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What is the smallest prime divisor of       $5^{23} + 7^{17}$

Powers of 5 are

5, 125, 625,  ..

Powers of 7 are

7,  49,  343, 2401, 16807  

7^1 ends in 7     

7^2 ends in 9      

7^3 ends in 3

7^4 ends in 1

7^5 ends in  7

and the pattern will then repeat.

7 9 3 1 7 9 3 1 7 9 3 1 7 9 3 1 7

So  5^23 ends in 5 and  7^17 ends in 7   

so

$5^{23} + 7^{17}$     ends in the last digit of  7+5   Which is 2

Any number ending in 2 is divisable by 2 so     the smallest prime divisor of that expression is 2  

Thank You!

Please do not do ALL of ANYONE'S homework for them!

Explain how they can do it themselves.

Nice answers and research from both of you. That is what I always like to see.   

Do not ask the same question multiple times, it is very rude.

If you do not understand an earlier answer then say so on the original thread!

Of couse Jasmin could make another post on this thread and ask for an explanation if she actually wants one. 

Do you want one Jasmin?

Jazmin may not have any idea what this notation really means.

Maybe one of you might like to explain it to her?  

Imheretosavetheday,

Guest has no idea what you are on about.   Maybe s/he did eventually work it out but you did not explain.

You have not even tried to teach guest a solitary thing!

Guest.

The graph is positive when the y value is positive, I mean the graph must be above the x axis.

You could change the wording to : "For what values of x will the function (graph) be positive."

YOU only refer to the x values.

so 

since it is above the x axis (where y is positive) between  -6 and -3   and then again for all values of x greater than +3

I can say 

The graph is positve from    (-6,-3)   and (+3, +infinity)      

You use ( or ) if the end point  IS NOT included

and you use [  or ]  if the end point IS included.

Please do not gimake a habit of giving answers like this.

You have just done guests homework for them but there is no reason to think that you have taught anyone anything.

A much better answer would be to explain what area and integrals have to do with each other etc etc.

Teaching answers take a lot longer to construct but they maybe  of learning value.

Please savetheday, save the day by teaching, not by giving final answers.

Also, if someone asks 2 or more questions only answer one of them.

We do not want this to be known as primarily a cheat site!

Thanks 

As you know, they are both right angled triangles.

For 45 degree ratios use a right angles isosceles triangle where the 2 equal sides are both 1 unit.

  Find the hypotenuse using pythagoras's theorem and then read off the ratio values from it.

 For both 60degrees and 30degrees.

Start with a equilateral triangle with side length 2 units.

Then cut it in half to get two congruent  triangles, you will only use one of them.

  it has a right angle, a 30degree angle and a 60 degree angle,

  the hypotenuse is 2 units and one side is 1 unit.

   Use pyth to find the 3rd side and that is it.

Look under this online page:  https://en.wikipedia.org/wiki/List_of_trigonometric_identities

Look under "Conversions of Common Angles"

1. $\text{We can use an integral: } \int^{2}_{-1}( x^3 + 2) dx = ? \text{ We now find the antiderivative of } x^3+2\\ \text{which turns out to be } \frac{1}{4} x^4 \text{. Then, we plug it in:} \frac{1}{4}(2)^4 - \frac{1}{4}(-1^4) = \frac{17}{4}.$

2. $\text{We use another integral: } \int^{3\pi/4}_{0} \sin x \: dx. \text{Take the antiderivative of }\sin x \text{, which is just }-\cos x. \text{Thus, the answer is }\\ -\cos(\frac{3\pi}{4}) - ( -\cos 0) = \frac{\sqrt{2}}{2} + 1.$

1) angle 2 and the 135deg angle are supplementary angles, their degree sum is 180 degrees.  Thus

2) angle 2 = 180deg - 135deg = 45deg

3) we know the angles in a triangle sum to 180 degrees.  The triangle shown has angle 1, angle 2=45deg, and a 90deg angle.

Thus angle 1 = 180-45-90 = 45deg

getting old or something, see answer #4

$20 = 2^2 5 \text{ so we'll want to use prime factors }\\ 2^4 \cdot 3^1 \cdot 5^1 = 240\\ \text{This has }(4+1)(1+1)(1+1)=20 \text{ divisors}$

I'm pretty sure for (2) we just find the smallest positive integer with 6 divisors and square it.

Proceeding as above

$6=2\cdot 3 \text{ so choose}\\ (2^2\cdot 3)^2 = 144$

Ok it is 11 i misunderstood CPhill

I think it's for nonzero digits. :)

Notice that this is just the sum of the first15 odd numbers. Because the sum of the first n is n^2, the answer is 15^2 = 225.

g(1) is f(2), and the sum of the coeffcients is g(1), so the answer is 128 - 24 + 2 = 106. :)