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sam asked in Science & MathematicsMathematics · 2 months ago

The sum of 6 consecutive integers is 153. What is their product?

Please help!!! i need yo understand this for a test

10 Answers

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  • Como
    Lv 7
    2 months ago

    x  + (x  + 1) + (x. + 2) + (x  + 3) + (x  + 4)  + (x  + 5) = 153

    6x  + 15    = 153

    6x   = 138

    x   = 23

    Product   =  23 x 24 x 25 x 26 x 27 x 28

    Product   =  22 604 400

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  • 2 months ago

    Let x = start number.

    Equation:

    x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 = 153

    6x + 15 = 153

    6x = 138

    x = 23

    Their product:

    = 23 * 24 * 25 * 26 * 27 * 28

    = 271,252,800

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  • 2 months ago

    Let the 6 numbers be

    x-2, x-1, x , x+1, x+2, x+3.

    their sum is

    6x+3=153=>x=25

    Thus their product=

    23*24*25*26*27*28=

    271252800.

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  • Philip
    Lv 6
    2 months ago

    Put the integers = (n-2), (n-1), n, (n+1). (n+2), (n+3). Sum = 6n+3 = 153, n = 25.

    Product = n(n+3)(n^2-1)(n^2-4) = 25*28*624*621 = 700*387,504 = 271,252,800.

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  • 153 = m - 2.5 + m - 1.5 + m - 0.5 + m + 0.5 + m + 1.5 + m + 2.5

    153 = 6m

    25.5 = m

    (m - 2.5) * (m - 1.5) * (m - 0.5) * (m + 0.5) * (m + 1.5) * (m + 2.5) =>

    (m^2 - 6.25) * (m^2 - 2.25) * (m^2 - 0.25) =>

    (1/4) * (4m^2 - 25) * (1/4) * (4m^2 - 9) * (1/4) * (4m^2 - 1) =>

    (1/64) * (4m^2 - 25) * (4m^2 - 9) * (4m^2 - 1) =>

    (1/64) * (4 * (51/2)^2 - 25) * (4 * (51/2)^2 - 9) * (4 * (51/2)^2 - 1) =>

    (1/64) * (4 * 2601/4 - 25) * (4 * 2601/4 - 9) * (4 * 2601/4 - 1) =>

    (1/64) * (2601 - 25) * (2601 - 9) * (2601 - 1) =>

    (1/64) * 2576 * 2592 * 2600 =>

    (1/64) * 26 * 100 * (2584 - 8) * (2584 + 8) =>

    (1/32) * 13 * 100 * 4 * (646 - 2) * 4 * (646 + 2) =>

    (1/2) * 13 * 100 * 4 * (323 - 1) * (323 + 1) =>

    2 * 13 * 100 * (323 - 1) * (323 + 1) =>

    26 * (323^2 - 1) * 100 =>

    26 * ((300 + 23)^2 - 1) * 100 =>

    26 * (90000 + 6900 + 6900 + 529 - 1) * 100 =>

    26 * (90000 + 13800 + 528) * 100 =>

    26 * (103800 + 528) * 100 =>

    26 * 104328 * 100 =>

    (25 + 1) * 104328 * 100 =>

    (25 + 1) * 4 * 26082 * 100 =>

    100 * 100 * 26082 + 4 * 100 * 26082 =>

    260,820,000 + 10,432,800 =>

    270,000,000 + 1,252,800 =>

    271,252,800

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  • Elaine
    Lv 7
    2 months ago

    Here is the algebraic way of solving the problem. Method works every time. 

    Your consecutive integers are: a, a + 1, a + 2, a+ 3, a +4, a+5 = 153

    6a + 15 = 153

    6a = 153 - 15

    6a = 138

    a = 23

    Your integers are 23, 24, 25 , 26, 27, 28

    Check 

    23 + 24 +25 +26 +27 + 28 = 153

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  • 2 months ago

    The sum of the numbers is 153.

    There are 6 of them.

    So the average is 153/6 = 25.5

    If you think about it, the average would be right between the middle two integers, so they must be 25 and 26. And from there we can figure out all 6.

    23, 24, 25, 26, 27, 28

    Now find their product by multiplying them together:

    23 * 24 * 25 * 26 * 27 * 28

    = 271,252,800

    UPDATE:

    If you wanted to be a little more formal, you could use algebra.

    Let the first integer be n.

    Let the next integer be n+1.

    etc.

    The sum is 153:

    n + n+1 + n+2 + n+3 + n+4 + n+5 = 153

    6n + 15 = 153

    6n = 138

    n = 138/6

    n = 23

    So the first number is 23 and then the others are 24, 25, etc. Like before, multiply them together to get the product.

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  • 2 months ago

    x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) = 153

    6x + 15 = 153

    6x = 138

    x = 23.  So the consecutive integers are 23, 24, 25, 26, 27, and 28.

    23 * 24 * 25 * 26 * 27 * 28 = 271,252,800.  Final.  (Very tedious without the use of a calculator, but it's obtainable.)

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  • 2 months ago

    47 is the answer

    • Puzzling
      Lv 7
      2 months agoReport

      Clearly since 47 is prime it can't be the product of 6 integers, consecutive or otherwise.

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  • Fuhr
    Lv 6
    2 months ago

    Seven              .

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