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

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

### 10 Answers

- ComoLv 72 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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- Jun AgrudaLv 72 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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- PinkgreenLv 72 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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- PhilipLv 62 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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- 2 months ago
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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- ElaineLv 72 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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- PuzzlingLv 72 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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- JohnathanLv 72 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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- Canada > usaLv 62 months ago
47 is the answer

- PuzzlingLv 72 months agoReport
Clearly since 47 is prime it can't be the product of 6 integers, consecutive or otherwise.

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