Sainik School Entrance Exam 2019 Solved Paper

Question : 9 of 150

Marks: +1, -0

[1 X 2 Y 6 Z]

is a number divisible by 9 , then the least value of

X+Y+Z

is:

Solution:

(a) A number is divisible by 9 if the sum of digit is divisible by 9

∴ {1+X+2+Y+6+Z}/{9}={(X+Y+Z)+9}/{9}

∴

Least value of

X+Y+Z=0

∵{0+9}/{9}=1 .

Go to Question:

Prev Question

Next Question