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
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A natural number is divisible by 9
if and only if some of it's digits is
divisible by 9 .
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It is given that ,
46a7b2
i ) sum of the digits
= 4 + 6 + a + 7 + b + 2
= 19 + a + b
Least value of a + b = 8 ,
19 + a + b = 19 + 8 = 27(divisible by 9 )
ii ) maximum value of a + b = 17
19 + a + b
= 19 + 17
= 36( divisible by 9 )
Therefore ,
Least value of a + b = 8
Maximum value of a + b = 17
I hope this helps you.
: )
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 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 someone