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Question Number 21738 by Joel577 last updated on 02/Oct/17
What is the last digit from the sum of 1 . 2^1 + 2 . 2^2 + 3 . 2^3 + ... + 50 . 2^(50) ?
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{last}\:\mathrm{digit}\:\mathrm{from}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{1}\:.\:\mathrm{2}^{\mathrm{1}} \:+\:\mathrm{2}\:.\:\mathrm{2}^{\mathrm{2}} \:+\:\mathrm{3}\:.\:\mathrm{2}^{\mathrm{3}} \:+\:…\:+\:\mathrm{50}\:.\:\mathrm{2}^{\mathrm{50}} \:? \\ $$
Answered by $@ty@m last updated on 02/Oct/17
S_(50) =1 . 2^1 + 2 . 2^2 + 3 . 2^3 + ... + 50 . 2^(50) 2.S_(50) = 1.2^2 + 2.2^3 +......+49.2^(50) +50.2^(51) ______________________________ Subtracting −S_(50) =2^1 +2^2 +2^3 +......+2^(50) −50.2^(51) −S_(50) =((2(2^(50) −1))/(2−1))−50.2^(51) −S_(50) =2^(51) −2−50.2^(51) S_(50) =49.2^(51) +2 −−−(1) we observe that last digit in 2^1 =2 last digit in 2^2 =4 last digit in 2^3 =8 last digit in 2^4 =6 last digit in 2^5 =2 . . ... and so on. ⇒last digit in 2^(4n+3) =8 ⇒last digit in 2^(4×12+3) =8 ⇒last digit in 49.2^(51) +2=4 Ans.
$${S}_{\mathrm{50}} =\mathrm{1}\:.\:\mathrm{2}^{\mathrm{1}} \:+\:\mathrm{2}\:.\:\mathrm{2}^{\mathrm{2}} \:+\:\mathrm{3}\:.\:\mathrm{2}^{\mathrm{3}} \:+\:…\:+\:\mathrm{50}\:.\:\mathrm{2}^{\mathrm{50}} \\ $$$$\mathrm{2}.{S}_{\mathrm{50}} =\:\:\:\:\:\:\:\:\:\mathrm{1}.\mathrm{2}^{\mathrm{2}} +\:\:\mathrm{2}.\mathrm{2}^{\mathrm{3}} +……+\mathrm{49}.\mathrm{2}^{\mathrm{50}} +\mathrm{50}.\mathrm{2}^{\mathrm{51}} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${Subtracting} \\ $$$$−{S}_{\mathrm{50}} =\mathrm{2}^{\mathrm{1}} +\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{3}} +……+\mathrm{2}^{\mathrm{50}} −\mathrm{50}.\mathrm{2}^{\mathrm{51}} \\ $$$$−{S}_{\mathrm{50}} =\frac{\mathrm{2}\left(\mathrm{2}^{\mathrm{50}} −\mathrm{1}\right)}{\mathrm{2}−\mathrm{1}}−\mathrm{50}.\mathrm{2}^{\mathrm{51}} \\ $$$$−{S}_{\mathrm{50}} =\mathrm{2}^{\mathrm{51}} −\mathrm{2}−\mathrm{50}.\mathrm{2}^{\mathrm{51}} \\ $$$${S}_{\mathrm{50}} =\mathrm{49}.\mathrm{2}^{\mathrm{51}} +\mathrm{2}\:−−−\left(\mathrm{1}\right) \\ $$$${we}\:{observe}\:{that} \\ $$$${last}\:{digit}\:{in}\:\mathrm{2}^{\mathrm{1}} =\mathrm{2} \\ $$$${last}\:{digit}\:{in}\:\mathrm{2}^{\mathrm{2}} =\mathrm{4} \\ $$$${last}\:{digit}\:{in}\:\mathrm{2}^{\mathrm{3}} =\mathrm{8} \\ $$$${last}\:{digit}\:{in}\:\mathrm{2}^{\mathrm{4}} =\mathrm{6} \\ $$$${last}\:{digit}\:{in}\:\mathrm{2}^{\mathrm{5}} =\mathrm{2} \\ $$$$. \\ $$$$. \\ $$$$…\:{and}\:{so}\:{on}. \\ $$$$\Rightarrow{last}\:{digit}\:{in}\:\mathrm{2}^{\mathrm{4}{n}+\mathrm{3}} =\mathrm{8} \\ $$$$\Rightarrow{last}\:{digit}\:{in}\:\mathrm{2}^{\mathrm{4}×\mathrm{12}+\mathrm{3}} =\mathrm{8} \\ $$$$\Rightarrow{last}\:{digit}\:{in}\:\mathrm{49}.\mathrm{2}^{\mathrm{51}} +\mathrm{2}=\mathrm{4}\:{Ans}. \\ $$
Commented by Joel577 last updated on 03/Oct/17
Thanks. But my friend got 8 He used mod to find the last digit
$${Thanks}.\:{But}\:{my}\:{friend}\:{got}\:\mathrm{8} \\ $$$${He}\:{used}\:{mod}\:{to}\:{find}\:{the}\:{last}\:{digit} \\ $$
Commented by $@ty@m last updated on 03/Oct/17
In that case either he or I am wrong. Try to find where is the mistake.
$${In}\:{that}\:{case}\:{either}\:{he}\:{or}\:{I}\:{am}\:{wrong}. \\ $$$${Try}\:{to}\:{find}\:{where}\:{is}\:{the}\:{mistake}. \\ $$

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