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We will answer this by considering the numbers which contribute towards trailing zeros and by calculating how many such numbers there are within 100!. The solution should enlighten the student on the use of prime numbers in number theory.

Answered by Waseem A.

Studies Physics at Oxford Alumni

This looks pretty daunting, but the key to this sort of a problem, which has come up a couple of times in previous PAT tests, is to recognise that this expression is actually the difference of two squares. It can be generalised to the problem of evaluating:

x^{2}-y^{2}

which has as its solution:

x^{2}-y^{2} = (x+y)(x-y)

so in this case:

3070^{2}-3069^{2} = (3070+3069)(3070-3069)=(6139)(1)=6139

Answered by Sam C.

Studies MEng Engineering Science at Oxford, New College

This is the sum of a geometric series with an infinite number of terms.

First, find the common ratio:

(-2/9)/(2/3) = -1/3 , (2/27)/(-2/9) = -1/3

Therefore, the common ratio is -1/3

-1<(-1/3)<1 therefore the series will converge to a finite number.

The general formula for a sum of an infinite geometric series is a/(1-r) where a is the first number in the sequence and r is the common ratio.

So, substituting in the numbers, the sum of the series = (2/3)/(1-[-1/3]) = 1/2

Energy will be lost due to the plastic deformation of the mass as the bullet enters the mass. However momentum will always be conserved, so you can use this to find the speed of the mass when the bullet enters.

Answered by Dominic D.

Studies Physics at Oxford, Keble College

Like with many questions in the PAT test once you know what to do it is quite easy! To solve you can use a difference of two squares i.e. (2007-2006)(2007+2006). As the first bracket gives 1 and the second gives 4013 (by arithmetic). Therefore you get 4013*1 = 4013

Answered by Dominic D.

Studies Physics at Oxford, Keble College

Although this might look like a smart trick at the beginning, the truth is that as a physicist this kind of approximations turn out to be useful very often. This is quite likely to come up in the PAT as well!

It is important to get used to understanding symbolic expressions (no numbers!), so I will first derive the maths and then apply the resulting expression to a couple of illustrative examples.

Our aim is to obtain an approximate value for the inexact square root x of an arbitrary number n, i.e. to solve x = n^{1/2}. In other words, we want to solve the equation f(x) = x^{2} = n. Suppose we know the exact root x_{0} of another number n_{0} which is very close to n, i.e. f(x_{0}) = x_{0}^{2} = n_{0} (this is key!). We can do a Taylor series expansion of f(x) about x_{0} up to first order (first derivative f'(x)), which is simply:

f(x) = f(x_{0}) + f'(x_{0})(x - x_{0}).

The first derivative f'(x) = 2x, which evaluated at x_{0} is f'(x_{0}) = 2x_{0}. Substituting f(x) = n and f(x_{0}) = n_{0} leads to:

n = n_{0} + 2x_{0}(x - x_{0}).

We are almost done! We know all the values of n, n_{0} and x_{0} and just need to solve for x. Rearranging:

x = x_{0} + (n - n_{0})/(2x_{0})

That's it! Remember that this expression is just an approximation for the actual square root of n, but a quite good one the closer n_{0} is to n.

Let's plug in some numbers. Suppose you were asked to approximate the square root of n = 66. Despite the solution being irrational, there is a close number n_{0} = 64 with a nice exact square root x_{0} = 8. Using the expression above we find:

x = 8 + (66 - 64)/(2*8) = 8 + 2/16 = 8 + 1/8 = 8 + 0.125 = 8.125

How good is this? It's just 0.01% off the actual value 8.12404...!

What if we chose n_{0} = 81 instead, the square root of which is x_{0} = 9? This would give x = 8.167 - worse, but still a reasonable result considering how far 81 is from 66. Never forget to quote both the positive and negative square roots in your final solution!

Would you be able to derive a similar expression for approximating a cubic root?

Answered by Sergio H.

Studies Physics with Theoretical Physics at Imperial College London

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