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Question

If 307=x+1y+1z where x,y,z are positive integers then value of (x+y+z) equals

A
9
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B
11
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C
13
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D
30
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Solution

The correct option is A 9

Given 307=x+1y+1z
given=x,y,z are positive integer.
On solving R. H. S. we will get
307=x×y×z+x+zy×z+1 ....... equ(1)
on comparing denominator and numerator of both side
We will get
30=x×y×z+x+z ........... equ(2)
7=y×z+1 .......... equ(3);
y×z=6
because y and z are positive integer
Hence there are 4 possibilities arise
1. y=6 and z=1
2.y=3 and z=2
3. y=2 andz=3
4.y=1 and z=6
now from equation(2)
30=6×x+x+z; .....because [yz=6]
7×x+z=30 ...... equ(4)
Because we know x is also positive integer
hence from the above 4 possibilities only z=2 satisfied equ(4)
hence putting the value of z=2 in equ(4) we will get x=4
and on solving equ(3) we get y=3

sox=4,y=3 and z=2
So x+y+z=4+3+2=9


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