 3n - 3

Adding 3 to both sides:

99 = 3n

Dividing both sides by 3:

n = 33

So, there are 33 integers between 100 and 200 that are divisible by 3.

Now, we need to consider the numbers that are divisible by both 2 and 3, which are divisible by 6. Since every other number between 100 and 200 is divisible by 2, we divide the count of numbers divisible by 3 by 2:

33 / 2 = 16.5

However, we cannot have a fraction of a number, so we can conclude that there are 16 integers between 100 and 200 that are divisible by 6.