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Question Number 75505 by naka3546 last updated on 12/Dec/19

Commented by mr W last updated on 12/Dec/19

i don′t think such m and n exist.  the last digit from right of 2019^m   is 1 or 9. 5 times of it is always 5,  but this can not be 2019^n , since  the last digit of 2019^n  is also 1 or 9.

$${i}\:{don}'{t}\:{think}\:{such}\:{m}\:{and}\:{n}\:{exist}. \\ $$$${the}\:{last}\:{digit}\:{from}\:{right}\:{of}\:\mathrm{2019}^{{m}} \\ $$$${is}\:\mathrm{1}\:{or}\:\mathrm{9}.\:\mathrm{5}\:{times}\:{of}\:{it}\:{is}\:{always}\:\mathrm{5}, \\ $$$${but}\:{this}\:{can}\:{not}\:{be}\:\mathrm{2019}^{{n}} ,\:{since} \\ $$$${the}\:{last}\:{digit}\:{of}\:\mathrm{2019}^{{n}} \:{is}\:{also}\:\mathrm{1}\:{or}\:\mathrm{9}. \\ $$

Commented by MJS last updated on 12/Dec/19

I think it′s meant like this  2019^n =......abcde  2019^m =......fghij  a=5f

$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\mathrm{meant}\:\mathrm{like}\:\mathrm{this} \\ $$$$\mathrm{2019}^{\mathrm{n}} =......{abcde} \\ $$$$\mathrm{2019}^{{m}} =......{fghij} \\ $$$${a}=\mathrm{5}{f} \\ $$

Answered by MJS last updated on 12/Dec/19

2019^4 =16 616 719 002 321  2019^7 =136 758 469 738 531 852 205 739  m=4∧n=7  m+n=11

$$\mathrm{2019}^{\mathrm{4}} =\mathrm{16}\:\mathrm{616}\:\mathrm{719}\:\mathrm{002}\:\mathrm{321} \\ $$$$\mathrm{2019}^{\mathrm{7}} =\mathrm{136}\:\mathrm{758}\:\mathrm{469}\:\mathrm{738}\:\mathrm{531}\:\mathrm{852}\:\mathrm{205}\:\mathrm{739} \\ $$$${m}=\mathrm{4}\wedge{n}=\mathrm{7} \\ $$$${m}+{n}=\mathrm{11} \\ $$

Commented by MJS last updated on 12/Dec/19

if it should be non−zero  m=20∧n=36  m+n=56

$$\mathrm{if}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{non}−\mathrm{zero} \\ $$$${m}=\mathrm{20}\wedge{n}=\mathrm{36} \\ $$$${m}+{n}=\mathrm{56} \\ $$

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