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MATHMAA

# KVPY2013PartIISolutions

KVPY2013PartIISolutions contains the solutions of part II which carry two marks each. It has 5 questions , totally 10 marks for this part.

## Part II 61-65 solutions

61) Let a, b, c, d, e be natural numbers in an arithmetic progression such that a+b+c+d+e is the cube of an integer and b+c+d is square of an integer. The least possible value of the number of digits of c is

A) 2   B) 3  C) 4   D)5

Let c= x and the common difference be y then

a = x-2y, b = x-y, c = x, d = x+y, e = x+ 2y. Let m, n be the integers such that

a+b+c+d+e = $$m^{3}$$ and b+c+d = $$n^{2}$$

x-2y + x - y + x + x+ y + x + 2y =$$m^{3}$$ and x - y + x + x + y = $$n^{2}$$

5x =$$m^{3}$$ and 3x = $$n^{2}$$

From these two equations we get x = $$\frac{m^{3}}{5}$$ = $$\frac{n^{2}}{3}$$ .

The least values of m and n which satisfy the equations are

m= 5*3 so that $$\frac{m^{3}}{5}$$ = $$5^{2}*3^{3}$$ =  675and

n = 5*3*3 then $$\frac{n^{2}}{3}$$= $$5^{2}*3^{3}$$ = 675.

Therefore c = x = 675  and has three digits.

Hence option (B) is correct answer.

62) On each face of a cuboid, the sum of its perimeter and its area is written. Among the six numbers so written, there are three distinct numbers and they are 16, 24 and 31. The volume of cuboid lies between

A) 7 and 14    B) 14 and 21  C) 21 and 28   D) 28 and 35.

Solution :

Let length , width and height of cuboid be x, y, z . It has three different areas and perimeters .

Top and bottom has same area and perimeter, front and back has same area and perimeter and right , left has same area and perimeter.

According to the given,

2(x+y)+ x*y = 16 ---------------(1)

2(y+z) + y*z = 24 rearranging this we get z = $$\frac{24-2y}{2+y}$$ -----(2)

2(z+x) + z*x = 31  rearranging this we get z = $$\frac{31-2x}{2+x}$$--------(3)

from (2) and (3) we get

(24-2y)(2+x)=(31-2x)(2+y) solving this we get,

4x = 2+ 5y

x = $$\frac{2+5y}{4}$$ -----(4)

from (1) and (4) we get

2( $$\frac{2+5y}{4}+y$$ + $$\frac{2+5y}{4}$$*y = 16

simplifying this we get quadratic equation $$y^{2}$$ + 4y -12 = 0

(y+6)(y-2)=0

y=2 or y=-6 (ignore as width cannot be negative).

y=2 then x= 3 (plug y value in (4) )

from (2)  z= 5

x = 3, y=2, z=5 then volume =xyz =30 which lies between 28 and 35 .

Hence option (D) is the solution.

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