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Suppose that the least common multiple of the first 25 positive integers is equal to

26A7114B4C0.

Find 100*A + 10*B + C.

\(\begin{array}{|rcll|} \hline && \text{lcm}(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25) \\ &=& 26{\color{red}7}7114{\color{red}4}4{\color{red}0}0 \\ && 26{\color{red}A}7114{\color{red}B}4{\color{red}C}0 \\\\ && A=7 \\ && B=4 \\ && C= 0\\ && 100*A + 10*B + C \\ &=& 100*7 + 10*4 + 0 \\ &=& 700 + 40 + 0 \\ &\mathbf{=}& \mathbf{740} \\ \hline \end{array} \)

Find 100*A + 10*B + C.

[100*7] + [10*4 + 0] =

700        +    40            = 740

The answer is 4, 64, 324, 1024. 

Could you start with the amount of options there are for a square to be chosen?  Would that matter?

Suppose x⁴+k=(a₁*x²+b₁*x+c₁)*(a₂*x²+b₂*x+c₂). this means that a₁*a₂*x⁴=x⁴, therefore a₁*a₂=1. to make things simpler, we can multiply the first trinomial by a₂ and the second trinomial by a₁ (this doesn't change the result, because a₁*a₂=1) and get:

x⁴+k=(a₂*a₁*x²+a₂*b₁*x+a₂*c₁)*(a₁*a₂*x²+a₁*b₂*x+a₁*c₂)=(1*x²+a₂*b₁*x+a₂*c₁)*(1*x²+a₁*b₂*x+a₁*c₂​)=

(x²+a₂*b₁*x+a₂*c₁)*(x²+a₁*b₂*x+a₁*c₂​) 

(I did this because i wanted to get rid of the coefficients of x² in the trinomials. we can now conclude that x⁴+k is a product of two trinomials with integer coefficients if and only if  it is a product of two trinomials with integer coefficients where the coefficient of x² in each of the trinomials is 1. Now we can assume that the coefficients of x² in each of the trinomials)

suppose x⁴+k=(x²+a*x+b)*(x²+c*x+d)=x⁴+a*x³+c*x³+b*x²+d*x²+(a*c)*x²+(a*d)*x+(c*b)*x+b*d=

x⁴+x³*(a+c)+x²*(b+d+a*c)+x*(a*d+c*b)+b*d. From this we can conclude that a+c=0, meaning that c=-a.

Let's substitute c with -a:

x⁴+k=x⁴+x³*(a+(-a))+x²*(b+d+a*(-a))+x*(a*d+(-a)*b)+b*d=x⁴+x²*(b+d-a²)+x*(a*(d-b))+b*d

If a=0:

x⁴+k=x⁴+x²*(b+d-a²)+x*(a*(d-b))+b*d=x⁴+x²*(b+d)+b*d. the coefficient of x² has to be 0, therefore d=-b. Let's substitute d with -b:

x⁴+k=x⁴+x²*(b+d)+b*d=x⁴+x²*0+b*(-b)=x⁴-b². this means that k=-b². but -b² is a nonpositive number (because b² is nonnegative) meaning that k has to be nonpositive. But k has to be positive, so that is a contradiction.

Therefore 'a' CANNOT be 0.

Now that we know we can divide by 'a':

x⁴+k=x⁴+x²*(b+d-a²)+x*(a*(d-b))+b*d. The coefficient of x is supposed to be 0, therefore a*(d-b)=0, meaning that d-b=0, and that d=b. Let's substitute d with b:

x⁴+k=x⁴+x²*(b+b-a²)+x*(a*(b-b))+b*b=x⁴+x²*(2*b-a²)+b². the coefficient of x² is supposed to be 0, meaning that 2*b-a²=0, and that 2*b=a². this means that b=a²/2. We also know that k=b², meaning that k=a⁴/4. This means that if an integer k satisifies the condition that is specified in the question, there exist a positive integer a such that a⁴/4=k. b=a²/2, meaning that 'a' has to be an even number. a=2*n, meaning that n⁴*4=k. so k satistifes the conditions if there exist a positive integer n such that n⁴*4=k. 1⁴*4=4, 2⁴*4=64, 3⁴*4=324. We need to find the number of nonnegative integers n such that n⁴*4<=2000.

n⁴*4<=2000 if and only if n⁴<=500 if and only if n<=500^1/4≈4.72870804502. Therefore, the largest positive integer n that satisfies n⁴*4<=2000 is 4, and there are exactly 4 positive integers that satisfy the equation- 1, 2, 3 and 4. This means that there are 4 values of k that are positive and <=2000 that have a positive integer n that satisfies n⁴*4<=2000-

k=1⁴*4=4, k=2⁴*4=64, k=3⁴*4=324, and k=4⁴*4=4⁵=1024. So, There are exactly 4 values of k that satisfy the following conditions:

1. k is a positive integer between 1 and 2000

2. x⁴+k can be factored into 2 distinct trinomials, with integer coefficients

x⁴+4=(x²+2*x+2)*(x²-2*x+2)

x⁴+64=(x²+4*x+8)*(x²-4*x+8)

x⁴+324=(x²+6*x+18)*(x²-6*x+18)

x⁴+1024=(x²+8*x+32)*(x²-8*x+32)

Personal opinion: Of the people who have looked at your problem, nobody seems to be able to solve it so far. It is not because nobody wants to help you. The problem seems genuinely to be a difficult one.

This is the updated question of what I made before

This is the problem. Now can someone help me please?

Determine all positive integers \(k\le 2000\)  for which \(x^4 + k\)  can be factored into two distinct trinomial factors with integer coefficients.

We'd like to help.....but your question is incomplete.  Can you repost it?

Please, someone help me!!!!

Heureka,

thank you kindly...makes sooo much sense!!!..

Alan proved that there are an infinite number of solutions for x^2 + y^2 =N. But, I'm not sure if that is what the question is!!. It sounds to me that there is a number N such that there are at least 2005 solutions for x^2 + y^2.

For example: x^2 + y^2 =2005^5 has 144 solutions for x^2 + y^2, so that 2005^23 will give you over 2005 solutions for x^2 + y^2, according to my CAS. 

I have no idea why the answer is what you say it is...How you get to it....thanx anyways...

\(\huge{ \begin{array}{|lrlrlrlrlrl|} \hline & a_{\color{red}1} &;& a_{\color{red}2}&;& a_{\color{red}3} &;& a_{\color{red}4} &;& \ldots &;& a_{\color{red}n} \\ \hline & \dfrac{1}{3}&;& \dfrac{8}{9}&;& 1&;& \dfrac{64}{81}&;& \ldots &;& \\\\ \Rightarrow & \dfrac{1^1}{3^1}&;& \dfrac{2^3}{3^2}&;& 1&;& \dfrac{4^3}{3^4}&;& \ldots &;& \\\\ \Rightarrow & \dfrac{{\color{red}1}^1}{3^{\color{red}1}}&;& \dfrac{{\color{red}2}^3}{3^{\color{red}2}}&; & \dfrac{{\color{red}3}^3}{3^{\color{red}3}}&;& \dfrac{{\color{red}4}^3}{3^{\color{red}4}}&;& \ldots &;&\dfrac{{\color{red}n}^3}{3^{\color{red}n}} \\ \hline \end{array} }\)

if we consider this sequence:

SAMKELLY

SAMKELLY

SAMKELLY....

The smiley is part of the sequence.

In what position will the 100th "L" be?

\(\begin{array}{|lrcll|} \hline &\boxed{ L_{2n} = 8+(n-1)\cdot 9 } \\\\ n=50: & L_{2\cdot 50}=L_{100} &=& 8+(50-1)\cdot 9 \\ & &=& 8+49\cdot 9 \\ & &=& 8+441\\ & &=& 449 \\ \hline \end{array}\)

The 100th "L" will be in position 449

If you have difficulty understanding what heureka did, just use this summation of the sequence which will give you the same answer:

∑[ 3 F_n mod 2, n, 1, 2016] =1,344, where F_n =the nth Fibonacci number.

Can someone prove this to me?

Good heavens Heureka,

this is WWAAAAAYYYYY over my comprehension!!!..thank you soo much!!

Hi Alan,

thanx, I'll sit with it a bit to try and understand what you did...thanx a lot..

Hi Alan,

thanx, I'll sit with it a bit to try and understand what you did...thanx a lot..

Hi Heureka,

I have no idea why the answer is what you say it is...How you get to it....thanx anyways...

Hi Heureka,

I have no idea why the answer is what you say it is...How you get to it....thanx anyways...

(3n)^2 + (4n)^2 = (5n)^2

Just substitute ANY number for n, from n=1 to infinity!. So, if you want to sub 2005, this is what you will get:

(3*2005)^2 + (4*2005) =(5*2005)^2

6015^2  +  8020^2 = 10025^2

36,180,225  +  64,320,400 =100,500,625

100,500,625 =100,500,625. As you can see, you can sub ANY number from 1, 2, 3, 4........2005, 2006, 2007...etc.

A sequence of 2016 terms is formed as follows: The first two terms are equal to 3, from the 3rd term onwards, each consecutive term is formed by finding the sum of the previous two terms.

If each of the terms of the sequence 3,3,6,9....are now devided by 2,

and the remainders are added,

what will the sum of the first 2016 remainders be?

\(\begin{array}{|rcll|} \hline && \text{Let } \mathcal{F} \text{ are the Fibonacci numbers }\\\\ \mathcal{F}_1 &=& 1\\ \mathcal{F}_2 &=& 1\\ \mathcal{F}_3 &=& 2\\ \mathcal{F}_4 &=& 3\\ \mathcal{F}_5 &=& 4\\ \mathcal{F}_6 &=& 5\\ \mathcal{F}_7 &=& 13\\ \mathcal{F}_8 &=& 21\\ \mathcal{F}_9 &=& 34\\ \mathcal{F}_{10} &=& 55\\ \mathcal{F}_{11} &=& 89\\ \mathcal{F}_{12} &=& 144\\ \ldots \\ \hline \end{array} \)

The sequence:

\(\begin{array}{|rccr|l|} \hline a_1 &=& &=& 1\cdot 3 \\ a_2 &=& &=& 1\cdot 3 \\ a_3 &=& 1\cdot 3 + 1\cdot 3 &=& 2\cdot 3 \\ a_4 &=& 1\cdot 3 + 2\cdot 3 &=& 3\cdot 3 \\ a_5 &=& 2\cdot 3 + 3\cdot 3 &=& 5\cdot 3 \\ a_6 &=& 3\cdot 3 + 5\cdot 3 &=& 8\cdot 3 \\ a_7 &=& 5\cdot 3 + 8\cdot 3 &=& 13\cdot 3 \\ a_8 &=& 8\cdot 3 + 13\cdot 3 &=& 21\cdot 3 \\ a_9 &=& 13\cdot 3 + 21\cdot 3 &=& 34\cdot 3 \\ \ldots \\ a_n &=& && \mathcal{F}_{n} \cdot 3 \\ \hline \end{array} \)

\(\begin{array}{|r|rcl|c|} \hline \text{cycle} && && \text{remainder after divided by } 2 \\ \hline 1 & a_1 &=& 1\cdot 3 & 1 \\ 1 & a_2 &=& 1\cdot 3 & 1 \\ 1 & a_3 &=& 2\cdot 3 & 0 \\ \hline 2 & a_4 &=& 3\cdot 3 & 1 \\ 2 & a_5 &=& 5\cdot 3 & 1 \\ 2 & a_6 &=& 8\cdot 3 & 0 \\ \hline 3 & a_7 &=& 13\cdot 3 & 1 \\ 3 & a_8 &=& 21\cdot 3 & 1 \\ 3 & a_9 &=& 34\cdot 3 & 0 \\ \hline \ldots \\ \hline 672 & a_{2014} &=& \mathcal{F}_{2014} \cdot 3 & 1 \\ 672 & a_{2015} &=& \mathcal{F}_{2015} \cdot 3 & 1 \\ 672 & a_{2016} &=& \mathcal{F}_{2016} \cdot 3 & 0 \\ \hline &&&& \text{sum }=672*2 = 1344 \\ \hline \end{array} \)

The sum of the first 2016 remainders will be 1344

what is a committee forming argument? what is a block walking argument?