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Question Number 184877 by Shrinava last updated on 13/Jan/23
Find the sum of the last ten digits:  5^(23)  ∙ 2^(2022)  − 2023
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{last}\:\mathrm{ten}\:\mathrm{digits}: \\ $$$$\mathrm{5}^{\mathrm{23}} \:\centerdot\:\mathrm{2}^{\mathrm{2022}} \:−\:\mathrm{2023} \\ $$
Answered by Rasheed.Sindhi last updated on 14/Jan/23
5^(23)  ∙ 2^(2022)  − 2023  5^(23) ∙2^(23) ∙2^(1999) −2023  2^(1999) .10^(23) −2023  2^(1999) ∙100000...0000_(23 zeros) −2023     =    ...a000...00_(23 zeroes)  −2023     a, the 24th digit from right,  is a  non-zero digit  We need borrow in order to subtract 2023  After borrowing:  Last 24 digits will be changed:  (a−1)99999...9 9 9_(22 nines)  (10)^(−)                                −2 0 2  3         _(−)              9 9 9...9 9 9 7 9 7 7_(Last 23 numbers)   Last ten digits:  9999997977
$$\mathrm{5}^{\mathrm{23}} \:\centerdot\:\mathrm{2}^{\mathrm{2022}} \:−\:\mathrm{2023} \\ $$$$\mathrm{5}^{\mathrm{23}} \centerdot\mathrm{2}^{\mathrm{23}} \centerdot\mathrm{2}^{\mathrm{1999}} −\mathrm{2023} \\ $$$$\mathrm{2}^{\mathrm{1999}} .\mathrm{10}^{\mathrm{23}} −\mathrm{2023} \\ $$$$\mathrm{2}^{\mathrm{1999}} \centerdot\mathrm{1}\underset{\mathrm{23}\:{zeros}} {\underbrace{\mathrm{00000}…\mathrm{0000}}}−\mathrm{2023} \\ $$$$\:\:\:=\:\:\:\:…{a}\underset{\mathrm{23}\:{zeroes}} {\underbrace{\mathrm{000}…\mathrm{00}}}\:−\mathrm{2023}\:\:\: \\ $$$${a},\:{the}\:\mathrm{24}{th}\:{digit}\:{from}\:{right},\:\:{is}\:{a} \\ $$$${non}-{zero}\:{digit} \\ $$$${We}\:{need}\:{borrow}\:{in}\:{order}\:{to}\:{subtract}\:\mathrm{2023} \\ $$$${After}\:{borrowing}: \\ $$$${Last}\:\mathrm{24}\:{digits}\:{will}\:{be}\:{changed}: \\ $$$$\overline {\left({a}−\mathrm{1}\right)\underbrace{\mathrm{99999}…\mathrm{9}\:\mathrm{9}\:\mathrm{9}}\:\left(\mathrm{10}\right)} \\ $$$$\underset{−} {\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}\:\mathrm{0}\:\mathrm{2}\:\:\mathrm{3}\:\:\:\:\:\:\:\:\:} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\underset{{Last}\:\mathrm{23}\:{numbers}} {\underbrace{\mathrm{9}\:\mathrm{9}\:\mathrm{9}…\mathrm{9}\:\mathrm{9}\:\mathrm{9}\:\mathrm{7}\:\mathrm{9}\:\mathrm{7}\:\mathrm{7}}} \\ $$$${Last}\:{ten}\:{digits}:\:\:\mathrm{9999997977} \\ $$
Commented by Shrinava last updated on 15/Jan/23
thank you dear professor cool
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{cool} \\ $$

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