1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to solve this real numbers problem?

  1. Jul 10, 2013 #1
    1. The problem statement, all variables and given/known data
    An eight digit number is a multiple of 73 and 137. If the second digit from the left of the number is seven, find the 6th digit from the left of the number.


    2. Relevant equations
    N.A.


    3. The attempt at a solution
    I don't know any clear method for solving this problem. I raised 73 to the 4th power and got an eight digit number, but it didn't have 7 as the second digit. Since 73 is very small compared to an eight digit number, it is impractical to write down all the eight digit numbers which are multiples of 73 and then finding one which is common with 137. Moreover, this problem is from an Olympiad and needs to be solved within roughly 72 seconds. I thought about LCM next but once I got the LCM(10001), I had no idea what to do with it. The HCF is 1(both numbers being prime). I substituted the numbers for variables. We get a7cdefgh. We need to find f. The problem is that I just don't see a way to solve this problem.
     
  2. jcsd
  3. Jul 10, 2013 #2
    What is 10001 x 1, x 2, x 3, .. x 12, x 13 etc. Do you see a pattern between the second digit and the sixth digits from the left of the number?
     
  4. Jul 10, 2013 #3
    Oh i see it now. Thank you. So is a multiple of the lcm of two numbers divisible by each of the two numbers as a rule?
     
  5. Jul 10, 2013 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    What do you think is meant by"lcm"?
     
  6. Jul 10, 2013 #5

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Yes, the least common multiple of two numbers is (by definition) always divisible by the two numbers. Thus every multiple is divisible by the two numbers too.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: How to solve this real numbers problem?
Loading...