Yahoo Answers: Answers and Comments for The sum of 6 consecutive integers is 153. What is their product? [Mathematics]
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From sam
enUS
Wed, 22 Jan 2020 21:31:03 +0000
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Yahoo Answers: Answers and Comments for The sum of 6 consecutive integers is 153. What is their product? [Mathematics]
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From Puzzling: The sum of the numbers is 153.
There are 6 of ...
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Wed, 22 Jan 2020 21:38:31 +0000
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.

From Pinkgreen: Let the 6 numbers be
x2, x1, x , x+1, x+2, x...
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Fri, 24 Jan 2020 16:12:52 +0000
Let the 6 numbers be
x2, x1, 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.

From Como: x + (x + 1) + (x. + 2) + (x + 3) + (x + 4)...
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Sun, 26 Jan 2020 09:30:42 +0000
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

From Jun Agruda: Let x = start number.
Equation:
x + x + 1 + x ...
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Sat, 25 Jan 2020 19:04:22 +0000
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

From Philip: Put the integers = (n2), (n1), n, (n+1). (n+...
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Thu, 23 Jan 2020 07:25:37 +0000
Put the integers = (n2), (n1), n, (n+1). (n+2), (n+3). Sum = 6n+3 = 153, n = 25.
Product = n(n+3)(n^21)(n^24) = 25*28*624*621 = 700*387,504 = 271,252,800.

From Captain Matticus, LandPiratesInc: 153 = m  2.5 + m  1.5 + m  0.5 + m + 0.5 + ...
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Wed, 22 Jan 2020 22:09:56 +0000
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

From Elaine: Here is the algebraic way of solving the probl...
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Wed, 22 Jan 2020 21:44:17 +0000
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

From Johnathan: x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x...
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Wed, 22 Jan 2020 21:37:12 +0000
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.)

From Fuhr: Seven .
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Wed, 22 Jan 2020 21:32:39 +0000
Seven .

From Anonymous: 47 is the answer
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Wed, 22 Jan 2020 21:34:01 +0000
47 is the answer