The series, \(1^{1} + 2^{2} + 3^{3} + ... + 10^{10} = 10405071317\).
Find the last ten digits of the series, \(1^{1} + 2^{2} + 3^{3} + ... + 1000^{1000}\).
# R
1000**1000
## [1] Inf
# To manage very large numbers you have to use gmp
library(gmp)
##
## Attaching package: 'gmp'
## The following objects are masked from 'package:base':
##
## %*%, apply, crossprod, matrix, tcrossprod
sum_n_power_last_digit <- function(n, last_digit){
sum_series = sum(as.bigz(seq(1, n))**(seq(1, n)))
sum_n = stringr::str_split(as.character(sum_series),'')[[1]]
last_digits = sum_n[(length(sum_n)-last_digit+1):length(sum_n)]
return(last_digits)
}
t = Sys.time()
sum_n_power_last_digit(1000,10)
## [1] "9" "1" "1" "0" "8" "4" "6" "7" "0" "0"
Sys.time() - t
## Time difference of 0.008686066 secs# Python
import numpy as np
import time
# default numpy dtype may not be appropriate to deal with very large number, python can do it, you can specify to numpy to use it as a python object
np.power(np.arange(1,15),np.arange(1,15))
## array([ 1, 4, 27,
## 256, 3125, 46656,
## 823543, 16777216, 387420489,
## 10000000000, 285311670611, 8916100448256,
## 302875106592253, 11112006825558016])
np.power(np.arange(1,15, dtype="object"),np.arange(1,15, dtype="object"))
## array([1, 4, 27, 256, 3125, 46656, 823543, 16777216, 387420489,
## 10000000000, 285311670611, 8916100448256, 302875106592253,
## 11112006825558016], dtype=object)
# In a function with python object
def sum_n_power_last_digit(n,last_digit):
sum_series = np.sum(np.power(np.arange(1,n+1, dtype="object"),np.arange(1,n+1, dtype="object")))
last_digits = str(sum_series)[-last_digit:]
return last_digits
t = time.time()
sum_n_power_last_digit(1000,10)
## '9110846700'
time.time() - t
## 0.0056819915771484375If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 10000.
get_sum_multiples_below1 <- function(below,multiple_1,multiple_2){
sum <- 0
for(i in 1:(below-1)){
if((i %% multiple_1 == 0)|(i %% multiple_2 == 0)){
sum <- sum + i
}
}
return(sum)
}
t <- Sys.time()
get_sum_multiples_below1(1000000,3,5)
## [1] 233333166668
Sys.time() - t
## Time difference of 0.4065418 secs
get_sum_multiples_below2 <- function(below,multiple_1,multiple_2){
seq <- seq(1,(below-1))
multiples <- sapply(seq,FUN = function(x){if((x %% multiple_1 == 0) | (x %% multiple_2 == 0)){x}else{0} })
return(sum(multiples))
}
t <- Sys.time()
get_sum_multiples_below2(1000000,3,5)
## [1] 233333166668
Sys.time() - t
## Time difference of 1.117711 secs
get_sum_multiples_below3 <- function(below,multiple_1,multiple_2){
seq <- seq(1,(below-1))
multiples <- seq[which((seq %% multiple_1 == 0) | (seq %% multiple_2 == 0))]
return(sum(multiples))
}
t <- Sys.time()
get_sum_multiples_below3(1000000,3,5)
## [1] 233333166668
Sys.time() - t
## Time difference of 0.008708954 secsimport time
import numpy as np
def get_sum_multiples_below1(below,mutiple_1,multiple_2):
sum = 0
for i in range(below):
if (i % mutiple_1 == 0) or (i % multiple_2 == 0):
sum+= i
return sum
t = time.time()
get_sum_multiples_below1(1000000,3,5)
## 233333166668
time.time() - t
## 0.025700807571411133
def get_sum_multiples_below2(below,mutiple_1,multiple_2):
multiples = [i if ((i % mutiple_1 == 0) or (i % multiple_2 == 0)) else 0 for i in range(below)]
return sum(multiples)
t = time.time()
get_sum_multiples_below2(1000000,3,5)
## 233333166668
time.time() - t
## 0.028688907623291016
def get_sum_multiples_below3(below,mutiple_1,multiple_2):
seq = list(range(below))
multiples = map(lambda i: i if ((i % mutiple_1 == 0) or (i % multiple_2 == 0)) else 0, seq)
return sum(multiples)
t = time.time()
get_sum_multiples_below3(1000000,3,5)
## 233333166668
time.time() - t
## 0.04091191291809082
def get_sum_multiples_below4(below,mutiple_1,multiple_2):
seq = np.arange(below)
seq = seq[((seq % mutiple_1 == 0) | (seq % multiple_2 == 0))]
return sum(seq)
t = time.time()
get_sum_multiples_below4(1000000,3,5)
## np.int64(233333166668)
time.time() - t
## 0.01642894744873047By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10 001st prime number?
nth_prime <- function(nth){
primes = c(2,3)
i = 3
while(length(primes)<nth){
i <- i+1
is_prime = TRUE
for(j in 2:(round(sqrt(i))+1)){
if(i%%j == 0){
is_prime = FALSE
}
if(is_prime == FALSE){
break
}
}
if(is_prime == TRUE){
primes = c(primes,i)
}
}
return(primes[nth])
}
t <- Sys.time()
nth_prime(10001)
## [1] 104743
Sys.time() - t
## Time difference of 0.7728629 secs
nth_prime2 <- function(nth){
primes = c(2,3)
i = 3
while(length(primes)<nth){
i <- i+1
ratio = i%%seq(2,round(sqrt(i))+1) == 0
if(!any(ratio)){
primes = c(primes,i)
}
}
return(primes[nth])
}
t <- Sys.time()
nth_prime2(10001)
## [1] 104743
Sys.time() - t
## Time difference of 0.582736 secs# Python
import numpy as np
def nth_prime(nth):
primes = [2,3]
i = 3
while len(primes) < nth:
i+=1
is_prime = True
for j in range(2,(int(np.sqrt(i))+1)):
if i%j == 0:
is_prime = False
break
if is_prime == True:
primes.append(i)
print(primes[-1])
t = time.time()
nth_prime(10001)
## 104743
time.time() - t
## 0.07397699356079102
def nth_prime2(nth):
primes = [2,3]
i = 3
while len(primes)<nth:
i+=1
ratio = np.mod(i,np.arange(2,int(np.sqrt(i)+1))) == 0
if not any(ratio):
primes.append(i)
print(primes[nth-1])
t = time.time()
nth_prime2(10001)
## 104743
time.time() - t
## 0.16478991508483887A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
Find the largest palindrome made from the product of two 3-digit numbers.
# Python
#
t = time.time()
three_digits = np.arange(100,1000)
products = np.repeat(three_digits, len(three_digits))*np.tile(three_digits,len(three_digits))
palindromes = [x for x in products if str(x)==str(x)[::-1]]
print(max(palindromes))
## 906609
time.time() - t
## 0.08670210838317871