Question

. The difference between a three-digit number and the
number formed by reversing its digits is divisible by
(A) 9
(B) 11
(C) Both 9 and 11
(D) Neither 9 nor 11​

Answer 1

Answer:

$\boxed{ \purple{ \bf \: option \: (c) \: is \: correct.}}$

Step-by-step explanation:

$\tt \: \: Let \: us \: consider \: a \: three \: digit \: number \: as \: abc$

☆ Now, abc can be rewritten as

$\tt \:  \longrightarrow \: abc \: = 100a + 10b + c$

☆ On reversing the digits, we get

$\tt \:  \longrightarrow \: three \: digit \: number \: as \: \bf{cba}$

☆ Now, cba can be written as

$\tt \:  \longrightarrow \: cba = 100c + 10b + a$

$\tt \:  \longrightarrow \: \bold \red{Now}$

$\tt \:  \longrightarrow \: abc \: - \: cba$

$\tt \:  \longrightarrow \: 100a + 10b + c - (100c + 10b + a)$

$\tt \:  \longrightarrow \: 100a + 10b \: + c - 100c - 10b - a$

$\tt \:  \longrightarrow \: 99a - 99c$

$\tt \:  \longrightarrow \: 99(a \: - \: c)$

$\tt\implies \:abc \: - \: cba \: is \: divisible \: by \: 99$

$\tt\implies \:abc - cba \: is \: divisible \: by \: both \: 9 \: and \: 11$