Proofs (1 Viewer)

kractus

Member

How would you do both these questions? I don't know where to start

011235

Active Member
Q14 (a)

Let n=2k+1 (odd) where k is a positive integer.

Let n consecutive numbers be a+1, a+2, a+3, ..., a+n for some integer a.

The sum of these is;

$\bg_white a+1+a+2+a+3+\dots+a+n = \\ na+(1+2+3+\dots+n) = \\ (2k+1)a+(1+2+3+\dots+2k+2k+1) = \\(2k+1)a+(1+2k) + (2+2k-1) + (3+2k-2) + \dots + (k+k+1) + (2k+1) = \\ (2k+1)(a + k + 1) = \\ n(a+k+1)$

Hence the sum is divisible by n.

Now try using the same logic for (b)

jimmysmith560

Le Phénix Trilingue
Moderator
Would the following working help with Question 15?

Let the digits of a 4-digit number n be a, b, c, d.

That is,

$\bg_white n=1000a + 100b +10c +d$

$\bg_white =(999a+99b+9c)+(a+b+c+d)$

$\bg_white =3(333a+33b+3c)+(a+b+c+d)$

Part 1:

If n is divisible by 3, then $\bg_white n=3m$ for some integer m.

Therefore:

$\bg_white 3m=3(333a+33b+3c)+(a+b+c+d)$

$\bg_white (a+b+c+d)=3(m-333a-33b-3c)$

Therefore the sum of the digits (𝑎 + 𝑏 + 𝑐 + 𝑑) is a multiple of 3.

Part 2:

If the sum of the digits is a multiple of 3, then (𝑎 + 𝑏 + 𝑐 + 𝑑) = 3𝑘 for some integer 𝑘.

Therefore:

$\bg_white n=3(333a+33b+3c)+3k$

$\bg_white =3(333a+33b+3c+k)$

Therefore 𝑛 is divisible by 3. We have now proved the result in both directions, so a 4-digit number is divisible by 3 if and only if the sum of its digits is divisible by 3.