Question 1202386: a,b,c are three natural numbers in ascending order lying between 30 and 40 . Given (a+b), (b+c)and (c+a) are not divisible by 3 . Find all triples (a,b,c) . Find the sum in each case . Which of the sum is divisible by 3
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52848)   (Show Source):
You can put this solution on YOUR website! .
a,b,c are three natural numbers in ascending order lying between 30 and 40 .
Given (a+b), (b+c)and (c+a) are not divisible by 3 . Find all triples (a,b,c) .
Find the sum in each case . Which of the sum is divisible by 3
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H I N T #1 : Make a super-heroic effort and find ONE such triple (a,b,c).
H I N T #2 : After successful completing step 1, make next super-heroic effort and find next such triple.
After successfully completing step1 and step 2, you should see the WAY.
As soon as you see the way, follow it to the end and do not stop until you complete everything on your own.
After that, you may report me about your achievements.
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
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.
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.
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).
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....
(1): (0,0,x)
We can't have two of the numbers each divisible by 3, because then the sum of those two would be divisible by 3.
(2): (0,1,1)
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:
{33,34,37}; {31,33,34}; {31,33,37}; {31,36,37}; {34,36,37}
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.
(3): (0,1,2)
This won't work, because the sum of the last two is divisible by 3.
(4): (0,2,2)
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:
{33,35,38}; (32,33,35}; {32,33,38}; {32,36,38}; {35,36,38}
(5) (1,1,1)
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}
(6) (2,2,2)
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}
We now have the complete answer to the problem....
a b c a+b+c a+b+c divisible by 3?
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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
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