## Missing digit – short puzzle

The number has digits, all different; which digit is missing?

*Mathematical Mind-Benders, Peter Winkler*

## Project Euler Problem 285

Another quite nice problem from Project Euler is number 285. The result of the problem depends on the computation of a certain probability, which in turn is related to the computation of a certain area. Below is an illustration of the computation for k equal to 10.

To save you some time, here’s a picture of the case k=1 which I ignored and spent quite a lot of time debugging… Plus, it only affects the last three digits or so after the decimal point…

Here’s a Matlab code which can construct the pictures above and can compute the result for low cases. To solve the problem, you should compute explicitly all these areas.

function problem285(k) N = 100000; a = rand(1,N); b = rand(1,N); ind = find(abs(sqrt((k*a+1).^2+(k*b+1).^2)-k)<0.5); plot(a(ind),b(ind),'.'); axis equal M = k; pl = 1; for k=1:M if mod(k,100)==0 k end r1 = (k+0.5)/k; r2 = (k-0.5)/k; f1 = @(x) (x<=(-1/k+r1)).*(x>=(-1/k-r1)).*(sqrt(r1^2-(x+1/k).^2)-1/k).*(x>=0).*(x<=1); f1 = @(x) f1(x).*(f1(x)>=0); f2 = @(x) (x<=(-1/k+r2)).*(x>=(-1/k-r2)).*(sqrt(r2^2-(x+1/k).^2)-1/k).*(x>=0).*(x<=1); f2 = @(x) f2(x).*(f2(x)>=0); if k == pl thetas = linspace(0,pi/2,200); hold on plot(-1/k+r1*cos(thetas),-1/k+r1*sin(thetas),'r','LineWidth',2); plot(-1/k+r2*cos(thetas),-1/k+r2*sin(thetas),'r','LineWidth',2); plot([0 1 1 0 0],[0 0 1 1 0],'k','LineWidth',3); hold off axis off end A(k) = integral(@(x) f1(x)-f2(x),0,1); end xs = xlim; ys = ylim; w = 0.01; axis([xs(1)-w xs(2)+w ys(1)-w ys(2)+w]); sum((1:k).*A)

## Some of the easy Putnam 2016 Problems

Here are a few of the problems of the Putnam 2016 contest. I choose to only list problems which I managed to solve. Most of them are pretty straightforward, so maybe the solutions posted here may be very similar to the official ones. You can find a complete list of the problems on other sites, for example here.

**A1.** Find the smallest integer such that for every polynomial with integer coefficients and every integer , the number

that is the -th derivative of evaluated at , is divisible by .

**Hints.** Successive derivatives give rise to terms containing products of consecutive numbers. The product of consecutive numbers is divisible by . Find the smallest number such that . Prove that does not work by choosing . Prove that works by working only on monomials…

## Monotonic bijection from naturals to pairs of natural numbers

This is a cute problem I found this evening.

Suppose is a bijection such that if and , then .

Prove that if then .

*Proof:* The trick is to divide the pairs of positive integers into families with the same product.

Note that the -th column contains as many elements as the number of divisors of . Now we just just use a simple observation. Let be on the -th column (i.e. ). If then cannot be on one of the first columns. Indeed, the monotonicity property implies . The fact that is a bijection assures us that cover the first columns. Moreover, one element from the -th column is surely covered, namely . This means that

where we have denoted by the number of positive divisors of .

## Necessary condition for the uniform convergence of a certain series

Let be a decreasing sequence of positive numbers such that the series

is uniformly convergent. Then must satisfy .

Note that this result implies that the series is not uniformly convergent on . It is surprisngly similar to the following result:

Suppose that is a decreasing sequence of positive real numbers such that the series is convergence. Then . It is no surprise that the proofs of these two results are similar.

## Nice inequality similar to IMO 2011

Prove that

*Note the similarity to the IMO 2011 inequality, and *surprisingly* the same method works.*

## Sleek proof that the harmonic series diverges

Suppose that the harmonic series does converge. Then the sequence of integrable functions functions is bounded above by an integrable function .