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Exercise 6.1 · Q5

Q.Find the LCM of

(i) 7!7! and 8!8!
(ii) 6!6! and 9!9!
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Find the LCM of two factorial pairs — since factorials are nested (each divides every larger one), the LCM of k!k! and m!m! (with m>km>k) is simply the larger factorial m!m!.

For positive integers k<mk<m: m!=m×(m−1)×⋯×(k+1)×k!m! = m\times(m-1)\times\cdots\times(k+1)\times k!, so k!∣m!k! \mid m! (k! divides m! exactly, with quotient m×(m−1)×⋯×(k+1)m\times(m-1)\times\cdots\times(k+1)). Consequently:

lcm(k!, m!)=m!whenever m>k\text{lcm}(k!,\,m!) = m!\quad\text{whenever } m > k

because m!m! is itself a multiple of k!k!, so no larger common multiple is needed.

(i) LCM of 7!7! and 8!8!

  1. Note 8!=8×7!8! = 8\times7!, so 7!7! divides 8!8! exactly (quotient 88).
  2. Since 7!7! is a factor of 8!8!, the least common multiple is just the larger number: lcm(7!,8!)=8!\text{lcm}(7!,8!) = 8!.
  3. Compute 8!=8×7×6×5×4×3×2×18! = 8\times7\times6\times5\times4\times3\times2\times1: 8×7=568\times7=56, ×6=336\times6=336, ×5=1680\times5=1680, ×4=6720\times4=6720, ×3=20160\times3=20160, ×2=40320\times2=40320.

(ii) LCM of 6!6! and 9!9! …

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