An algebra problem by Ritam Podder

Algebra Level 1

(999+999-1998)^999=?

Challange: Do it faster than a calculator.


The answer is 0.

Ritam Podder by Ritam Podder , 21, India

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5 solutions

( 999 + 999 − 1998 ) 999 = ? (999+999-1998 )^{999} = ?

[ ( 999 + 999 ) − 1998 ] 999 [(999+999)-1998]^{999}

= ( 1998 − 1998 ) 999 =(1998-1998)^{999}

= 0 999 = 0 =0^{999} = \boxed{0}

Yes, too easy. You can calculate faster than a calculator because of x 999 x^{999} is overload for the calculator.

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Azadali Jivani
Aug 14, 2015

0^n = 0 (Ans.)

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Ritam Podder
Aug 12, 2015

(999+999-1998)^999 =(1998-1998)^999 =0^999 =0

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998+998 is not equal to 1998... Please update your solution.

Joceleonard Tonido - 5 years, 10 months ago

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Thank You.

Ritam Podder - 5 years, 10 months ago
Edwin Gray
Apr 3, 2019

999 + 999 - 1998 = 0. 0^n = 0.(n.ne.0)

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Anything power 0 is always 0.

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No, not necessarily. For example, 0 0 = indeterminate ≠ 0 0^0=\textrm{indeterminate}\neq0

Aaron Tsai - 5 years ago

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