A palindromic number is similar to a palindrome. It is the same read both left to right, and right to left.

Project Euler tells us, that the largest palindrom made from the product of two 2-digit numbers is 9009. That number is made by multiplying 91 and 99.

I must now find the largest palindrome, made from the product of two 3-digit numbers.

What is given, is that the three digit numbers cannot end with a zero.

There are probably other restrictions as well.

I’ll need a function that tests if a given number is palindromic.

palindromic <- function(x){
  sapply(x, function(x) (str_c(rev(unlist(str_split(as.character(x),""))), collapse="")==as.character(x)))
}

The function part converts x to character, splits it in individual characters, unlists the result, reverses that, and concatenates it to a string. Then it is compared to the original x – converted to a character.
The sapply part kinda vectorises it. But it is still the slow part.

If I could pare the number of numbers down, that would be nice.

One way would be to compare the first and last digits in the number.

first_last <- function(x) { 
  x %/% 10^(floor(log10(x))) == x%%10
}

This function finds the number of digits – 1 in x. I then modulo-divide the number by 10 to the number of digits minus 1. That gives me the first digit, that I compare with the last. If the first and the last digit is the same – it returns true.

Now I am ready. Generate a vector of all three-digit numbers from 101 to 999. Expand the grid to get all combinations. Convert to a tibble,
filter out all the three-digit numbers that end with 0. Calculate a new column as the multiplication of the two numbers, filter out all the results where the first and last digit are not identical, and then filter out the results that are not palindromic. Finally, pass it to max (using %$% to access the individual variables), and get the result.

library(dplyr)
library(magrittr)

res <- 101:999 %>% 
  expand.grid(.,.) %>% 
  as_tibble() %>% 
  filter(Var1 %% 10 != 0, Var2 %% 10 != 10) %>% 
  mutate(pal = Var1 * Var2) %>% 
  filter(first_last(pal)) %>% 
  filter(palindromic(pal)) %$% 
  max(pal)

There are probably faster ways of doing this…