6a-3b+6=a+b
a what's the least
a ve b natural number
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6
Answer:
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.
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