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14. 8a34b91 is divisible by 9, where a and b are whole numbers. What can be
the minimum value of a + b? Also, write the maximum value of a + b.
Answers
Given :-
- 8a34b91 is divisible by 9
To Find :-
- Minimum and Maximum value of ( a + b )
Solution :-
❑ As we know that , according to the divisibility rule of ' 9 ' if sum of all digits of a no. is divisible by 9 then that full no. is divisible by 9
★ No. given to us is ( 8a34b91 )
★ Where value of ( a + b ) should be in digits ( 0 – 18 )
➼ Finding the minimum value
According to the rule we can write it as ::
➨ 8 + a + 3 + 4 + b + 9 + 1 should be a multiple of 9
➨ 25 + a + b should be a multiple of 9
➨ 27 is the nearest multiple to 25 , So :
➺ 25 + a + b = 27
➺ a + b = 27 – 25
➺ a + b = 2
Therefore , minimum value of ( a + b ) is 2
➼ Finding the maximum value
➨ 25 + a + b = multiple of 9
➨ Here , max. value of ( a + b ) as digits in this no. can be 18 because a digit is b/w ( 0 to 9 ) and we’re given two digits
✒ Let’s take value of ( a + b ) be 10
➺ 25 + 10 = not divisible
✒ Let , ( a + b ) be 11
➺ 25 + 11 = 36 ( divisible )
Note :-
This was hit and trial method and if you will try to add 12,13,14,15,16,17,18 ( the max. sum of two digits ) they will not be divisible
Therefore , maximum value of ( a + b ) is 11
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