Question

2022-01-27
Suppose $a_{n}$ as the sum of the numbers in which the quotient and the remainder are equal when divided by $n.$ For example, $a_{5}=6+12+18+24=60.$ Find the least $n$ such that $a_{n}\ \textgreater \ 500.$​

Answer 1

Answer:

• n = 11

Step-by-step explanation:

We can observe general terms of n:

• aₙ = n + 1 + 2n + 2 + 3n + 3 + ... + (n - 1)n + (n - 1)

Rewrite it as:

• aₙ = (n + 1) + 2(n + 1) + 3(n + 1) + ... (n - 1)(n + 1) =
•       (n + 1)(1 + 2 + 3 + ... + n - 1) =
•       (n + 1)(1 + n - 1)(n - 1)/2 =
•       (n - 1)n(n + 1)/2

We need to find the least n for which aₙ > 500:

• (n - 1)n(n + 1)/2 > 500 ⇒ (n - 1)n(n + 1) > 1000

If n = 10:

• a₁₀ = 9*10*11 = 990 < 1000

If n = 11:

• a₁₁ = 10*11*12 = 1320 > 1000

So the least n is 11

Answer 2

If I'm understanding the construction of this sequence correctly, we have

• a₁ = 0

(the sum is empty since there is no positive integer b such that b•1 + b = 1)

• a₂ = 3

(since 3 = 1•2 + 1)

• a₃ = 4 + 8 = 12

(since 4 = 1•3 + 1 and 8 = 2•3 + 2)

• a₄ = 5 + 10 + 15 = 30

(since 5 = 1•4 + 1, 10 = 2•4 + 2, and 15 = 3•4 + 3)

and so on.

Notice that for n ≥ 2, aₙ is simply the sum of the first n - 1 multiples of n + 1. So

$\displaystyle a_n = \sum_{i = 1}^{n - 1} (n+1)i$

Recall that

$\displaystyle \sum_{i=1}^n i = \frac{n(n+1)}2$

Then

$\displaystyle a_n = (n+1) \sum_{i=1}^{n-1} i = (n+1) \cdot \frac{(n-1)n}2 = \frac{n(n^2-1)}2$

Solve for n such that aₙ > 500 :

n (n² - 1)/2 > 500

n (n² - 1) > 1000

n³ - n > 1000

We can solve this by inspection. Noticing that 10³ = 1000, if we replace n = 10 we get

10³ - 10 = 990 > 1000

which is false, but the difference is quite small. So we move up to n = 11 and find

11³ - 11 = 1320 > 1000

which is true, so n = 11 is the least number such that aₙ > 500.