Bertie

Mistype I think, A is the same as C.

$$\displaystyle \\1754 \equiv 71 \mod99\\1457\equiv 71 \mod 99 \\ 368 \equiv 71 \mod 99$$

Nit-picker !

But you're correct, I was careless, I should have stated that k was positive.

It's a geometric series with first term 1 and common ratio 2, so

$$S_{n}=2^{n-1}-1,$$

and

$$S_{n+1}=2^{n}-1.$$

Suppose that both of these are exactly divisible by some integer k (say), then there will be integers say s and t such that

$$s\times k = 2^{n-1}-1, \; \text{ and }\; t\times k = 2^{n}-1.$$

Multiply the first of these equations by 2 and subtract that from the second one and you have

$$(t\times k)-(2\times s\times k) = 1,$$

or,

$$k(t-2s)=1.$$

Since k, t and s are integers, it follows that

$$k = 1 \; (\text{and }\; t - 2s=1).$$

Hi Melody

As Alan has said $$r\angle \theta$$ is a shorthand notation used mainly by electrical engineers.

It's actually shorthand for the complex number $$r(\cos \theta + \imath \sin \theta).$$

Written out in full, we would have

$$r_{1}\angle \theta_{1}\times r_{2}\angle\theta_{2}\\=r_{1}r_{2}(\cos\theta_{1}+\imath\sin\theta_{1})(\cos\theta_{2}+\imath\sin\theta_{2})\\=r_{1}r_{2}(\cos\theta_{1}\cos\theta_{2}-\sin\theta_{1}\sin\theta_{2}+\imath(\sin\theta_{1}\cos\theta_{2}+\cos\theta_{1}\sin\theta_{2}))\\=r_{1}r_{2}(\cos(\theta_{1}+\theta_{2})+\imath\sin(\theta_{1}+\theta_{2}))\\=r_{1}r_{2}\angle(\theta_{1}+\theta_{2}).$$

I'll do the middle one, you can do the other two.

f(x) = 3x - 4,

so

f(1.5) = 3(1.5) - 4 = 4.5 - 4 = 0.5.

?

It comes from the word implied.

There is an implied relationship  between the two variables.

It may or may not be possible (or desirable) to convert it into an explicit relationship.

Thanks for the comments Melody.

Notice that the 6n plus or minus 1 thing is not unique, 2n or 4n plus or minus 1 share the same property.

I chose 6n because the 6 happened to fit in with the question.

All primes greater than 3 can be expressed in the form 6n+1 or 6n-1.

For example, 5 = 6 - 1,

7 = 6 + 1,

11 = 2.6 - 1,

13 = 2.6 + 1,

17 = 3.6 - 1

19 = 3.6 + 1,

23 = 4.6 - 1,

and so on.

There are a whole load of numbers for which 6n plus or minus 1 is not prime, but the important point is that 6n plus or minus one covers all of the primes greater than 3. 

Consider then

$$N = 7^{\, 6n+1}-6^{\,6n+1}-1=(7\times7^{\,6n})-(6\times6^{\,6n})-1.$$

Now

$$7^{\,6}=117649=2736\times 43 + 1\equiv1(\bmod 43),$$

so

$$(7^{\,6})^{2}, (7^{\,6})^{3},\:\text{and in general}\:(7^{\,6})^{n}=7^{\,6n}\equiv 1(\bmod 43),$$

in which case

$$7\times7^{\,6n}\equiv7\times 1\equiv7(\bmod 43).$$

Similarly, since

$$6^{\,6}=46656=1085\times 43 + 1\equiv1(\bmod 43),$$

$$6\times6^{\,6n}\equiv 6(\bmod43).$$

It follows that

$$N\equiv7-6-1\equiv 0(\bmod 43),$$

that is , there is a remainder of zero if N is divided by 43.

To cover the 6n - 1 case, consider

$$42N =7\times 6(7^{\,6n-1}-6^{\,6n-1}-1)=(6\times7^{\,6n})-(7\times6^{\,6n})-42.$$

Proceeding as above,

$$42N\equiv6-7+1\equiv0(\bmod43),$$

implying that 42N is divisible by 43, and since 42 isn't, it follows that N is.

Also Melody, it's often advantageous to write the constant of integration as

$$\ln(k).$$

That allows the result to be written as

$$\ln(\sin y) + \ln(k) = \ln(k \sin y)$$

making later simplification easier.