Is Only One Number in Any 100 Consecutive Natural Numbers Divisible by 100?

[mathsfail]

True or false with answer written in my book.

(c)Out of any 100 consecutive natural numbers,exactly 1 natural number is divisible by 100.(true/false).

Ans-True= 100/100= 1

(d)Out of any 101 consecutive natural numbers,exactly 1 natural number is divisible by 100.(true/false).

Ans-False.There may be one or two numbers divisible by 100.

I don't get the answer of (c).If we take numbers from 100-200.there should be 2 numbers divisible by 2.Only from 1-100,we get 1 number divisible by 2.This should be false i think and the answer should be same as d.Correct be if i am wrong.


Second problem

How many natural numbers from 200-500(including both the limits) will be divisible by 3.How to solve this.

 

[chiro]

Hey mathsfail and welcome to the forums.

The answer formally is to show that in your collection of numbers there exists a number n = 100*q for some number q.

All numbers can written as n = pq + r where in this case p = 100 and r is an integer from 0 to 99 inclusive. If you can show that there exists a number n in your set of numbers such that r = 0 then you have done the proof.

Using this, and the fact that you have all numbers a through to a + 99, can you now prove this?

 

[coolul007]

There are 101 numbers, 100-200

 

[HallsofIvy]

True or false with answer written in my book.

(c)Out of any 100 consecutive natural numbers,exactly 1 natural number is divisible by 100.(true/false).

Ans-True= 100/100= 1

(d)Out of any 101 consecutive natural numbers,exactly 1 natural number is divisible by 100.(true/false).

Ans-False.There may be one or two numbers divisible by 100.

I don't get the answer of (c).If we take numbers from 100-200.there should be 2 numbers divisible by 2.Only from 1-100,we get 1 number divisible by 2 This should be false i think and the answer should be same as d.Correct be if i am wrong.

First, you mean "divisibe by 100" not "divisible by 2", 100 to 200 is 101 numbers, not 100

Second problem

How many natural numbers from 200-500(including both the limits) will be divisible by 3.How to solve this.

Did you give this any thought at all? 200= 66*3+ 2 and 2/3 but 201= 67*3 is divisible by 3. 500= 166*3+ 2 but 498= 166*3 is divisible by 3. There are 499- 201= 298 numbers between 201 and 498, including those numbers, and 1/3 of them are divisible by 3.