sam asked in Science & MathematicsMathematics · 2 months ago

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

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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

• 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

• 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.

• 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.

• 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

• 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

• 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.

• 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.)

• 2 months ago

• Puzzling
Lv 7
2 months agoReport

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