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Algebra Level 3

1 # ( 2 # ( 3 # ( 4 # ⋯ ( 98 # 99 ) ) ) ⋯ ) 1 \# ( 2\# (3\# (4\# \cdots (98 \# 99 )))\cdots )

Let # \# denote a binary operator such that a # b = a + b + a b \color{#D61F06}{a}\# \color{#3D99F6}{b} =\color{#D61F06}{a}+\color{#3D99F6}{b}+\color{#D61F06}{a}\color{#3D99F6}{b} . Then what is the value of the expression above?

Notation : n ! n! denote the factorial of n n .

99 ! + 1 99! + 1 100 ! − 1 100! - 1 99 × 99 ! 99\times 99! 100 ! − 99 ! 100! - 99!

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Snehan Jayakumar by Snehan Jayakumar , 18, India

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2 solutions

We note that a # b = a + b + a b = ( a + 1 ) ( b + 1 ) − 1 a\#b = a+b+ab = (a+1)(b+1)-1 . This implies that:

1 # ( 2 # ( 3 ( # 4... ( 98 # 99 ) ) ) . . . ) = 1 # ( 2 # ( 3 ( # 4...97 # ( 99 ˙ 100 − 1 ) ) ) . . . ) = 1 # ( 2 # ( 3 ( # 4...96 # ( 98 [ 99 ˙ 100 − 1 + 1 ] − 1 ) ) ) . . . ) = 1 # ( 2 # ( 3 ( # 4...96 # ( 98 ˙ 99 ˙ 100 − 1 ) ) ) . . . ) = 1 # ( 2 # ( 3 ( # 4...95 # ( 97 ˙ 98 ˙ 99 ˙ 100 − 1 ) ) ) . . . ) = 100 ! − 1 \begin{aligned} 1\#(2\#(3(\#4...(98\#99)))...) & = 1\#(2\#(3(\#4...97\#(99\dot{}100-1)))...) \\ & = 1\#(2\#(3(\#4...96\#(98[99\dot{}100-1+1]-1)))...) \\ & = 1\#(2\#(3(\#4...96\#(98\dot{}99\dot{}100-1)))...) \\ & = 1\#(2\#(3(\#4...95\#(97\dot{}98\dot{}99\dot{}100-1)))...) \\ & = \boxed{100! - 1} \end{aligned}

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Pulkit Gupta
Dec 9, 2015

The complete solution has been posted so I shall just post a trick method for fast computation ( the irony to the title :D )

Lets take 1#2 which computes to 5. Taking number of terms to be 2, this is nothing but (n+1) ! -1

Now, taking n = 99 yields 100! - 1

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