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MATHMAA

# KVPY2013solutions

Here are the solutions with explanations of KVPY2013solutions examination held on 27-October-2013.

## KVPY2013solutions

1) Let x,y,z be three non negative integers such that x+y+z=10. The maximum possible value of xyz+xy+yz+zx is
A) 52           B)64           C) 69        D) 73

Solution:-
First let us write all the possible value of x,y,z which satisfy the condition x+y+z=10. The possible value are as follows

(1,1,8), (1,2,7), (1,3,6), (1,4,5), (2,3,5), (2,4,4), (2,6,2), (3,3,4)

Now we have to check these numbers by multiplying them and see that which has the largest value.

(x,y,z) = xyz + xy + yz + zx
(1,1,8) = 8  + 1  + 8  + 8    = 25.
(1,2,7) = 14 + 2  + 14 + 7    = 37
(1,3,6) = 18 + 3  + 18 + 6    = 45
(1,4,5) = 20 + 4  + 20 + 5    = 49
(2,3,5) = 30 + 6  + 15 + 10   = 61
(2,4,4) = 32 + 8  + 16 + 8    = 64
(2,6,2) = 24 + 12 + 12 + 4    = 52
(3,3,4) = 36 + 9  + 12 + 12   = 69

Hence, we see that the ordered pair (3,3,4) has the maximum value of 69 so the correct option is (C) .

Here we have short cut method for this question.

Here we have to find the maximum possible value of xyz+xy+yz+zx so, if we multiply the numbers in the triad and check which one gets the largest value for xyz then that ordered pair has the maximum possible value.
We have the largest value of xyz that is if we multiply the values in the triad we see that triad (3,3,4) is largest of all with value 36 so it is the correct triad. Now if we plug x=3,y=3,z=4 in the equation xyz+xy+yz+zx we get the value 69. Hence the correct option is (c).

That was the solution for the first question in detail. Click NEXT to get the second solution in detail. Please write your review on these solutions which I provide here. I am really happy to get the emails from the students who need the solutions for KVPY. So again I started working on this to provide further solutions. To go through the questions of Part II click BACK.