Show that the number 122^n − 102^n − 21^n is always one less than a multiple of 2020. For every positive integer n.


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Question Number 63674 by Tawa1 last updated on 07/Jul/19

$Show that the number 122^{n} - 102^{n} - 21^{n} is always one less than a$ $multiple of 2020 . For every positive integer n .$
Commented by Prithwish sen last updated on 07/Jul/19
$(101+21)^n −(101+1)^n −21^n ={ 101^n +n101^(n−1) 21+......+n101.(21)^(n−1) +21^n }−{101^n +n101^(n−1) +.....+n101+1)−21^n = n101^(n−1) (21−1)+ ........+n101(21^(n−1) −1) −1 = 101×20{101^(n−2) +.......n(21^(n−2) +21^(n−2) +...+1)} −1 =2020×m −1 m=integer which is always 1 less than multiple of 2020 . Proved$
${(101 + 21)}^{n} - {(101 + 1)}^{n} - 21^{n}$ $= {101^{n} + {n 101}^{n - 1} 21 + . . . . . . + n 101 . {(21)}^{n - 1} + 21^{n}} - {101^{n} + {n 101}^{n - 1} + . . . . . + n 101 + 1) - 21^{n}$ $= {n 101}^{n - 1} (21 - 1) + . . . . . . . . + n 101 (21^{n - 1} - 1) - 1$ $= 101 \times 20 {101^{n - 2} + . . . . . . . n (21^{n - 2} + 21^{n - 2} + . . . + 1)} - 1$ $= 2020 \times m - 1 m = integer$ $which is always 1 less than multiple of 2020 . Proved$
Commented by Tawa1 last updated on 07/Jul/19

$God bless you sir$
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