코딩 연습

Some positive integers \(n\) have the property that the sum \([n+{\rm reverse}(n)]\) consists entirely of odd (decimal) digits. For instance, \(36+63=99\) and \(409+904=1313\). We will call such numbers reversible; so \(36, \; 63, \; 409\), and \(904\) are reversible. Leading zeroes are not allowed in either \(n\) or \({\rm reverse}(n)\). There are \(120\) reversible numbers below on-thousand. H..

In laser physics, a "white cell" is a mirror system that acts as a delay line for the laser beam. The beam enters the cell, bounces around on the mirrors, and eventually works its way back out. The specific white cell we will be considering is an ellipse with the equation \(4^x+y^2=100\). The section corresponding to \(-0.01 \le x \le 0.01\) at the top is missing, allowing the light to enter and..

Let \(\rm ABC\) be a triangle with all interior angles being less than \(120\) degrees. Let \(\rm X\) be any point inside the triangle and let \({\rm XA}=p,\; {\rm XC}=q\), and \({\rm XB}=r\). Fermat challenged Torricelli to find the position of \(\rm X\) such that \(p+q+r\) was minimized. Torricelli was able to prove that if equilateral triangles \(\rm AOB, \; BNC\) and \(\rm AMC\) are construc..

Find the smallest \(x+y+z\) with integers \(x>y>z>0\) such that \(x+y\), \(x-y\), \(x+z\), \(x-z\), \(y+z\), \(y-z\) are all perfect squares. \(x>y>z>0\) 을 만족하는 자연수 \(x, \; y\; z\) 에 대해서 \(x+y\), \(x-y\), \(x+z\), \(x-z\), \(y+z\), \(y-z\) 가 모두 완전제곱수가 되는 \(x+y+z\) 의 최솟값을 구하시오. 늘 그렇듯이 처음엔 무식하게 하나하나 완전제곱수인 것을 모두 확인하는 방법을 사용했지만, 아니나 다를까 시간이 너무 오래 걸린다. 그래서 다음의 방법을 생각 !! \[x+y=a^2 \label{a}\tag{1}\] ..

A positive integer, \(n\), is divided by \(d\) and the quotient and remainder are \(q\) and \(r\) respectively. In addition \(d, \;q\) and \(r\) are consecutive positive integer terms in a geometric sequence, but not necessarily in that order. For example, \(58\) divided by \(6\) has quotient \(9\) and remainder \(4\). It can also be seen that \(4, \;6,\;9\) are consecutive terms in a geometric ..

Consider the infinite polynomial series \({\rm A_G}(x) = x{\rm G_1} + x^2 {\rm G_2} + x^3 {\rm G_3} + \cdots \), where \({\rm G}_k\) is the \(k\)th term of the second order recurrence relation \({\rm G}_k = {\rm G}_{k-1} + {\rm G}_{k-2}\), \({\rm G_1}=1\), and \({\rm G}_2=4\); that is, \(1,\; 4,\; 5,\; 9,\; 14,\; 23, \; \cdots\). For this problem we shall be concerned with values of \(x\) for wh..

Let \((a, \;b, \;c)\) represent the three sides of a right angle triangle with integral length sides. It is possible to place four such triangles together to form a square with length \(c\). For example, \((3, \;4, \;5)\) triangles can be placed together to form 1 \(5\) by \(5\) square with a \(1\) by \(1\) hole in the middle and it can be seen that the \(5\) by \(5\) square can be tiled with tw..

Consider the isosceles triangle with base length, \(b=16\), and legs, \(L=17\). By using the Pythagorean theorem it can be seen that the height of the triangle, \(h=\sqrt{17^2-8^2}=15\), which is one less than the base length. With \(b=272\) and \(L=305\), we get \(h=273\), which is one more than the base length, and this is the second smallest isosceles triangle with the property that \(h=b \pm..

Consider the infinite polynomial series \({\rm A_F}(x)=x{\rm F}_1 + x^2 {\rm F}_2 + x^3 {\rm F}_3 + \cdots\), where \({\rm F}_k\) is the \(k\)th term in the Fibonacci sequence: \(1, \;1, \;3, \;5, \; 8,\; \cdots\); that is, \({\rm F}_k = {\rm F}_{k-1} + {\rm F}_{k-2},\;\; {\rm F}_1 = 1,\) and \({\rm F}_2=1\). For this problem we shall be interested in values of \(x\) for which \({\rm A_F}(x)\) i..

The positive integers, \(x, \;y,\) and \(z\), are consecutive terms of an arithmetic progression. Given that \(n\) is a positive integer, the equation, \(x^2 - y^2 - z^2 =n\), has exactly one solution when \(n=20\):\[13^2-10^2-7^2=20\] In fact there are twenty-five values of \(n\) below one hundred for which the equation has a unique solution. How many values of \(n\) less than fifty million hav..