Question

If 4xy7 is exactly divisible by 3 , then the least value of ( x + y ) is ?

Answer 1

least value of (x+y) should be 1.

let x=1 nd y=0

required no is 4107

4107 is exactly divisibe by 3 and quotient is 1369.

And if x=0 y=1

requird no is 4017

4017 is exactly divisible by 3 and quotient is 1339.

so , least value of (x+y) is 1

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Answer 2

Given: 4xy7 is exactly divisible by 3

Find: Least value of ( x + y )

• For a number to be divisible by 3, its sum of digits should be divisible by 3.
• Here sum of digits = 4+x+y+7

                                       =11+x+y

• We have to minimize x+y so the nearest number to 11 divisible by 3 is 12.

11+x+y=12

∴Least value of (x+y)=1