Melody

Hi Ginger, it is nice to see you back again. :)

What is the average of all positive integers that have four digits when written in base 3, but two digits when written in base 6? Write your answer in base 10.

4 digits base 3

smallest 1000 = 3^3 = 27

biggest  2222 = 2*27+2*9+2*3+2 = 80

3 digits base 6

smallest 100 = 36

biggest  1000-1 = 6^3-1 = 4095

So the intersection of these is    36 to 80  there are  80-36+1 = 45 of them

sum of numbers from 36 to 80 = (45/2)(72+44) = 45/2*116= 45*58 = 2610

average = 2610/45 =58

A more straight forward approach would be to say that of there are 45 numbers the median on is in the 46/2 = 23rd position.  In this case the median and the mean will be the same.

36+23-1 = 58   Just as expected.

So the answer is 58 base 10

It has just been pointed out to me that I have answered for 3 digits base 6 instead of 2 digits.

You will need to adjust what i have done to get the correct answer.  

I am here to help you learn, not to help you meet your deadlines.

The polynomial f(x) has degree 3. If f(-1) = 15, f(0)= 0, f(1) = -5, and f(2) = 12 , then what are the $x$-intercepts of the graph of f? Explain.

\(f(x)=ax^3+bx^2+cx+d\\ f(0)=d=0 \qquad \text{so d =0}\\~\\ f(-1)=-a+b-c=15\qquad (1a)\\ f(1)=a+b+c=-5\qquad (2a)\\ f(2)=8a+4b+2c=12\qquad(3a)\\~\\ b-15=a+c\qquad (1b)\\ a+c=-b-5\qquad (2a)\\ so\\ b-15=-b-5\\ 2b=10\\ b=5\\ so\ \)

\(-a+5-c=15\qquad (1c)\\ a+c=-10 \qquad(1d)\\ 8a+20+2c=12\qquad(3c)\\ 4a+c=2\qquad(3d)\\\)

\(a+c=-10 \qquad(1d)\\ 4a+c=2\qquad(3d)\\ Solve\;\; simultaneously\\ etc\)

Please let this guest finish it and check for errors all by him/herself !

Is   (11^p+7)        \(11^p+7 \qquad\text { or is it } \qquad11^{p+7}\qquad ?\)

so   ((11^p)+7) or   (11^(p+7))

Thanks Chris,

Yes it is nifty.

Sometimes is works more easily than other times. This was a nice example. Not totally trivial but not very hard either.

The fact  that it equals one makes it easier.

I spent some time practicing questions like this a while back.

It is pretty cool. Euclid was a smart cookie   

lim_(x->∞) (sqrt(4 x^2 + x) + (sqrt(9 x^2 - x) - 5 x)) = 1/12

\( \displaystyle\lim_{x\rightarrow \infty}\;\;[\sqrt{4 x^2 + x}+ \sqrt{9 x^2 - x} - 5 x]= \frac{1}{12}\)

There is nothing to solve.  This is either true or not true and it is not true

Have you written the question incorrectly?

Thanks Heureka   

164x+37y=1

164=4*37+16

37=2*16+5

16=3*5+1

now substituting backwards

1=16-3*5

1=16-3*[37-2*16]

1= -3*37 + 7*16

1= -3*37 + 7*[164-4*37]

1= -3*37 + 7*164-28*37

1= 7*164-31*37

1= 164(7) + 37(-31)

An answer is x= 7 and y= -31

I know it is not needed but now I will find the general solution.

1= 164(7) + 37(-31)

1= 164(7-37t) + 37(-31+164t)

x=7-37t   and  y=164t-31     where t is an integer.

The maximum is 7 and the minimum is 0 so  product of these is 0

Heureka, could you please check if what I have done is valid    

I am not sure about the rules for modulus but this is my attempt

What is the remainder when 129^(34)+96^(38) is divided by 11?

129^(34)+96^(38) [mod11]

= (-3)^34+(-3)^38 [mod 11]

=(-3)^34(1+(-3)^4) [mod11]

= 9^17 (1+81) [mod11]

= (-2)^17(82) [mod11]

=( -32*32*32*4)(5) [mod11]

= (-1*-1*-1*-1 *4)(5) [mod11]

=20 mod11

=9  mod 11