18 Aug, 02:51PM in sunny Singapore!

IP Year 1 Maths Questions

Subscribe to IP Year 1 Maths Questions 10 posts

Please Login or Signup to reply.
  • Seowlah's Avatar
    897 posts since Jul '09
    • Question 1

      What is the greatest number that will divide 740, 608 and 861, leaving the same remainder in each case

    • Question 2

      Form a three digit number with the following conditions

       

      The digits in the number must be different

       

      All the digits in the number must be prime numbers.

       

      The number formed must be the product of four prime numbers.

       

      What is the three digit number ?

      Edited by Seowlah 16 Feb `17, 5:13PM
  • Sgkaypoh's Avatar
    4 posts since Feb '17
    • Question 1

      11 x 7 + 3 = 740
      11 x 5 +3 = 608
      11 x 8 + 3 = 861

      Ans is 11

      Question 2

      315

      https://www.facebook.com/kaypoh.sg.7?lst=100015334178518%3A100015334178518%3A1487228777

  • Seowlah's Avatar
    897 posts since Jul '09
  • Moderator
    eagle's Avatar
    23,320 posts since Aug '01
    • Originally posted by Seowlah:

      So, answer cannot be 11

      I think he meant

      11*67 + 3 = 740

      11*55 + 3 = 608

      11*78 + 3 = 861

  • Sgkaypoh's Avatar
    4 posts since Feb '17
  • Seowlah's Avatar
    897 posts since Jul '09
    • But this method of getting the answer of 11 is like using guess and check method.

       

      This question is under the topic of HCF and LCM. Need to use these concepts to find the answer of 11.

      Edited by Seowlah 18 Feb `17, 7:33PM
  • Sgkaypoh's Avatar
    4 posts since Feb '17
    • Actually the solution is not so difficult to arrive at. Since the remainer is the same for all the numbers, the differences between the numbers should be perfectly divisible by the HCF. In particular 861-740 = 121, which is 11^2, readily identifying 11 as the HCF. Applying this to the 3 numbers confirms the solution.. 

      alternatively you can subtract 3 from each number and write down all the factors for each, thereby identifying the HCF. For 608 it would have been very easy because 608 - 3 =  605, telling you 5 is one of the factors. Dividing 605 by 5 = 121, which is 11^2, thereby identifying 11 as the HCF.

      Hope this helps.

       

  • Lindamerrillme's Avatar
    2 posts since Apr '17
Please Login or Signup to reply.