math

posted by .

WHAT IS THE LEAST NUMBER THAT IS
DIVISIBLE BY 2,3,4,5,6,9AND 10
EXPLAIN HOW YOU FOUND YOUR ANSWER

  • math -

    Please see the answer posted by Reiny to this question.

    http://www.jiskha.com/display.cgi?id=1200617520

  • math -

    Similarly:

    Use the digits 0-9 once, and only once, to create a 10 digit number in which the number created by the first digit is divisible by one, the number created by the first two digits is divisible by two, the number created by the first three digits is divisible by three, and so on, until the number created by the tenth (last) digit is divisible by ten. Is there more than one solution to this problem? >>

    Lets take the easy road and see where it takes us.

    Divisibility rules:
    Odd numbers can only be divided evenly by another odd number.
    Even numbers can be evenly divided by either odd or even numbers.
    2---Numbers that end in 0, 2, 4, 6, or 8 are evenly divisible by 2.
    3---If the sum of a number's digits is evenly divisible by 3, the number is divisible by
    4---If the last two digits are both zero or they form a two digit number evenly divisible by 4, then the whole number is evenly divisible by 4.
    5--Any number ending in 5 or 0 is evenly divisible by 5.
    6--If the number is even and the sum of the digits is evenly divisible by 3, the whole number is divisible by 3.
    7---Double the last digit and subtract from the number without the last digit. If the result is evenly divisible by 7, so is the original number.
    8---If the last 3 digits are zero or if they form a number that is evenly divisible by 8, then the whole number is evenly divisible by 8.
    9---If the sum of the digits is evenly divisible by 3, the number is evenly divisible by 9.
    10---Any number ending in 0 is divisible by 10.
    11--- Add up the odd position digits. Add up the even position digits. Calculate the difference between the two sums. If the difference is divisible by 11, the original number is divisible by 11

    Lets portray our number, with the digits identified, by

    _.._.._.._.._.._.._.._.._.._
    1..2..3..4..5..6..7..8..9.10

    Clearly, the last digit must be 0 in order for the whole number to be divisible by 10.

    Similarly, the 5th digit must be 5 in order for the 5 digit number to be divisible by 5.

    This makes our number so far _.._.._.._..5.._.._.._.._..0
    .............................................1..2..3..4..5..6..7..8..9.10

    The sum of the digits 1 - 9 is 45, which makes any nine digit number using all nine digits divisible by 9. Thus, we need not concern ourselves with the 9th digit.

    By the definition of our problem, all even numbers of digits must be divisible by an even number and all odd digit numbers must be divisible by an odd number, making the even digits even, and the odd digits odd.

    The 2nd, 4th, 6th, and 8th digits can then only be 2, 4, 6, or 8 and the 1st, 3rd, 5th, 7th, and 9th digits can only be 1, 3, 5, 7, or 9.

    For the 8 digit number to be divisible by 8, the last 3 digits must be divisible by 8, making these three 3 digit numbers start, and end, with 2, 4, 6, or 8. These three digits therefore form numbers in the 200+, 400+, 600+ or 800+ family, each of which is, itself, divisible by 8, thus forcing the 7th and 8th digits together to be evenly divisible by 8. The only possibilities for these two digits then becomes 16, 32, 72 and 96.

    This makes our possible numbers so far....._.._.._.._..5.._..1..6.._..0
    .............................................................. _.._.._.._..5.._..3..2.._..0
    .............................................................. _.._.._.._..5.._..7..2.._..0
    .............................................................. _.._.._.._..5.._..9..6.._..0
    ..............................................................1..2..3..4..5..6..7..8..9.10

    From our divisibility rules, a number is divisible by 4, if, and only if, the last 2 digits are divisible by 4 forcing our 3rd plus 4th digit number to be divisible by 4 with the 3rd digit being odd. The possible 2 digit numbers are therefore 12, 16, 32, 36, 52, 56, 72, 76, 92 and 96. Since all of these candidates end in 2 or 6, the 4th digit can be safely said to be 2 or 6.

    This makes our possible numbers so far....._.._.._..2..5.._..1..6.._..0
    .............................................................. _.._.._..6..5.._..3..2.._..0
    .............................................................. _.._.._..6..5.._..7..2.._..0
    .............................................................. _.._.._..2..5.._..9..6.._..0
    ..............................................................1..2..3..4..5..6..7..8..9.10

    From our divisibility rules, a number is divisible by 3 if the sum of its digits is divisible by 3. Similarly, a number is divisible by 6 if the sum of its digits is divisible by 3. SInce the first 3 digits must be divisible by 3, as must the first 6 digits, the number formed by the 4th, 5th, and 6th digits must be divisible by 3. Having the 4th and 5th digits, and knowing that the 6th digit must be even, we know that this 6th digit can only be 4 or 8.

    This makes our possible numbers so far... _..8.._..2..5..4..1..6.._..0
    ..............................................................._..4.._..2..5..8..1..6.._..0
    .............................................................. _..8.._..6..5..4..3..2.._..0
    .............................................................. _..4.._..6..5..8..3..2.._..0
    .............................................................. _..8.._..6..5..4..7..2.._..0
    .............................................................. _..4.._..6..5..8..7..2.._..0
    .............................................................. _..8.._..2..5..4..9..6.._..0
    .............................................................. _..4.._..2..5..8..9..6.._..0
    ..............................................................1..2..3..4..5..6..7..8..9.10

    However, the digits in 254 and 658 do not add up to a number divisible by 3.

    This makes our possible numbers so far.. _..4.._..2..5..8..1..6.._..0 A
    .............................................................. _..8.._..6..5..4..3..2.._..0 B
    .............................................................. _..8.._..6..5..4..7..2.._..0 C
    .............................................................. _..4.._..2..5..8..9..6.._..0 D
    ..............................................................1..2..3..4..5..6..7..8..9.10

    We know that any number placed in the 9th digit will make the total number divisible by 9 so lets work with the 1st and 3rd digits.

    The numbers remaining in the 4 cases are 3-7-9 for case A, 1-7-9 for case B, 1-3-9 for case C, and 1-3-7 for case D.

    We have already pointed out from our divisibility rules that a number is divisible by 3 if the sum of its digits is divisible by 3. Which of the three candidate groups of 3 remaining digits will make our first three digits divisible by 3?

    I think this final screening will lead you to the remaining possibilities, with the divisibility by 7 forcing the final answer to fall out.

    Many thanks and much appreciation to MathMat for initially unearthing this clever solution.

  • math -

    Insert operation signs to make this correct: 2 _5_3 = 17

  • math -

    I need to make up ten numbers that are divisible by 2,3,4,5,6,8,9,10. Please help.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. algebra

    how would i find least common multiple of 20, 50 and 3 ?
  2. Math

    I need some help please Determine the least four digit number that is divisible by 2,3, and 5. Explain your reasoning. Explain how you know that 123 123 123 900 is divisible by 2,3,5,9, and 10. How do you know that it is also divisible …
  3. Math/Algebra

    if you add one number divisible by n and one number not divisble by n will the result be divisible by n?
  4. teaching math

    Indicate whether the deductive reasoning used is an example of affirming the hypothesis or denying the conclusion. If a number is divisible by 3, then the sum of the digits of that number is divisible by 3. The sum of the digits of …
  5. Math

    The question is this: You know that a number is divisible by 6 if it is divisible by both 3 and 2. So why isn't a number divisible by 8 if it is divisible by both 4 and 2?
  6. Math

    This question confuses me... Identify a number thatis divisible by 17, 3, and 9, which is not divisible by 7, 18, 5, 16, and 14.Describe how you found this answer.
  7. math

    The following conjectures were made by students. Are they true or false?
  8. Math

    Which number can be used as a counter example to, "Any number that is divisible by 3 is divisible by 9"?
  9. Math PLEASE HELP!!!

    Which of the following is false? a. A number that is divisible by 3 and 20 is divisible by 60. b. A number that is divisible by 4 and 15 is divisible by 60. c. A number that is divisible by 5 and 12 is divisible by 60. d. A number
  10. Maths

    Is this statement true? Give an example to support your answer. "If two numbers are divisible by another number then their difference is also divisible by that number". I think that its like this: 3 and 4 is divisible by 12 then 4-3=1

More Similar Questions