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.

 

[CellCoree]

The answer for (c) is true because if we take any set of 100 consecutive natural numbers, there will always be exactly one number that is divisible by 100. This can be seen by dividing the set into groups of 100 (e.g. 1-100, 101-200, etc.) and noting that the last number in each group will always be divisible by 100.

For the second problem, to find the number of natural numbers divisible by 3 between 200-500, we can use the formula (last number - first number)/divisor + 1. In this case, it would be (500-200)/3 + 1 = 101. So there are 101 natural numbers between 200-500 that are divisible by 3.