Waste less time on Facebook — follow Brilliant.
×

Help please!

My teacher gave me a problem. Let \(1+4+10+20+35+..........+n=an^3-bn^2+cn-d\). If \(\frac{b}{a}+ \frac{d}{c}= \frac{e}{f}\) and \(gcd(e,f)=1\). What is the value of \(gcd(e+f,171)\)?

Note by Fatin Farhan
4 years ago

No vote yet
2 votes

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

My answer is gcd(e + f, 171) = 1.

Solution: First, we must know that 1, 4, 10, 20, 35, ... form a kind of sequence. How do I know this? The sequence has a common difference in the third iteration or third operation of finding their common difference. (Sorry, I don't know how to type in Latex so I have to save time typing instead of typing and adjusting the texts). This sequence is ruled by the polynomial P(n) = (1/6) n^3 + (1/2) n^2 + (1/3) n.

Now to attack the question, since we have a sum of this sequence, therefore we have to sum up P(n). (Sorry again for cheating by using the formulas, I don't have much knowledge and mastery of Abel's Formula for Derivation.)

Since summation of n = (1/2)(n)(n+1) Summation of n^2 = (1/6)(n)(n+1)(2n+1) Summation of n^3 = ((1/2)(n)(n+1))^2

To simplify this up, Summation of P(n) = (1/24)(n^4 + 6 n^3 + 11 n^2 + 6 n) Equating this to an^3-bn^2+cn-d, we have to make equivalences since the summation of P(n) has degree of 4 while the given has 3.

This will yield: a = n/24 b = -n/4 c = 11n/24 d = -n/4 And checking these values will be equivalent to P(n).

Now to evaluate b/a+d/c=e/f (-n/4)(24/n) + (-n/4)(24/11n) = e/f e/f = -72/11 gcd (-72, 11) = 1 So e + f = -61 or 61 (This depends on where you will assign the negative sign but it does not have an effect on the gcd of these numbers.)

Final Answer: gcd(-61, 171) = gcd(61, 171) = 1

John Ashley Capellan - 4 years ago

Log in to reply

Further information regarding my solution: 1. The common difference that I am referring to is that when you get the differences of each terms in the sequence. In the first operation, the differences (for the first 5 terms) are 3, 6, 10, and 15. In the second operation, we obtain differences of 3, 4, and 5. And finally, in the third operation, we get the uniform difference of 1.

  1. The summations are governed by the rule of summation wherein the summation of sums, we can distribute their summations each term. And the other rule such that the summation of a product of a constant and a variable is equivalent to the constant times the summation of the variable.

John Ashley Capellan - 4 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...