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Please Help me understanding these

  1. Aug 9, 2012 #1
    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.
     
  2. jcsd
  3. Aug 9, 2012 #2

    chiro

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    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?
     
  4. Aug 9, 2012 #3
    There are 101 numbers, 100-200
     
  5. Aug 9, 2012 #4

    HallsofIvy

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

    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.
     
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