Question

6a-3b+6=a+b



a what's the least
a ve b natural number​

Answer 1

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.

Answer 2

Answer:

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