Question

Find the least value of b in 9b6 to make it divisible by 3 and 9 both.​

Answer 1

Answer:

b=3

936

Step-by-step explanation:

9+3+6=18

therefore , is divisible by 3 and 9 because the sum of the three numbers are  the multiple of 3 and 9

Answer 2

$\mathfrak{\huge{Answer:}}$

$\mathbb{GIVEN}$

Number : 9b6

$\mathbb{TO\:FIND}$

Such a value of b, which makes the given number divisible by 3 and 9 both.

$\mathbb{METHOD}$

What we'll do is utilize the information we know about the divisibility of numbers by 3 and by 9. This method is very similar to the trial and error method.

$\tt{CASE1:}$ When b = 1

Sum of the digits = 9 + 1 + 6 = 16

This number is neither divisible by 3 nor 9.

Therefore, b >< 1

$\tt{CASE2:}$ When b = 2

Sum of the digits = 9 + 2 + 6 = 17

Even this number isn't divisible neither by 3 nor by 9.

Therefore, b >< 2

$\tt{CASE3:}$ When b = 3

Sum of the digits = 9 + 3 + 6 = 18

This number is divisible both by 3 as well as 9.

Thus, b = 3

That's your answer ! $\huge{\sf{b = 3}}$