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Hint: what is the square root of 49?

what is the question?

N mod 4 = 3
N mod 5 = 2
N mod 7 = 6 

4A+3=5B+2=7C+6
A = 6, B = 5, C = 3
Therefore:
4 x 6 + 3 = 27

Since LCM of [4, 5, 7] = 140, then:
N =140D + 27, where D =0, 1, 2, 3.......etc.

How do we find out the value of side y? We only know x...

I totally agee with you, @Guest. That's the only way to evaluate it! 

f(x) = ax + b means that f(f(x)) = a(ax+b) + b

So f(f(x)) = a²x + ab + b

If this equals 9x + 8 then we must have a² = 9, so a = 3 (or -3) and ab + b = 8

With a=3:  3b + b = 8, so b = 2

With a=-3:  -3b + b = 8, so b = -4

.

In the sequence {2001, 2002, 2003,...} each term after the third is found

by subtracting the previous term from the sum of the two terms that precede that term.

For example, the fourth term is 2001+2002-2003=2000.

What is the value of the 2004th term? 

2. Solution:

\(\huge{ \begin{equation} a_n = \begin{cases} 2000+n \text{,} & \text{if } \ n \ \text{ is odd} \\ 2004-n \text{,} & \text{if } \ n \ \text{ is even} \\ \end{cases} \end{equation} }\)

\(\large{ \begin{array}{|rcll|} \hline a_{2004} &=& 2004 - 2004 \\ &=& 0 \\ \hline \end{array} }\)

Somebody could correct me on this, but I don't think you can "evaluate it efficiently" primarily due to two different bases(3, 4),  and the PLUS sign between them. It can be written like this:(3^1024) + (4^15625). When evaluated fully, it has 9,408 digits!!

Hi Heureka,

I am wondering if it can still be done easily if it is made more complicated ?

eg

Find the coefficient of x^3 y^3 z^2 in the expansion of (3x+5y+7z)^8.

In the sequence {2001, 2002, 2003,...} each term after the third is found

by subtracting the previous term from the sum of the two terms that precede that term.

For example, the fourth term is 2001+2002-2003=2000.

What is the value of the 2004th term? 

\(\begin{array}{|lcll|} \hline a_1 &=& 2001 \\ a_2 &=& 2002 \\ a_3 &=& 2003 \\ a_4 &=& -1\cdot a_3 + 1\cdot a_2 + 1\cdot a_1 \\ a_5 = -1\cdot a_4 + 1\cdot a_3 + 1\cdot a_2 &=& +2\cdot a_3 + 0\cdot a_2 - 1\cdot a_1 \\ a_6 = -1\cdot a_5 + 1\cdot a_4 + 1\cdot a_3 &=& -2\cdot a_3 + 1\cdot a_2 + 2\cdot a_1 \\ a_7 = -1\cdot a_6 + 1\cdot a_5 + 1\cdot a_4 &=& +3\cdot a_3 + 0\cdot a_2 - 2\cdot a_1 \\ a_8 = -1\cdot a_7 + 1\cdot a_6 + 1\cdot a_5 &=& -3\cdot a_3 + 1\cdot a_2 + 3\cdot a_1 \\ a_9 = -1\cdot a_8 + 1\cdot a_7 + 1\cdot a_6 &=& +4\cdot a_3 + 0\cdot a_2 - 3\cdot a_1 \\ a_{10} = -1\cdot a_9 + 1\cdot a_8 + 1\cdot a_7 &=& -4\cdot a_3 + 1\cdot a_2 + 4\cdot a_1 \\ a_{11} = -1\cdot a_{10} + 1\cdot a_{9} + 1\cdot a_{8} &=& +5\cdot a_3 + 0\cdot a_2 - 4\cdot a_1 \\ a_{12} = -1\cdot a_{11} + 1\cdot a_{10} + 1\cdot a_{9} &=& -5\cdot a_3 + 1\cdot a_2 + 5\cdot a_1 \\ a_{13} = -1\cdot a_{12} + 1\cdot a_{11} + 1\cdot a_{10} &=& +6\cdot a_3 + 0\cdot a_2 - 5\cdot a_1 \\ a_{14} = -1\cdot a_{13} + 1\cdot a_{12} + 1\cdot a_{11} &=& -6\cdot a_3 + 1\cdot a_2 + 6\cdot a_1 \\ \ldots \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{a_n} & \mathbf{=} & \mathbf{ -(-1)^n\lfloor \dfrac{n-1}{2}\rfloor \cdot a_3 + \lfloor\dfrac{1+(-1)^n}{2}\rfloor \cdot a_2+(-1)^n\lfloor \dfrac{n-2}{2} \rfloor \cdot a_1 } \\\\ a_{2014} &=& \small{ -(-1)^{2014}\lfloor \dfrac{2014-1}{2}\rfloor \cdot 2003 + \lfloor\dfrac{1+(-1)^{2014}}{2}\rfloor \cdot 2002 +(-1)^{2014}\lfloor \dfrac{2014-2}{2} \rfloor \cdot 2001 } \\\\ &=& -\lfloor \dfrac{2013}{2}\rfloor \cdot 2003 + \lfloor\dfrac{1+1}{2}\rfloor \cdot 2002 +1\cdot \lfloor \dfrac{2012}{2} \rfloor \cdot 2001 \\\\ &=& -\lfloor \dfrac{2013}{2}\rfloor \cdot 2003 + \lfloor\dfrac{1+1}{2}\rfloor \cdot 2002 +1\cdot \lfloor \dfrac{2012}{2} \rfloor \cdot 2001 \\\\ &=& -\lfloor \dfrac{2013}{2}\rfloor \cdot 2003 + \lfloor\dfrac{2}{2}\rfloor \cdot 2002 +1\cdot \lfloor \dfrac{2012}{2} \rfloor \cdot 2001 \\\\ &=& -1006 \cdot 2003 + 1 \cdot 2002 +1\cdot 1006 \cdot 2001 \\\\ &=& -2015018 + 2002 + 2013006 \\\\ &=& -2013016 + 2013006 \\\\ &=& 0 \\ \hline \end{array} \)

The value of the 2004th term is 0

Yes, that does sound like a good analogy.     :)

RP’s desperation is an imaginary misunderstanding. Like the child who panics because she thinks she’s drowning, but the water is only three feet deep, so all she has to do is stand up.   

Cool! I get three points for butting in.

Maybe I should log in to claim my copyright and points.  

"An arithmetic sequence with first term 1 has a common difference of 6. A second sequence begins with 4 and has a common difference of 7. In the range of 1 to 100, what is the largest number common to both sequences?"

We must have:

1 + 6n = 4 + 7m    where m and n are integers

This can be rearranged as:

6n - 7m = 3  or

2n - 7m/3 = 1

m must be a multiple of 3, and, because 2n is always even, 7m/3 must be odd to give a difference of 1.  This means m must be an odd multiple of 3.

m           n           2n-7m/3          4+7m          1+6n

3            4               1                    25               25

9            11             1                    67               67

15          18             1                  109             109

So 67 is the largest number in the range 1 to 100 that is common to both sequences.

Changing it to fractions: \(\frac{5}{19}=\frac{x}{250}\), so we get 1250/19.

Thank you

Thanks Heureka, I can't remember ever seeing this type of solution before. 

So I am very pleased that you have shown me.   

LOL   That is a lot of butting in    

Do I have to demerit myself as well since I butted in too ?

RP does not seem to need our help anyway she nearly always say she is desperate for help and then works it out within minutes by herself.   

This forum must teach by osmosis LOL 

Find the coefficient of x^3 y^3 z^2 in the expansion of (x+y+z)^8.

Trinomial Coefficient:

\(\begin{array}{|rcll|} \hline \dbinom{8}{3,3,2} &=& \dfrac{8!}{3!3!2!} \\\\ &=& \dfrac{3!4\cdot 5 \cdot 6 \cdot 7 \cdot 8}{3!3!2!} \\\\ &=& \dfrac{4\cdot 5 \cdot 6 \cdot 7 \cdot 8}{3!2!} \quad & | \quad 3! = 6 \quad 2! = 2 \\\\ &=& \dfrac{4\cdot 5 \cdot 6 \cdot 7 \cdot 8}{6\cdot 2} \\\\ &=& \dfrac{4\cdot 5 \cdot 7 \cdot 8}{2} \\\\ &=& 2\cdot 5 \cdot 7 \cdot 8 \\\\ &=& 10 \cdot 56 \\\\ &\mathbf{=}& \mathbf{ 560 } \\ \hline \end{array}\)

The trinomial coefficient of \(x^3 y^3 z^2\) in the expansion of \((x+y+z)^8\) is 560

I hope this helped,

Gavin

-1 is a demerit, for butting in.

i² = -1

That’s a negation, so it didn’t happen.

If I square my imagination, then it becomes real:

(You BUTTED in on Melody’s post!)²