Find the smallest positive integer that is a multiple of 35 whose digits are all the same.
(In reply to
Some Tips by K Sengupta)
We can easily long divide 10, 100, 1000, ... until we reach 1000,000 by 7 to obtain a remainder of 1 or find the said remainder by modular calculations to arrive at the same end result.. Neither of these are shown here.
So, (10^x-1)=0 (mod 7) , when the minimum value of x is 6.
=> 10^6-1 = 0 mod 6
-> I must be excused... There is no provision of the equivalent symbol to show up in this online comment document.
Since gcd(7,9)=1, it follows that:
(10^9-1)/9= 111,111 is evenly divisible by 7
Since, gcd(5,7)=1, it follows that the minimum integer comprising entire of 5s and divisible by 35 must be 111,111*5=555,555
Edited on June 4, 2022, 12:28 am
Puzzle Solution
