A 100-digit positive integer is divisible by 72...

((10^100) - 64) mod 72 =0 

9999999999999999999999999 9999999999999999999999999 9999999999999999999999999 9999999999999999999999936 =100 digits long!

Sum =[98 x 9] + 3 + 6 =891

Mod is short for "Modulus". When you divide a number by the "modulus" it tells you right away if it is divisible by the modulus EVENLY, when the result =0. If it returns any other number, which is the REMAINDER, that means the number does not divide the modulus EVENLY.  Examples: 72 mod 7 =2 The remainder. But 77 mod 7=0, which means it divides the modulus evenly with no remainder.

Guest, I know what mod is but I do not follow your logic. You lost me right at the beginning.

Where did 64 come from?  

I did get the same asnwer as you, so I guess your logic is sound but I cannot follow it.

Here are my thoughts.

A 100-digit positive integer is divisible by 72. What is the greatest possible value of the sum of the number’s digits?

72=9*8    So it must be divisable by both 9 and by 8

if a number is divisable by 9 the digits have to add to a multiple of 9

1000 is divisable by 8, any number ending in 000 will be divisable by 8.

so the first   97 digits could be 9

now I have to think about the units, 10s and 100 places.

The sum of the 3 digits must be add to 9,18 or 27.    (in order for the number to be disable by 9)

27 is no good because the number would end in 9 and so not be divisable by 8

How about 18     

864  that works    

so I am going to say that the greatest possible value for the sum of the digits is

9*97+18 = 891

Hi Melody: He/she wants a 100-digit number which is divisible by 72.

I simply tried: 1E100 mod 72 =64 as the remainder. So, if I subtracted it from 1E100, the number would consist of all 9s and 36 and the end and be divisible by 72 with no remainder, which is the case. At the same time, it gives him/her the largest sum of the 100 digits, 98 of them being 9s and the last two being 36.