Find three different natural numbers, larger than \(100\) such that each of them is divisible by the difference of the other two numbers? The values of differences also have to be different from each other.
We know that the product \(c \times d\) is divisible by a prime \(p\). Show that either \(c\) or \(d\) must be divisible by \(p\).
Find one way to encrypt letters of Latin alphabet as sequences of \(0\)s and \(1\)s, each letter corresponds to a sequence of five symbols.
Using the representation of Latin alphabet as sequences of \(0\)s and \(1\)s five symbols long, encrypt your first and last name.
Decipher the following quote from Alice in Wonderland:
Lw zrxog eh vr qlfh li vrphwklqj pdgh vhqvh iru d fkdqjh.
The same letters correspond to the same in the phrase, different letters correspond to different. We know that no original letters stayed in place, meaning that in places of e,r,h there was surely something else.
After mastering the Caesar shift cypher one may wonder how to generalize it. One possible way is to use Affine cypher. The difference between these two methods can be described as follows:
In case of Caesar cypher we took a letter with position \(n\) from \(1\) to \(26\) and added to its position a number \(d\) obtaining the number \(n+d\), then we compute its residue modulo \(26\).
In case of affine cypher we take a letter with position \(n\) and consider a number \(nx + d\) modulo \(26\).
To decipher such code we need to know values \(x\) and \(d\), then if we have a letter in the code with position \(m\), we can find \(n\) as \(n= (m-d)x^{-1}\) modulo \(26\). Here we have to explain what is \(x^{-1}\): for a number \(x < 26\) we are looking for such a number \(y\), that \(26\) divides \(xy-1\).
Does there always exist a number \(x^{-1}\) modulo \(26\) for any \(x\)?
Using data \(x=3\), \(d=8\) encrypt the word "SOLUTION".
Prove the magic trick for the number \(1089 = 33^2\): if you take any \(3\)-digit number \(\overline{abc}\) with digits coming in strictly descending order and subtract from it the number obtained by reversing the digits of the original number \(\overline{abc} - \overline{cba}\) you get another \(3\)-digit number, call it \(\overline{xyz}\). Then, no matter which number you started with, the sum \(\overline{xyz} + \overline{zyx} = 1089\).
Recall that a number \(\overline{abc}\) is divisible by \(11\) if and only if \(a-b+c\) also is.
Find the last two digits of the number \[33333333333333333347^4 - 11111111111111111147^4\]
Replace all stars with ”+” or ”\(\times\)” signs so the equation holds: \[1*2*3*4*5*6=100\] Extra brackets may be added if necessary. Please write down the expression into the answer box.
In how many ways can one change \(\pounds 2\) into coins worth \(50\)p, \(20\)p and \(10\)p? One does not necessarily need to use all available coin types, i.e. having \(5\) coins of \(20\)p and \(10\) coins of \(10\)p is allowed.