How many integers bet 90 and 990 are divisible by 7?

is there any short and simple way of finding the answer?

thanks

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- Dec 22nd 2012, 06:33 AMrcsnumber of integers bet 90 and 990
How many integers bet 90 and 990 are divisible by 7?

is there any short and simple way of finding the answer?

thanks - Dec 22nd 2012, 06:43 AMPlatoRe: number of integers bet 90 and 990
- Dec 22nd 2012, 06:50 AMILikeSerenaRe: number of integers bet 90 and 990
Hi rcs! :)

The answer is 990 / 7 - (90 - 1) / 7, where "/" denotes a division that is rounded down.

I am assuming that you would count 990 and 90 if they would have been divisible by 7 (which they are not anyway).

To explain:

(90 - 1) / 7 is the number of integers divisible by 7 below and excluding 90.

990 / 7 is the number of integers divisible by 7 below and including 990.

The difference is the total number.

Suppose you are in the middle of a road with poles alongside the road.

And suppose you want to know the number of poles from where you are to the end of the road.

To find out you count the number that you can see from where you stand to the beginning of the road.

And if you are next to a pole you do not count that one.

Subtract from the total number of poles and there you are. - Dec 22nd 2012, 07:09 AMrcsRe: number of integers bet 90 and 990
thanks Prof . Plato

- Dec 22nd 2012, 07:12 AMrcsRe: number of integers bet 90 and 990
- Dec 22nd 2012, 07:33 AMPlatoRe: number of integers bet 90 and 990
- Dec 22nd 2012, 07:43 AMrcsRe: number of integers bet 90 and 990
- Dec 22nd 2012, 07:50 AMPlatoRe: number of integers bet 90 and 990
Sorry that was a typo. It should be

**NOT CORRECT** - Dec 22nd 2012, 07:52 AMrcsRe: number of integers bet 90 and 990
ahhh i see... well then thank you again Sir. God Bless

- Dec 23rd 2012, 10:36 PMibduttRe: number of integers bet 90 and 990
Suggest we find the multiple of 7 just greater than 90 and just less than 990.

We will get 91 and 987. Now all the multiples of 7 between these two numbers will form an AP with first term 91, common difference 7 and last term 987.

Question now simply is to find n such that the nth term is 987.

If a is the first term, common difference d then the nth term is given by

an = a + ( n-1) d

Thus we have 91 + ( n – 1 ) x 7 = 987

OR 91 + 7n – 7 = 987

7n = 987 – 91 + 7 = 903

n = 129 Hence there are 129 integers between 90 and 990 which are divisible by 7