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Serlea-Shows-that-the-sum-of-the-digit-of-100-25-25-is-divisible-by-4-




Question Number 85184 by Serlea last updated on 19/Mar/20
Serlea  Shows that the sum of the digit of                  100^(25) −25  is divisible by 4.
$$\mathrm{Serlea} \\ $$$$\mathrm{Shows}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digit}\:\mathrm{of} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{100}^{\mathrm{25}} −\mathrm{25} \\ $$$$\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{4}. \\ $$
Commented by mr W last updated on 19/Mar/20
100^(25) =10^(50) =100...000_(50 zeros)   100^(25) −25=100...000_(50 zeros) −25=99..9_(48 9s) 75  sum of digits=48×9+7+5=444=111×4  which is divisible by 4.
$$\mathrm{100}^{\mathrm{25}} =\mathrm{10}^{\mathrm{50}} =\mathrm{1}\underset{\mathrm{50}\:{zeros}} {\mathrm{00}…\mathrm{000}} \\ $$$$\mathrm{100}^{\mathrm{25}} −\mathrm{25}=\mathrm{1}\underset{\mathrm{50}\:{zeros}} {\mathrm{00}…\mathrm{000}}−\mathrm{25}=\underset{\mathrm{48}\:\mathrm{9}{s}} {\mathrm{99}..\mathrm{9}75} \\ $$$${sum}\:{of}\:{digits}=\mathrm{48}×\mathrm{9}+\mathrm{7}+\mathrm{5}=\mathrm{444}=\mathrm{111}×\mathrm{4} \\ $$$${which}\:{is}\:{divisible}\:{by}\:\mathrm{4}. \\ $$
Commented by Serlea last updated on 26/Mar/20
Great
$$\mathrm{Great} \\ $$

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