코딩 연습

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..

Given the positive integers \(x,\; y\), and \(z\), are consecutive terms of an arithmetic progression, the least value of the positive integers \(n\), for which the equation, \(x^2 - y^2 - z^2 =n\), has exactly two solutions is \(n=27\):\[34^2-27^2-20^2=12^2-9^2-6^2=27\] It turns out that \(n=1155\) is the least value which has exactly ten solutions. How many values of \(n\) less than one millio..

Consider the consecutive primes \(p_1=19\) and \(p_2=23\). It can be verified that \(1219\) is the smallest number such that the last digits are formed by \(p_1\) whilst also being divisible by \(p_2\). In fact, with the exception of \(p_1-3\) and \(p_2=5\), for every pair of consecutive primes, \(p_2 > p_1\), there exist values of \(n\) for which the last digits are formed by \(p_1\) and \(n\) ..

A number consisting entirely of ones is called a repunit. We shall define \(R(k)\) to be a repunit of length \(k\); for example, \(R(6)=111111\). Let us consider repunits of the form \(R \left (10^n \right )\). Although \(R(10),\; R(100)\), or \(R(1000)\) are not divisible by \(17\), \(R(10000)\) is divisible by \(17\). Yet there is no value of \(n\) for which \(R \left ( 10^n \right )\) will di..

A number consisting entirely of ones is called a repunit. We shall define \(R(k)\) to be a repunit of length \(k\). For example \(R(10)=1111111111=11\times 41 \times 271 \times 9091\), and the sum of these prime factors is \(9414\). Find the sum of the first fort prime factors of \(R \left ( 10^9 \right ) \). 숫자 \(1\) 로만 이루어진 수를 repunit 이라고 한다. \(R(k)\) 를 \(k\) 개의 \(1\) 로 이루어진 수라고 하자. 예를 들어, \(R..

There are some prime values, \(p\), for which there exists a positive integer, \(n\), such that the expression \(n^3 + n^2 p\) is a perfect cube. For example, when \(p=19, \; 8^3 + 8^2 \times 19 = 12^3\). What is perhaps most surprising is that for each prime with this property the value of \(n\) is unique, and there are only four such primes below one-hundred. How many primes below one million ..