Question

Find the smallest positive integer n such that the digit sum of n is divisible by 5, and the digit sum of n +1 is also divisible by 5. Note: The digit sum of 1440 is 1+4+4+0 = 9.​

Answer 1

Answer:

139,999

Step-by-step explanation:

If the digit sum of n is divisible by 5, the digit sum of n+1 can't physically be divisble by 5, unless we utilise 9's at the end, this way whenever we take a number in the tens (i.e. 19), the n+1 will be 1 off being divisble, so if we take a number in the hundreds, (109, remember it must have as many 9's at the end as possible) the n+1 will be 2 off being divisble, so continuing this into the thousands being three, tenthousands being 4, the hundred thousands will be 5 off (or also divisble by 5). So if we stick a 1 in the beginning (for the lowest value), and fill the last digits with 9's, we by process of elimination realise that the tenthousands digit must be 3 such that the digit sum is divisible by 5, therefore we get 139,999