Repetition of digits

Algebra Level 3

555555 … 5 ⏟ 100 5’s × 999999 … 9 ⏟ 200 9’s − 555555 … 5 ⏟ 300 5’s + 555555 … 5 ⏟ 200 5’s + 555555 … 5 ⏟ 100 5’s = ? \underbrace{555555\ldots5}_{\text{100 5's}} \times \underbrace{999999\ldots9}_{\text{200 9's}} - \underbrace{555555\ldots5}_{\text{300 5's}} + \underbrace{555555\ldots5}_{\text{200 5's}} + \underbrace{555555\ldots5}_{\text{100 5's}} =\, ?

99999999 0 500000 555555555

Mohamed Aboalamayem by mohamed aboalamayem , 57, Egypt

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1 solution

Mohamed aboalamayem formula for repetition digits : ( X )n . ( 9 )m = x - ( x )n - ( x )m . So : ( x )n . ( 9 )m - x + ( x )n + ( x )m = 0 , each of n , m is integer greater than (0) . So we have : ( 5 )100 . ( 9 )200 - 5 + ( 5 )200 + ( 5 )100 = 0 . thus there are many my formulas of mine for repetition of digits .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

blue ( x ) = ( x )(n+m) , blue ( 5 ) = ( 5 )300 , as we know .

mohamed aboalamayem - 5 years, 2 months ago

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