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Worked Examples · Example 7

Q.Show that, 41!+141! + 1 is not divisible by any number from 2 to 41.

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Every kk from 22 to 4141 divides 41!41! exactly (it is one of its factors), so 41!+141!+1 always leaves remainder 11 when divided by such a kk.

41!=1×2×3×⋯×40×4141! = 1\times2\times3\times\cdots\times40\times41 contains every integer from 11 to 4141 as a factor. If a∣ba\mid b (a divides b exactly), then b≡0(moda)b \equiv 0 \pmod a.

  1. Let kk be any integer with 2≤k≤412\le k\le41.
  2. Since 41!=1×2×⋯×k×⋯×4141! = 1\times2\times\cdots\times k\times\cdots\times41, the number kk appears explicitly as one of the multiplied factors, so kk divides 41!41! exactly: 41!=k×m41! = k\times m for some integer mm (namely m=41!/km = 41!/k).
  3. Therefore 41!≡0(modk)41! \equiv 0 \pmod k.
  4. Adding 11 to both sides: 41!+1≡0+1≡1(modk)41!+1 \equiv 0+1 \equiv 1 \pmod k. …

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