#48 - Self-powers
The series, $1^1+2^2+3^3+\dots+10^{10}=10405071317$.
Find the last ten digits of the series, $1^1+2^2+3^3+\dots+1000^{1000}$.
The series, $1^1+2^2+3^3+\dots+10^{10}=10405071317$.
Find the last ten digits of the series, $1^1+2^2+3^3+\dots+1000^{1000}$.
The first two consecutive numbers to have two distinct prime factors are: \[\begin{aligned} 14 &= 2\times 7 \\ 15 &= 3\times 5 \end{aligned}\]
The first three consecutive numbers to have three distinct prime factors are: \[\begin{aligned} 644 &= 2^2\times 7\times 23 \\ 645 &= 3\times 5\times 43 \\ 646 &= 2\times 17\times 19 \end{aligned}\]
Find the first four consecutive integers to have four distinct prime factors each. What is the first of these numbers?
It was proposed by Christian Goldbach that every odd composite number that can be written as the sum of a prime and twice a square. \[\begin{aligned} 9 &= 7 + 2\times 1^2 \\ 15 &= 7 + 2\times 2^2 \\ 21 &= 3 + 2\times 3^2 \\ 25 &= 7 + 2\times 3^2 \\ 27 &= 19 + 2\times 2^2 \\ 33 &= 31 + 2\times 1^2 \end{aligned}\]
It turns out that the conjecture was false.
What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?
There are exactly ten ways of selecting three from five, 12345: \[123,\,\,124,\,\,125,\,\,134,\,\,135,\,\,145,\,\,234,\,\,235,\,\,245,\,\,\text{and }345\]
In combinatorics, we use the notation $_5 C_3=10$.
In general $_nC_r=\frac{n!}{r!(n-r)!}$, where $r\leq n, n!=n\times(n-1)\times\dots\times 3\times 2\times 1$, and $0!=1$.
How many, not necessarily distinct, values of $_nC_r$ for $1\leq n\leq 100$, are greater than one-million?
Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
Type Formula First terms Triangle $T_n=\frac{1}{2}n(n+1)$ 1, 3, 6, 10, 15, … Pentagonal $P_n=\frac{1}{2}n(3n-1)$ 1, 5, 12, 22, 35, … Hexagonal $H_n=n(2n-1)$ 1, 6, 15, 28, 45, … It can be verified that $T_{285}=P_{165}=H_{143}=40755$.
Find the next triangle number that is also pentagonal and hexagonal.
Pentagonal numbers are generated by the formula, $P_n=\frac{1}{2}n(3n-1)$. The first ten pentagonal numbers are: \[1,5,12,22,35,51,70,92,117,145,\dots\]
It can be seen that $P_4+P_7=22+70=92=P_8$. However, their difference, $70-22=48$, is not pentagonal.
Find the pair of pentagonal numbers, $P_j$ and $P_k$, for which their sum and difference are pentagonal and $D=\lvert P_j - P_k\rvert$.
The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.
Let $d_1$ be the 1st digit, $d_2$ be the 2nd digit, and so on. In this way, we note the following:
- $d_2d_3d_4=406$ is divisible by 2
- $d_3d_4d_5=063$ is divisible by 3
- $d_4d_5d_6=635$ is divisible by 5
- $d_5d_6d_7=357$ is divisible by 7
- $d_6d_7d_8=572$ is divisible by 11
- $d_7d_8d_9=728$ is divisible by 13
- $d_8d_9d_{10}=289$ is divisible by 17
Find the sum of all 0 to 9 pandigital numbers with this property.
The $n^{\text{th}}$ term of the sequence of triangle numbers is given by, $t_n=\frac{1}{2}n(n+1)$; so the first ten triangle numbers are: \[1,3,6,10,15,21,28,36,45,55,\dots\]
By converting each letter in a word to a number corresponding to its alphabetical position and adding these values we form a word value. For example, the word value for SKY is $19+11+25=55=t_{10}$. If the word value is a triangle number then we shall call the word a triangle word.
Using words.txt (right click and ‘Save Link/Target As…’), a 16K text file containing nearly two-thousand common English words, how many are triangle words?
An irrational decimal fraction is created by concatenating the positive integers:
0.123456789101112131415161718192021...
It can be seen that the 12th digit of the fractional part is 1.
If $d_n$ represents the $n^{\text{th}}$ digit of the fractional part, find the value of the following expression: \[d_1\times d_{10}\times d_{100}\times d_{1000}\times d_{10000}\times d_{100000}\times d_{1000000}\]
We shall say that an $n$-digit number is pandigital if it makes use of all the digits 1 to $n$ exactly once. For example, 2143 is a 4-digit pandigital and is also prime.
What is the largest $n$-digit pandigital prime that exists?