Bertie

Consider the infinite Geometric series \(\displaystyle \frac{1}{x}+\frac{1}{x^{2}}+\frac{1}{x^{3}}+\dots\)  .

Its sum to infinity (I think that there is a mistake in the question, there should be a plus sign and a series of dots after that last term), is \(\displaystyle \frac{1}{x-1}\).

Equate the two, differentiate both sides, and then substitute x = 3.

Tiggsy

If you happen to know, off the top of your head, that sqrt(5) is 2.236 (3dp) then you could write 0.05 = 5/100 which, on taking the sqr root would be 2.236/10 = 0.2236.

Admittedly, the answer is not unique, but

57131 mod 1324 = 199,

37139 mod 1324 = 67.

Try it.

The biggest factor they have in common is 2^2 * 331 = 1324.

I think that the answer is 19C5*14C5*9C3*6C3, whatever that works out to be.

This business of dividing by 2 and 3! when you have one group of five then the other group of five is determined, or for the groups of three it doesn't matter in what order they are chosen, is wrong.

There is a difference between splitting the 19 into two groups of five and three groups of three without distinction, and splitting them into groups where each group studies a different topic.

Consider, for example, a group of four students, call them A,B,C and D, split into two twos.

4C2 = 6, so we have 6 pairs, AB, AC, AD, BC, BD, CD.

However, when the first pairing is chosen, in effect the second pairing is also chosen, so we have just three (= 6/2) possible groupings, {AB, CD}, {AC, BD} and {AD, BC}. That's the situation if it doesn't matter which pairing comes first, that is, for example, {AB, CD} is the same as {CD, AB}.

Now suppose that the students are going into two different language classes, say the first pairing study French and the second pairing study German. Now, the order of the pairings makes a difference, {AB, CD} is different from {CD, AB}.

In the first case, A and B study French and in the second case they study German, (and the other way round for C and D).

If this is the situation, the original calculation of 4C2 = 6 stands.

(Notice btw that, for the 19 student case, as far as the calculation of the final number is concerned, it doesn't matter which group is chosen first, that is, for example, 19C5*14C5*9C3*6C3 = 19C3*16C3*13C5*8C5 = ... .)

BobP

Work from the back to the front, three applications of the difference between two squares leads you to sqrt(33^2 - 8).

I'm with Melody.

For the 20 ball first part, the number of three ball combinations is 20C3 = 1140.

The number of ways in which we miss with all three numbers is 17C3 = 680.

The number of ways in which we have 1 hit and three misses is 3*17C2 = 408.

The number of ways in which we have 2 hits and 1 miss is 3C2*17 = 51.

The number of ways in which we have three hits is 1.

(Check that all possibilities have been covered, 680 + 408 + 51 + 1 = 1140.)

That gets us the probabilities,

zero hits 680/1140 = 0.59649 (5 dp),

1 hit 408/ 1140       = 0.35789 (5 dp),

2 hits 51/1140        = 0.04474 (5 dp),

3 hits 1/1140          = 0.00088 (5 dp).

An alternative method is to calculate the probabilities directly.

Using H for a hit and M for a miss, the probability of three misses MMM is (17/20)*(16/19)*(15/18) = 0.59649.

The probability of the sequence HMM is (3/20)*(17/19)*(16/18) = 0.119298, and since both of the other 1 hit sequences MHM and MMH have the same probability, the probability of 1 hit will be 3*0.119298 = 0.35789 (5 dp).

The probabilty of the two hit sequence HHM is (3/20)*(2/19)*(17/18) = 0.014912, and since both of the other two hit sequences HMH and MHH have the same probability, the probability of two hits will be 3*0.014912 = 0.04474 (5 dp).

The probability of three hits will be (3/20)*(2/19)*(1/18) = 0.00088 (5 dp).

That means that the probability of at least two hits (two matches) is 0.04474 + 0.00088 = 0.04562.

The probability of not winning that way will be 0.59649 + 0.35789 = 0.95438.

The probability of winning with the superball is 0.1 and not winning with the superball is 0.9.

There are four possible outcomes,

(1) not getting at least two matches and not picking the superball : probability 0.95438* 0.9 = 0.85894,

(2) getting at least two matches, but not the superball : probability                    0.04562*0.9 = 0.04106,

(3) not getting at least two matches but do pick the superball : probability        0.95438*0.1  = 0.09544,

(4) getting at least two matches and picking the superball : probability              0.04562*0.1 = 0.00456,

(the probabilities summing  to 1, as they should).

That means that the probability of winning is 0.04106 + 0.09544 + 0.00456 (or 1 - 0.85894) = 0.14106.

Tiggsy.

How many balls do you get for your ticket ?

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.