a, b, c are three distinct integers from 2 to 10 (both inclusive)

Question

a, b, c are three distinct integers from 2 to 10, both inclusive. Exactly one of ab, bc and ca is odd. abc is a multiple of 4. The arithmetic mean of a and b is an integer and so is the arithmetic mean of a, b and c. How many such triplets are possible (unordered triplets)?

A. 8
B. 6
C. 2
D. 4
E. 10

Show Answer

Official Answer

D

A. 8
B. 6
C. 2
D. 4
E. 10

HAPPYatHARVARD · Answer

out of a , b, c
c has to be 4 or 8.

a and b has to be odd.
so they can be 3, 5, 7, 9
a+b+c is divisible by 3
hence only numbers possible are
 3 5 4
3 7 4
7 9 8
5 9 4
hence 4 such triplets are possible.

ayushkumar22941 · Answer

HAPPYatHARVARD 
you missed 7+3+8 here both 7 and 3 odd and 8 is also there as triplet

srini2117 · Answer

Hi.. 2 and 10 are included in the set . Should not the answer be 8 ?

HAPPYatHARVARD · Answer

Hi 2 and 10 are not divisible by 4.

sharathnair14 · Answer

EXPLANATION

This is more a puzzle than a challenging question. But is a very interesting puzzle nevertheless.
Exactly one of ab, bc and ca is odd => Two are odd and one is even.

abc is a multiple of 4 => the even number is a multiple of 4.
The arithmetic mean of a and b is an integer => a and b are odd.
and so is the arithmetic mean of a, b and c. => a + b + c is a multiple of 3.

c can be 4 or 8.
c = 4; a, b can be 3, 5 or 5, 9
c = 8; a, b can be 3, 7 or 7, 9
Four triplets are possible.

The question is "How many such triplets are possible (unordered triplets)?"

Hence the answer is "4"
Choice D is the correct answer.

CREDIT : https://iim-cat-questions-answers.2iim. ... rob4.shtml

lyuboslav_p · Answer

Isn't the question wrong? It is obvious that C is even, and for A & B we now they are either both even or both odd. None of the two choices works because if both are even -> there is no odd number among ab, bc, ca. If both are odd then we don't have exactly one odd among ab, bc, ca but 2???

adityashinde88 · Answer

A+B= even,  This means that A and B are odd, because Odd+ Odd= Even. And it says that the sum is even.
ab, bc, ca , exactly one is odd, for this to happen C should be even because only then bc and ca will become even.  Odd*Odd = Odd, Even*Odd= Even. 
Figuring out the options for A and B: We get Odd options:  3,5,7,9.  
And for C we get Even options:  2,4,6,8,10. 
It says abc/4 this means c has to be multiple of 4. Because None of the odd options which are A and B can be a divisible of 4, So C has to be a multiple of 4. So our options are 4,8 for c.
Making our triplets in such a way that they are divisible by 3, you need to make, A+B+C a multiple of 3, we can do this in 4 ways: 
Our options 3,5,7,9 for a and b, 4,8 for c
 3+5+4=12  5+7+4=16 7+9+4=20
3+7+4=14  5+9+4=18 
3+9+4=16
3+5+8=16 5+7+8=20
 3+7+8=18  5+9+8=22
3+9+8=20  7+9+8=24 
We have 4 triplets that satisfy the given conditions, as it says unordered triplets, if it had said ordered triplets, we would have 8 triplets

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a, b, c are three distinct integers from 2 to 10 (both inclusive)

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a, b, c are three distinct integers from 2 to 10 (both inclusive)

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