Question

what is the average of the smallest and the greatest three digit number that are divisible by 18. with explination

Answer 1

(1)

We know that Smallest 3-digit number = 100

To find the first 3 digit number divisible by 18, Divide the first 3 digit number by 18.

⇒ 100/18 = 5.5

Clearly the first three digit number is not divisible by 18.

Let us divide the second 3 digit number by 18.

⇒ 101/18 = 12.62.

Clearly the second three digit number is also not divisible by 18.

Let us divide the fifth 3 digit number by 18.

⇒ 105/18 = 5.8

Clearly the third three digit number is also not divisible by 18.

Let us divide the eight 3 digit number by 18.

⇒ 108/18 = 6.

So, 108 is the first three digit number exactly divisible by 18.

(ii)

We know that greatest three digit number = 999.

To find the last 3 digit number divisible by 18, Divide it by 18.

⇒ 999/18 = 55.5

Clearly the last 3 digit number number is not divisible by 18.

Let us divide the preceding 3 digit number 998 by 18.

⇒ 998/18 = 55.44

Clearly the last second e digit number is also not divisible by 18.

Let us divide the 5th preceding 3 digit number by 18.

⇒ 995/18 = 55.27.

Clearly it is not divisible by 18.

Let us divide the 9th preceding digit 990 by 18.

⇒ 990/18 = 55.

So, 990 is the last three digit number divisible by 18.

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⇒ Smallest three digit number divisible by 18 = 108.

⇒ Greatest three digit number divisible by 18 = 990.

Average of smallest and greatest three digit number:

⇒ (108 + 990)/2

⇒ 1098/2

⇒ 549.

Hope it helps!