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Given below are two statements:

Statement $I$: $5$ divides $n^5-n$ wherever $n$ is a nonnegative integer

Statement $II$: $6$ divides $n^3-n$ whenever $n$ is a nonnegative integer

In the light of the above statements, choose the correct answer from the options given below

  1. Both Statement $I$ and Statement $II$ are correct
  2. Both Statement $I$ and Statement $II$ are incorrect
  3. Statement $I$ is correct but Statement $II$ is incorrect
  4. Statement $I$ is incorrect but Statement $II$ is correct
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2 Answers

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in exam Best way to solve take few values and verify       

Statement  I: 5 divides n^5−n wherever n is a nonnegative integer

Statement II: 6 divides n^3−n whenever  n is a nonnegative integer

n=1  both 0  hence divisible

n=2  30 & 30 hence divisible

n=3  240 & 210 hence divisible

n=4 1020 & 60 hence divisible

n=10  99990 & 990 again divisible

ans is A

(however  standard method is induction)
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Let P(n) be " 5 divides n5 - n ", where n = 0, 1, 2, ...

  • Basis step: 5 devides  05 - 0 = 0 => P(0) is true.
  • Inductive step: Assume P(n) is true, i.e. 5 divides n5 - n.
      • Then (n+1)5 - (n+1)
        • = n5  + 5n4  + 10n3  + 10n2  + 5n + 1 - n - 1
          = (n5 - n)+ 5n4 + 10n3  + 10n2  + 5n
          = (n5 - n)+ 5*(n4 + 2n3  + 2n2  + n)
  • (n5 - n) can be divided by 5, apparently 5*(n4 + 2n3  + 2n2  + n) can be divided by 5. This means that P(n+1) is true. This completes the inductive step and completes the proof.

and statement 2 is also true as it is property in number system, you can also prove this by taking examples.

Answer:

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