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Exercise 6.1 · Q4

Q.If (n+2)!=60×(n−1)!(n+2)! = 60 \times (n-1)!, find nn.

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Solve the factorial equation (n+2)!=60×(n−1)!(n+2)! = 60\times(n-1)! for the positive integer nn by peeling (n+2)!(n+2)! down to (n−1)!(n-1)! and cancelling.

Any factorial (n+2)!(n+2)! can be unrolled down to a smaller factorial:

(n+2)!=(n+2)×(n+1)×n×(n−1)!(n+2)! = (n+2)\times(n+1)\times n\times(n-1)!

valid for n≥1n \ge 1 (so that (n−1)!(n-1)! is defined for non-negative integers).

  1. Start with the given equation: (n+2)!=60×(n−1)!(n+2)! = 60\times(n-1)!.
  2. Unroll the left side down to (n−1)!(n-1)!: (n+2)!=(n+2)(n+1)(n)(n−1)!(n+2)! = (n+2)(n+1)(n)(n-1)!.
  3. Substitute: (n+2)(n+1)(n) (n−1)!=60 (n−1)!(n+2)(n+1)(n)\,(n-1)! = 60\,(n-1)!.
  4. Since (n−1)!≠0(n-1)! \ne 0, divide both sides by (n−1)!(n-1)!: (n+2)(n+1)(n)=60(n+2)(n+1)(n) = 60.
  5. This is a product of three consecutive integers equal to 6060. Try small positive integers: for n=3n=3: (3+2)(3+1)(3)=5×4×3=60(3+2)(3+1)(3) = 5\times4\times3 = 60. ✓ This matches. …

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