$a b c$ is a three-digit number. This three-digit number is reversed and subtracted from the original number. By which number will the difference obtained be always divisible?

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Solution

𝒂𝒃𝒄 is a three-digit number
⇒ 𝒂𝒃𝒄 = 𝟏𝟎𝟎𝒂 + 𝟏𝟎𝒃 + 𝒄
After reversing the digits of 𝑎𝑏𝑐, the number becomes 𝒄𝒃𝒂
⇒ 𝒄𝒃𝒂 = 𝟏𝟎𝟎𝒄 + 𝟏𝟎𝒃 + 𝒂

Now,
𝒂𝒃𝒄 − 𝒄𝒃𝒂 = (𝟏𝟎𝟎𝒂 + 𝟏𝟎𝒃 + 𝒄) − (𝟏𝟎𝟎𝒄 + 𝟏𝟎𝒃 + 𝒂)
⇒ 𝒂𝒃𝒄 − 𝒄𝒃𝒂 = 𝟗𝟗(𝒂 − 𝒄)
⇒ 𝒂𝒃𝒄 − 𝒄𝒃𝒂 = 𝟗 × 𝟏𝟏 × (𝒂 − 𝒄)
Thus, the difference is divisible by 𝟗 as well as 𝟏𝟏.

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