Adding Fractions 4

Directions: Using the integers 1 to 10 at most one time each, fill in the boxes so that the sum is equal to 1.

Hint

What three unit fractions add up to exactly 1 whole? What numbers would you use for the denominators? Why?

Answer

There are many solutions. Here are some possible solutions:
1/6 + 4/8 + 3/9
3/6 + 2/8 + 1/4

Source: Joshua Nelson

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Add Fractions with Decimal Sums

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5 comments

  1. Our class came up with two different solutions:

    1.) 1/4 + 2/8 + 5/10

    2.) 3/6 + 1/4 + 2/8

  2. Sorry, I forget it.is “2.)”

  3. all 48 solutions

    1/4 + 2/8 + 3/6= 1
    1/4 + 2/8 + 5/10= 1
    1/4 + 3/6 + 2/8= 1
    1/4 + 5/10 + 2/8= 1
    1/5 + 2/4 + 3/10= 1
    1/5 + 3/10 + 2/4= 1
    1/5 + 3/10 + 4/8= 1
    1/5 + 4/8 + 3/10= 1
    1/6 + 2/4 + 3/9= 1
    1/6 + 3/9 + 2/4= 1
    1/6 + 3/9 + 4/8= 1
    1/6 + 3/9 + 5/10= 1
    1/6 + 4/8 + 3/9= 1
    1/6 + 5/10 + 3/9= 1
    1/10 + 2/5 + 3/6= 1
    1/10 + 2/5 + 4/8= 1
    1/10 + 3/6 + 2/5= 1
    1/10 + 4/8 + 2/5= 1
    2/4 + 1/5 + 3/10= 1
    2/4 + 1/6 + 3/9= 1
    2/4 + 3/10 + 1/5= 1
    2/5 + 1/10 + 3/6= 1
    2/5 + 1/10 + 4/8= 1
    2/5 + 3/6 + 1/10= 1
    2/5 + 4/8 + 1/10= 1
    2/8 + 1/4 + 3/6= 1
    2/8 + 1/4 + 5/10= 1
    2/8 + 3/6 + 1/4= 1
    2/8 + 5/10 + 1/4= 1
    3/6 + 1/4 + 2/8= 1
    3/6 + 1/10 + 2/5= 1
    3/6 + 2/5 + 1/10= 1
    3/6 + 2/8 + 1/4= 1
    3/9 + 1/6 + 2/4= 1
    3/9 + 1/6 + 4/8= 1
    3/9 + 1/6 + 5/10= 1
    3/10 + 1/5 + 2/4= 1
    3/10 + 1/5 + 4/8= 1
    3/10 + 2/4 + 1/5= 1
    3/10 + 4/8 + 1/5= 1
    4/8 + 1/5 + 3/10= 1
    4/8 + 1/6 + 3/9= 1
    4/8 + 1/10 + 2/5= 1
    4/8 + 2/5 + 1/10= 1
    4/8 + 3/10 + 1/5= 1
    5/10 + 1/4 + 2/8= 1
    5/10 + 1/6 + 3/9= 1
    5/10 + 2/8 + 1/4= 1

  4. All 48 solutions reduce to 9 unique solutions. (ignore the order)

    a/b+c/d+e/f=1

    let a<c<e, done.

    1/4 + 2/8 + 3/6= 1
    1/4 + 2/8 + 5/10= 1
    1/5 + 2/4 + 3/10= 1
    1/5 + 3/10 + 4/8= 1
    1/6 + 2/4 + 3/9= 1
    1/6 + 3/9 + 4/8= 1
    1/6 + 3/9 + 5/10= 1
    1/10 + 2/5 + 3/6= 1
    1/10 + 2/5 + 4/8= 1

    • Rudolf Österreicher

      If a/b + c/d + e/f = 1, then the following 6 equations (obtained by swapping summands (fractions) around, multiplying the number of solutions by 6) are also true:

      a/b + c/d + e/f = 1
      c/d + a/b + e/f = 1
      a/b + e/f + c/d = 1
      c/d + e/f + a/b = 1
      e/f + c/d + a/b = 1
      e/f + a/b + c/d = 1

      So for each solution, there are 5 equivalent ones, multiplying the number of solutions by 6.

      So there are 9 * 6 = 54 solutions that reduce down to these 9 unique ones.

      Your list of “all” 48 solutions…
      … has 2/4 + 1/6 + 3/9= 1, but is missing 2/4 + 3/9 + 1/6= 1
      … has 3/9 + 1/6 + 2/4= 1, but is missing 3/9 + 2/4 + 1/6= 1
      … has 3/9 + 1/6 + 4/8= 1, but is missing 3/9 + 4/8 + 1/6= 1
      … has 3/9 + 1/6 + 5/10= 1, but is missing 3/9 + 5/10 + 1/6= 1
      … has 4/8 + 1/6 + 3/9= 1, but is missing 4/8 + 3/9 + 1/6= 1
      … has 5/10 + 1/6 + 3/9= 1, but is missing 5/10 + 3/9 + 1/6= 1
      With these additional 6 solutions, you get all the 54 solutions.

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