Question 1202386
<br>
Consider the remainders when each of the three numbers a, b, and c is divided by 3 -- that is, look at the values of a, b, and c mod 3.<br>
The possible values mod 3 are 0, 1, and 2.  Consider the possible combinations of those three values mod 3 to see which will satisfy the condition that the sum of any two of the mod 3 values is not equal to 0 mod 3.<br>
Note that in looking at these combinations order does not matter, so for example if we have looked at the case (0,1,1) we don't need to look at the case (1,0,1).<br>
Note that in my response I will consider "between 30 and 40" to mean the smallest possible natural number is 31 and the largest possible is 39....<br>
(1): (0,0,x)<br>
We can't have two of the numbers each divisible by 3, because then the sum of those two would be divisible by 3.<br>
(2): (0,1,1)<br>
This combination works; adding any two of the numbers gives a sum that is not divisible by 3.  For this combination we have the following sets:<br>
{33,34,37}; {31,33,34}; {31,33,37}; {31,36,37}; {34,36,37}<br>
Note that in each of these sets the sum of all three numbers is equal to 2 mod 3, so these sums are not divisible by 3.<br>
(3): (0,1,2)<br>
This won't work, because the sum of the last two is divisible by 3.<br>
(4): (0,2,2)<br>
This combination also works; adding any two of the numbers gives a sum that is not divisible by 3.  For this combination we have the following sets:<br>
{33,35,38}; (32,33,35}; {32,33,38}; {32,36,38}; {35,36,38}<br>
(5) (1,1,1)<br>
This is another combination that works -- the sum of any two of the numbers is not divisible by 3.  Note, however, that in this case the sum of all three numbers IS divisible by 3.  For this combination we have only the single set {31,34,37}<br>
(6) (2,2,2)<br>
This is similar to case (5); the sum of any two of the three numbers is not divisible by 3, but the sum of all three numbers is.  For this case we again have a single set: {32,35,38}<br>
We now have the complete answer to the problem....<br><pre>
  a   b   c   a+b+c   a+b+c divisible by 3?
 -------------------------------------------
 33  34  37    104     no
 31  33  34     98     no
 31  33  37    101     no
 31  36  37    104     no
 34  36  37    107     no

 33  35  38    106     no
 32  33  35    100     no
 32  33  38    103     no
 32  36  38    106     no
 35  36  38    109     no

 31  34  37    102     yes

 32  35  38    105     yes</pre>