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Question Number 11267 by 786786AM last updated on 18/Mar/17

Sum of all 2 digit numbers which when  divided by 4 yield unity as remainder is

$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{2}\:\mathrm{digit}\:\mathrm{numbers}\:\mathrm{which}\:\mathrm{when} \\ $$$$\mathrm{divided}\:\mathrm{by}\:\mathrm{4}\:\mathrm{yield}\:\mathrm{unity}\:\mathrm{as}\:\mathrm{remainder}\:\mathrm{is} \\ $$

Answered by mrW1 last updated on 19/Mar/17

13+17+21+∙∙∙+97  =((13+97)/2)×22  =1210

$$\mathrm{13}+\mathrm{17}+\mathrm{21}+\centerdot\centerdot\centerdot+\mathrm{97} \\ $$$$=\frac{\mathrm{13}+\mathrm{97}}{\mathrm{2}}×\mathrm{22} \\ $$$$=\mathrm{1210} \\ $$

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