5 average from 56,675 reviews

Why limit yourself to someone who lives nearby, when you can choose from tutors across the UK?

By removing time spent travelling, you make tuition more convenient, flexible and affordable

We've combined live video with a shared whiteboard, so you can work through problems together

All your Online Lessons are recorded. Make the most out of your live session, then play it back after

You want to look at turning points, intersections and when the gradient is possitive or negative. It is also important to demonstrate what is going on with graphs when they are tending towards 0 (especially in the interval -1

Let the number we create = x
Then, x = abcabc
x = 100,000a + 10,000b + 1,000c +100a +10b +c
x = 100,100a + 10,010b + 1,001c
x = 13 (7700a + 770b + 77c)
Hence, x | 13.

Answered by Jordan R.

Studies Mathematics at Warwick

The problem seems to be hard as the equation involved both cosx and sinx But we can relate the two as (cosx)^2 + (sinx)^2 = 1, so we can rearrange this as 0 = 2(sinx)^2 - 7sinx + 3. This quadratic in sinx has solutions sinx = (7+5)/4 = 3 or sinx = (7-5)/4 = 0.5. As |sinx| is at most 1, we have sinx = 0.5, so that x = pi/6 or x = 5pi/6. (We should draw a graph of sinx to make sure we have all the solutions in the stated range.) So the answer is 2.

Answered by James R.

Studies Mathematics at Cambridge

The number of 0's at the end of a number is the same as the number of factors of 10 that the number has. So consider the prime factor decomposition of 100! and note that every factor of 10 has prime factors 2 __and__ 5. Because both 2 and 5 are needed, the question can be rephrased as "How many factors of 2 and 5 has 100!, and which is the smaller number?".
With this, we can start to work. Since 100! = 1 x 2 x... x 100, we see that prime factors of 2 come from 2, 4, 6, ..., 98, 100, and prime factors of 5 come from 5, 10, ..., 95, 100. Note that while we look for numbers divisible by 2, we must also remember that they can be divisible by 4, 8, and any other powers of 2, contributing more prime factors of 2. The same goes for factors of 5. With this, we can see that clearly it is the number of prime factors of 5 that will be smaller, and this the number of factors of 10. Now all that remains is to count the number of prime factors of 5. All multiples of 5 have a prime factor of 5, giving 20 (from 5, 10, 15, 20,... , 100). Also, recalling that multiples of 25 one extra factor of 5, we get 4 more factors of 5 (from 25, 50, 75, 100). Noting that 5^3 is larger than 100, we are done. Thus, 100! ends with 24 0's.

Answered by Henry R.

Studies MMath Mathematics at Christ Church, Oxford

Certainly. The nth student (Sn) changes the state of every nth locker - i.e. the multiples of n. If changed an odd number of times, the locker is open -- if even then closed.

i. how many closed after 3rd student? if closed then the locker is either left open by S2 and closed by S3 or closed by S2 and left closed by S3. i.e [ not(2 | n) & (3 | n) ] or [ not(3 | n) & (2 | n) ].

The first 6 lockers are OCCCOO (this pattern repeats). 1000=166*6+4 so there are 166*3+3=501 closed lockers. (166*3 C's from the 166 cycles, +3 from the incompleted cycle).

ii. After S3 the first 12 lockers are OCCC OOOC CCOO. S4 changes state of 4,8 and 12, leaving OCCO OOOO CCOC. 1000=83*12+4 so there are 83*5+2=417

iii. 100=2^2*5^2 is changed once for every factor. There are 3*3=9 factors (number of 2's in factor =0,1,2; of 5's =0,1,2). 9 is odd so locker open.

iv. We need to find number of factors of 1000 = 2^3*5^3 which are less than or equal to 100. Consider systematically number of 2's and 5's in factor.

1,2,4,8 ; 5,10,20,40 ; 25,50,100 (after this factor too large). So 11 factors (odd) so locker open.

Bonus questions:

1. Consider infinite closed lockers. What proportion will be open after infinite students pass?

i. how many closed after 3rd student? if closed then the locker is either left open by S2 and closed by S3 or closed by S2 and left closed by S3. i.e [ not(2 | n) & (3 | n) ] or [ not(3 | n) & (2 | n) ].

The first 6 lockers are OCCCOO (this pattern repeats). 1000=166*6+4 so there are 166*3+3=501 closed lockers. (166*3 C's from the 166 cycles, +3 from the incompleted cycle).

ii. After S3 the first 12 lockers are OCCC OOOC CCOO. S4 changes state of 4,8 and 12, leaving OCCO OOOO CCOC. 1000=83*12+4 so there are 83*5+2=417

iii. 100=2^2*5^2 is changed once for every factor. There are 3*3=9 factors (number of 2's in factor =0,1,2; of 5's =0,1,2). 9 is odd so locker open.

iv. We need to find number of factors of 1000 = 2^3*5^3 which are less than or equal to 100. Consider systematically number of 2's and 5's in factor.

1,2,4,8 ; 5,10,20,40 ; 25,50,100 (after this factor too large). So 11 factors (odd) so locker open.

Bonus questions:

1. Consider infinite closed lockers. What proportion will be open after infinite students pass?

Answered by Carlo A.

Studies Mathematics and Philosophy at Oxford, Hertford College

We can see that 104 = 2^3 * 13 = 2*2*26, 30 = 2 + 2 + 26, and 108 = 2*2 + 2*26 + 2*26, so the coefficients agree with the Vieta's formulas, so the roots of the equation above are 2, 2, 26. In conclusion, it has 2 distinct real roots.

Alternatively, we can try to factorise the polynomial. This can be done by (x-2)^2*(x-26), and so we can see that the equation has 2 distinct real roots.

Answered by Andreea I.

Studies Mathematics and Computer Science at Oxford, Merton College

mtw:mercury1:status:ok