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Exercise 6.2 · Q4

Q.If 16!+17!=x8!\dfrac{1}{6!} + \dfrac{1}{7!} = \dfrac{x}{8!}, find xx.

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The key idea is to rewrite all terms with a common factorial denominator (8!8!) by expanding factorials, then equate numerators. The value of xx is 64.

Why this approach works

The problem gives you an equation involving factorials of consecutive integers: 6!6!, 7!7!, and 8!8!. The trick is that factorials grow by multiplying by the next integer — so 7!=7×6!7! = 7 \times 6! and 8!=8×7!=8×7×6!8! = 8 \times 7! = 8 \times 7 \times 6!.

If you want to combine fractions like 16!+17!\frac{1}{6!} + \frac{1}{7!}, you need a common denominator. The most natural common denominator here is 8!8!, because it's the largest factorial in the equation and the right-hand side is already expressed in terms of 8!8!. Once you rewrite both left-hand terms as fractions over 8!8!, you simply add the numerators and compare with x8!\frac{x}{8!}.


Step-by-step solution

  1. Express 7!7! and 8!8! in terms of 6!6!

    7!=7×6!7! = 7 \times 6!

    8!=8×7!=8×7×6!8! = 8 \times 7! = 8 \times 7 \times 6!

  2. Rewrite 16!\frac{1}{6!} with denominator 8!8!

16!=8×78×7×6!=568!\frac{1}{6!} = \frac{8 \times 7}{8 \times 7 \times 6!} = \frac{56}{8!}

Because 8!=8×7×6!8! = 8 \times 7 \times 6!, multiplying numerator and denominator by 8×78 \times 7 gives the equivalent fraction.

  1. Rewrite 17!\frac{1}{7!} with denominator 8!8!

17!=88×7!=88!\frac{1}{7!} = \frac{8}{8 \times 7!} = \frac{8}{8!}

Since 8!=8×7!8! = 8 \times 7!, multiply numerator and denominator by 88.

  1. Add the two fractions

16!+17!=568!+88!=56+88!=648!\frac{1}{6!} + \frac{1}{7!} = \frac{56}{8!} + \frac{8}{8!} = \frac{56 + 8}{8!} = \frac{64}{8!}

  1. Compare with the given equation …

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