Problem: What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

Commentary: This problem again isn’t very difficult. Basically we just exploit the existence of prime factorizations and construct the least common multiple (lcm) in the most intuitive way possible. You might find something a bit strange in this solution, namely in the prime-y stuff; however, I don’t really want to get into that until Problem 7. So I won’t.

R Code:

ElapsedTime <- system.time({
##########################
###
# Function to get primes below specified n; option to get only those below sqrt(n)
PrimesBelow <- function(n,below.sqrt=FALSE){
if (below.sqrt==TRUE){
m <- ceiling(sqrt(n))
} else {
m <- n
}

primes <- c(2, 3)
i <- 3
while (i < m-1){
flag <- 0
i <- i+2
if (i%%6 == 3){
flag <- 1
}
if (flag == 0){
s <- sqrt(i)+1
possibleprimes <- primes[primes < s]
for (prime in possibleprimes){
if ((i%%prime == 0)){
flag <- 1
break
}
}
if (flag == 0){
primes <- c(primes,i)
}
}
}
primes
}

# Function to get the prime factorization of all integers below
PrimeFact <- function(n){
list <- numeric(0)
for (i in 2:n){
ps <- numeric(0)
for (prime in primes){
p <- 1
while (i%%(prime^p) == 0){
p <- p+1
}
p <- p-1
ps <- c(ps, p)
}
if (length(list) == 0){
list <- data.frame(primes, ps)
} else {
list <- data.frame(list, data.frame(ps))
}
}
list
}
###

n <- 20

primes <- PrimesBelow(n)

list <- PrimeFact(n)

expnts <- numeric(0)
for (nump in 1:length(primes)){
temp <- numeric(0)
for (i in 2:n){
temp <- c(temp, list[[i]][nump])
}
expnts <- c(expnts, sort(temp, decreasing=TRUE))
}

answer <- prod(primes^expnts)
##########################
})
ElapsedMins <- floor(ElapsedTime/60)
ElapsedSecs <- (ElapsedTime-ElapsedMins*60)
cat(sprintf("\nThe answer is:  %d\nTotal elapsed time:  %d minutes and %f seconds\n",
answer, ElapsedMins, ElapsedSecs))


Output: The answer is: 232792560 Total elapsed time: 0 minutes and 0.008600 seconds