An Eight digit no.

${The}$ ${maximum}$ ${possible}$ ${sum}$ ${of}$ ${eight}$ ${digits}$ = $2+ 3 + 4 + 5 + 6 + 7 + 8 + 9 = 44$
${The}$ ${ minimum}$ ${ possible}$ ${ sum}$ ${of }$ ${eight}$ ${ digits}$ = $0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28$

${For}$ ${ the }$ ${number}$ ${to}$ ${ be}$ ${ divisible}$ ${ by}$ ${ 9,}$ ${ the}$ ${ only}$ ${ possible}$ ${ sum}$ ${ of}$ ${ the}$ ${ digits}$ ${ is}$ $36$ .

${ If }$ ${we}$ ${pick}$ ${ the}$ ${set }$ ${of}$ ${digits}$ $\Rightarrow$ { ${1,2,3,4,5,6,7,8}$ }

${Their}$ ${sum}$ = $36$

${Total}$ ${ permutations}$ = $8!$

${Now,}$ ${if}$ ${we}$ ${remove}$ { ${1,8}$ } ${ and}$ ${replace}$ ${ them}$ ${ by}$ { ${0,9}$ } ${ the}$ ${sum}$ ${is}$ ${still}$ $36$ .

${In}$ ${this}$ ${case}$ ${total}$ ${possible}$ ${permutations}$ = $7\times(7\times6\times5\times4\times3\times2\times1) = 7\times7!$

${ Similarly,}$ ${3}$ ${more}$ ${cases}$ ${occur,}$ ${when}$ { ${2,7}$ } ,{ ${3,6}$ } { ${4,5}$ } ${are}$ ${replaced}$ ${by}$ { ${0,9}$ }.

${.:}$ ${Total}$ ${numbers}$ ${divisible}$ ${by}$ $9$ : $8! + 4\times(7\times7!) = 8\times7! + 28\times7! = 36\times7!$

${Comparing},{a=7},{b= 36}$

.: $a + b = 43$