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Given-N-in-base-10-is-158b687a-where-b-and-a-are-digits-a-lt-b-Given-N-2-a-4-1-Determinate-couples-a-b-such-that-11-divise-N-2-Determinate-couples-a-b-such-that-3-and-25-divise-N-thanks

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Question Number 121568 by mathocean1 last updated on 09/Nov/20
Given: N in base 10 is 158b687a where b and a are digits (a<b). Given: N≡2+a[4]. 1) Determinate couples (a;b) such that 11 divise N. 2) Determinate couples (a;b) such that 3 and 25 divise N. thanks.
$$\mathrm{Given}:\:\mathrm{N}\:\mathrm{in}\:\mathrm{base}\:\mathrm{10}\:\mathrm{is} \\ $$$$\mathrm{158b687a}\:\mathrm{where}\:\mathrm{b}\:\mathrm{and}\:\mathrm{a}\:\mathrm{are} \\ $$$$\mathrm{digits}\:\left(\mathrm{a}<\mathrm{b}\right).\:\mathrm{Given}: \\ $$$$\mathrm{N}\equiv\mathrm{2}+\mathrm{a}\left[\mathrm{4}\right]. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Determinate}\:\mathrm{couples} \\ $$$$\left(\mathrm{a};\mathrm{b}\right)\:\mathrm{such}\:\mathrm{that}\:\mathrm{11}\:\mathrm{divise}\:\mathrm{N}. \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Determinate}\:\mathrm{couples}\:\left(\mathrm{a};\mathrm{b}\right) \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{3}\:\mathrm{and}\:\mathrm{25}\:\mathrm{divise}\:\mathrm{N}. \\ $$$$\mathrm{thanks}. \\ $$
Answered by kolos last updated on 17/Nov/20
1)jf 11 divise N , then N≡0mod(11) 15806870+10000b+a≡0mod(11) 2+b+a≡omod(11) a+b≡9mod11 (a,b)=(0,9)(1,8)(2,7)...(8,1)(9,0) 2)if N is multiple of 25 , therefore the last two digits are 00 ,25,50 or 75 as the last two digits of N are 7a , then a=5 to be multiple of 3 , the sum of the digits are a multiple too then 1+5+8+b+6+8+7+5≡0mod3 b+40≡0mod3 b≡2mod3 b=2,5,8 (a,b)=(5,2)(5,5)(5,8) I hope i′ve helped :)
$$ \\ $$$$\left.\mathrm{1}\right){jf}\:\mathrm{11}\:{divise}\:{N}\:,\:{then}\:{N}\equiv\mathrm{0}{mod}\left(\mathrm{11}\right) \\ $$$$\mathrm{15806870}+\mathrm{10000}{b}+{a}\equiv\mathrm{0}{mod}\left(\mathrm{11}\right) \\ $$$$\mathrm{2}+{b}+{a}\equiv{omod}\left(\mathrm{11}\right) \\ $$$${a}+{b}\equiv\mathrm{9}{mod}\mathrm{11} \\ $$$$\left({a},{b}\right)=\left(\mathrm{0},\mathrm{9}\right)\left(\mathrm{1},\mathrm{8}\right)\left(\mathrm{2},\mathrm{7}\right)…\left(\mathrm{8},\mathrm{1}\right)\left(\mathrm{9},\mathrm{0}\right) \\ $$$$ \\ $$$$\left.\mathrm{2}\right){if}\:{N}\:{is}\:{multiple}\:{of}\:\mathrm{25}\:,\:{therefore} \\ $$$${the}\:{last}\:{two}\:{digits}\:{are}\:\mathrm{00}\:,\mathrm{25},\mathrm{50}\:{or}\:\mathrm{75} \\ $$$$ \\ $$$${as}\:{the}\:{last}\:{two}\:{digits}\:{of}\:{N}\:{are}\:\mathrm{7}{a}\:,\:{then} \\ $$$${a}=\mathrm{5} \\ $$$$ \\ $$$${to}\:{be}\:{multiple}\:{of}\:\mathrm{3}\:,\:{the}\:{sum}\:{of}\:{the} \\ $$$${digits}\:{are}\:{a}\:{multiple}\:{too} \\ $$$$ \\ $$$${then} \\ $$$$\mathrm{1}+\mathrm{5}+\mathrm{8}+{b}+\mathrm{6}+\mathrm{8}+\mathrm{7}+\mathrm{5}\equiv\mathrm{0}{mod}\mathrm{3} \\ $$$${b}+\mathrm{40}\equiv\mathrm{0}{mod}\mathrm{3} \\ $$$${b}\equiv\mathrm{2}{mod}\mathrm{3} \\ $$$${b}=\mathrm{2},\mathrm{5},\mathrm{8} \\ $$$$ \\ $$$$\left({a},{b}\right)=\left(\mathrm{5},\mathrm{2}\right)\left(\mathrm{5},\mathrm{5}\right)\left(\mathrm{5},\mathrm{8}\right) \\ $$$$ \\ $$$$\left.{I}\:{hope}\:{i}'{ve}\:{helped}\::\right) \\ $$$$ \\ $$
Commented by mathocean1 last updated on 19/Nov/20
thank you very much sir!
$${thank}\:{you}\:{very}\:{much}\:{sir}! \\ $$

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