SOLUTION: Let n=355​,813​,6de be a​ base-ten numeral with d and e its last two digits. Give all of the choices of the​ two-digit numbers de for which n is divisible by 12.

We form the smallest potential value for de which is 10, the smallest 
two-digit number.

If de were equal to 10 we would have an integer when we divided

n=355​813​610 by 12.  But we get 29651134.17, which is not an integer.

So we go to the next integer above 29651134.17, which is 29651135, and

multiply it by 12 and we get 355813620, which is divisible by 12.

So we have 355813620, so that's a choice of de = 20

We add 12 to that to get the next multiple of 12: 

355813620 + 12 = 355813632, so that's a choice of de = 32

We add 12 to that to get the next multiple of 12 

355813632 + 12 = 355813644, so that's a choice of de = 44

We add 12 to that to get the next multiple of 12 

355813644 + 12 = 355813656, so that's a choice of de = 56

We add 12 to that to get the next multiple of 12 

355813656 + 12 = 355813668, so that's a choice of de = 68

We add 12 to that to get the next multiple of 12 

355813668 + 12 = 355813680, so that's a choice of de = 80

We add 12 to that to get the next multiple of 12 

355813680 + 12 = 355813692, so that's a choice of de = 92

That's as far as we can go, for adding 12 to that would change
the 3rd digit from the right to a 7.  But it must remain 6.

So all the choices for de are 20, 32, 44, 56, 68, 80, and 92.

Edwin