An algebra problem by Budi Utomo

Algebra Level 3

If a a is positive integer such that a ⋅ a ! = 100 ( a − 2 ) ! a\cdot a! = 100(a-2)! , find 1 ⋅ 1 ! + 2 ⋅ 2 ! + 3 ⋅ 3 ! + ⋯ + a ⋅ a ! 1\cdot 1! + 2 \cdot 2! + 3 \cdot 3! +\cdots + a \cdot a! .


Notation: ! ! is the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1\times2\times3\times\cdots\times8 .

5039 119 719 23

Budi Utomo by Budi Utomo , 23, Indonesia

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2 solutions

Achal Jain
Mar 22, 2017

The equation on the left hand side can be easily broken down into a ( a ) ( a − 1 ) ( a − 2 ) ! \large a(a)(a-1)(a-2)! .

Now notice the ( a − 2 ) ! \large (a-2)! part gets cancelled . We remain with a 2 ( a − 1 ) = 100 \large a^{2}(a-1)=100 Now Since a a is an integer we can easily see that

25 × 4 = 100 \large 25\times 4=100 and Voila! \large \text{Voila!} a = 5 \large \boxed {a=5} . Plug in the value and you get your solution.

The problem shouldn't be level 4 \large 4

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Budi Utomo
Mar 21, 2017

<=> a . a ! a.a! = 100. ( a − 2 ) ! 100.(a-2)! <=> a . a . ( a − 1 ) . ( a − 2 ) ! a.a.(a-1).(a-2)! = 100. ( a − 2 ) ! 100.(a-2)! <=> a 2 . ( a − 1 ) a^{2}.(a-1) = 100 100 <=> a 3 − a 2 a^{3}-a^{2} = 100 100 <=> a 3 − a 2 − 100 a^{3}-a^{2}-100 = 0 0 [Remember a E Z+] <=> a = 5 a = 5

1.1! + 2.2! + ... + a.a! = (a+1)! - 1 = 6! - 1 = 719

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