46a7b2 is a number of 6 digits in which a and b are two digits.This number is divisible by 9. Find the least value of a+b .Also state the maximum value of a+b .
Answers
Answer:
8
Step-by-step explanation:
when a number is divisible by 9 the sum of the digits of that number are also divisible by 9
when 46a7b2 is divisible by 9
then 4+6+a+7+b+2 is a multiple of 9
19 + a + b is divisible by 9
27 is the next multiple of 9 after 19
so 19 + a + b = 27
a+b =8
hope this helps you
please mark it as the brainliest answer
Answer:
8
Step-by-step explanation:
For a no. To be divisible by 9 the sum of the no. Should be a multiple of 9.
In the no. 46A7B2, A & B are single-digit no.
So the maximum sum of A+B should be 18.
The numbers can't be negative too, but the A & B can be 0.
So the minimum value of A+B=0 and Maximum value of A+B=18.
THAT'S NOT CORRECT. It's a common mistake.
The sum of other digits equal 19 so the minimum value of A+B is equal to 19 + X = multiple of which is the closest to 19.
Over here X = A+B.
And the multiple has to be greater than 19 as X can't be negative. The closest such multiple is 27
Therefore,
X=27-19
X=8
This is the minimum value of A+B
The maximum value of A+B should be equal to or less than 18.
Since X is lesser than 18 the value of the multiple should be the closest multiple of 9 which is less than 37.
(19+18=37)
Such multiple is 36, therefore,
X=36-19
X=17
This should be the Maximum value of A+B
Hope this helps