For which k is it true that the sum of k consecutive integers is always divisible by k?

Mathematics

1 answer:

nevsk [136]2 years ago

5 0

Answer:

Step-by-step explanation:

We have two consecutive even integers:

a and b.

We know that we can write a even integer as:

a = 2*n

where n is an integer.

Then the consecutive integer to a, b, will be:

b = 2*n + 2.

Then the sum of a and b is:

a + b = 2n + 2n + 2 = 4*n + 2 = 2*(2n + 1) = k.

The next two consecutive integers greater than b will be:

c = 2n + 4

d = 2n + 6

then:

c + d = (2n + 4) + (2n + 6) = 4n + 10 = k + 4.

I hope i helped:D

You might be interested in

Can somebody help me with my geometry?

ahrayia [7]

The slope of AC is -0.4

Proof:
In triangles ABC and DBE,
∠DBE is common to both triangles.
AB = 2DB (D is the midpoint of the interval AB)
Also, BC = 2BE (E is the midpoint of the interval BC)

Thus triangles ABC and DBE are similar in the ratio 2:1

Since, they are similar, ∠BDE must equal ∠BAC (corresponding angles in similar triangles)

If ∠BDE = ∠BAC, DE must be parallel to AC (corresponding angles are equal along parallel lines)
Thus, the slope of AC = the slope of DE
Thus, the slope of AC is -0.4

3 0

4 years ago

How many times smaller is 2 × 10^-10 than 4 × 10^-5

natta225 [31]

Answer:

200,000

Step-by-step explanation:

To find how many times smaller is smaller number than greater number, divide greater number by smaller number.

Greater number $=4\times 10^{-5}$

Smaller number $=2\times 10^{-10}$

Then

$\dfrac{\text{Greater number}}{\text{Smaller number}}\\ \\=\dfrac{4\times 10^{-5}}{2\times 10^{-10}}\\ \\=\dfrac{4}{2}\times \dfrac{10^{-5}}{10^{-10}}\\ \\=2\times \dfrac{1}{10^{-5}}\\ \\=2\times 10^5\\ \\=2\times 100,000\\ \\=200,000$