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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 ..

A number consisting entirely of ones is called a repunit. We shall define \(R(k)\) to be a repunit of length \(k\); for example, \(R(6)=111111\). Given that \(n\) is a positive integer and \({\rm GCD}(n, \;10)=1\), it can be shown that there always exists a value, \(k\), for which \(R(k)\) is divisible by \(n\), and let \(A(n)\) be the least such value of \(k\); for example, \(A(7)=6\) and \(A(4..

A number consisting entirely of ones is called a repunit. We shall define \(R(k)\) to be a repunit of length \(k\); for example \(R(6)=111111\). Given that \(n\) is positive integer and \({\rm GCD}(n, \;10) =1\), it cane be shown that there always exists a value, \(k\), for which \(R(k)\) is divisible by \(n\), and let \(A(n)\) be the least such value of \(k\); for example, \(A(7)=6\) and \(A(41..

A hexagonal tile with number \(1\) is surrounded by a ring of six hexagonal tiles, starting at "12 o'clock" and numbering the tiles \(2\) to \(7\) in an anti-clockwise direction. New rings are added in the same fashion, with the next ring being numbered \(8\) to \(19\), \(20\) to \(37\), \(38\) to \(61\), and so on. The diagram below shows the first three rings. By finding the difference between..

The radical of \(n\), \({\rm rad}(n)\), is the product of distinct prime factors of \(n\). For example, \(504=2^3 \times 3^2 \times 7\), so \({\rm rad}(504)=2 \times 3 \times 7 = 42\). We shall define the triplet of positive integers \((a, \;b,\;c)\) to be an abc-hit if: 1. \({\rm GCD}(a, \;b)={\rm GCD}(a, \;c) = {\rm GCD}(b, \;c)=1\) 2. \(a

The minimum number of cubes to cover every visible face on a cuboid measuring \(3 \times 2 \times 1\) is twenty-two. If we then add a second layer to this solid it would require forty-six cubes to cover every visible face, the third layer would require seventy-eight cubes, and the fourth layer would require one hundred and eighteen cubes to cover every visible face. However, the first layer on a..

The palindromic number 595 is interesting because it can be written as the sum of consecutive squares: \(6^2 + 7^2 + 8^2 +9^2 + 10^2 + 11^2 + 12^2\). There are exactly eleven palindromes below one-thousand that can be written as consecutive square sums, and the sum of these palindromes is \(4164\). Note that \(1 = 0^2 + 1^2\) has not been included as this problem is concerned with the squares of..