Problem 1018
François, a mathematician fascinated by large numbers, found one endowed with an astonishing property: if we write a "1" on the left and a "1" on the right of this positive integer having less than 100 digits, we get a number 99 times bigger than itself.- 1A, 1B, 1C. What are the last 9 digits of the number of François? (in the same order as the number of François, the digit for units on the right, without space).
- 2A, 2B, 2C. What is its number of digits?
We start by noticing that if a is the number searched for, writing a "1" on the left and right is the number 10 × a + 1 + 10Number of digits(a) + 1. If this number is equal to 99 × a, we have the equality: 1 + 10Number of digits(a) + 1 = 89 × a.
It is thus sufficient to check, for each number of digits a between 1 and 100, if the expression 1 + 10Number of digits(a) + 1 is divisible by 89.
The program, written in python for its native support for arbitrarily large integers, is available here.
It is thus sufficient to check, for each number of digits a between 1 and 100, if the expression 1 + 10Number of digits(a) + 1 is divisible by 89.
The program, written in python for its native support for arbitrarily large integers, is available here.
- 1A. The number of François terminates by 797752809.
- 2A. 21. 2B. 65. The number has 21 or 65 digits.