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Let $a$ be the $81$ – digit number of which all the digits are equal to $1$. Then the number $a$ is,

1. divisible by $9$ but not divisible by $27$
2. divisible by $27$ but not divisible by $81$
3. divisible by $81$
4. None of the above

recategorized | 25 views

+1 vote

$a = 11111111.....81\ digits$

Divisibility by 9

If sum of digits of a number is divisible by $9$ then that number is divisible by $9$

So sum of digits of $a =81$ which is divisible by $9$

Hence $a$ is divisible by $9$

Divisibility by 27

Group the digits of $a$ in block of $3$ from right to left i.e. $111 , 111, 111, ....27$ blocks

Now if the sum of these blocks are divisible by $27$ then the number is divisible by $27$

i.e. $111+111+111....27$ terms = $111*27$ which s divisible by $27$

Hence  $a$ is also divisible by $27$

Divisibility by 81

Generalized Divisibilty Rule says,

If the number is too large, you can also break it down into several strings with $e$ digits each, satisfying either $10^e = 1$ or $10^e = -1 (mod\ D)$. The sum (or alternate sum) of the numbers have the same divisibility as the original one. (where $D$ is the divisor)

$D=81$

$10^9 \ mod\ 81 = 1 \implies e=9$

So we make blocks of $9$ digits and add them

$111111111+ 111111111+ 111111111+111111111+111111111+111111111+111111111+111111111+111111111 = 999999999$

Now if $999999999$ is divisible by $81$ then $a$ is also divisible by $81$

For checking divisibility of $81$ one method is

Subtract 8 times the last digit from the rest, if the result obtained is divisible by $81$ then the number is also divisible by $81$

We apply this rule recursively to get the answer as follows.

$99999999 - 9*8 = 99,999,927$

$9999992 - 7*8 = 9,999,936$

$999993 - 6*8 = 999945$

$99994 - 5*8 = 99954$

$9995 - 4*8 = 9963$

$996 - 3*8 = 972$

$97 - 2*8 = 81$

Since $81$ is divisible by $81 \implies 999999999$ is divisible by $81 \implies$ $a$ is also divisible by $81$

$\therefore$ Option $C.$ is correct.

by Boss (23.8k points)

+1 vote