Math, asked by kaminigambhir84, 2 months ago

Please Explain

(b) 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

Answered by ahalsanam2007
0

Answer:Hi ,

A natural number is divisible by 9

if and only if some of it's digits is

divisible by 9 .

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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