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Question Number 84739 by Rio Michael last updated on 15/Mar/20
find the unit digit in the number 15^(1789) + 17^(1789) + 19^(1789)
$$\mathrm{find}\:\mathrm{the}\:\mathrm{unit}\:\mathrm{digit}\:\mathrm{in}\:\mathrm{the}\:\mathrm{number} \\ $$$$\:\mathrm{15}^{\mathrm{1789}} \:+\:\mathrm{17}^{\mathrm{1789}} \:+\:\mathrm{19}^{\mathrm{1789}} \\ $$
Answered by TANMAY PANACEA last updated on 15/Mar/20
last digit of 15^(1789) =5 17^1 →last digit =7 17^2 →last digit=9 17^3 →last digit=3 17^4 →last digit=1 so repetation of 7931 cyclic way 1789/4=447+(1/4) so last digit of 17^(1789) =7 19^(1789) 19^1 →l.d=9 19^2 →l.d=1 cyclic of 9,1 last digit of 19^(1789) 1789/2=894+(1/2) last digit of 19^(1789) is=9 so last digit of 15^(1789) +17^(1789) +19^(1789) is(5+7+9=21 so last digit of given exoression=1
$${last}\:{digit}\:{of}\:\mathrm{15}^{\mathrm{1789}} =\mathrm{5} \\ $$$$\mathrm{17}^{\mathrm{1}} \rightarrow{last}\:{digit}\:=\mathrm{7} \\ $$$$\mathrm{17}^{\mathrm{2}} \:\rightarrow{last}\:{digit}=\mathrm{9} \\ $$$$\mathrm{17}^{\mathrm{3}} \rightarrow{last}\:{digit}=\mathrm{3} \\ $$$$\mathrm{17}^{\mathrm{4}} \rightarrow{last}\:{digit}=\mathrm{1} \\ $$$$\:{so}\:\:{repetation}\:{of}\:\mathrm{7931}\:{cyclic}\:{way} \\ $$$$\mathrm{1789}/\mathrm{4}=\mathrm{447}+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{last}}\:\boldsymbol{{digit}}\:\boldsymbol{{of}}\:\mathrm{17}^{\mathrm{1789}} =\mathrm{7} \\ $$$$\mathrm{19}^{\mathrm{1789}} \\ $$$$\mathrm{19}^{\mathrm{1}} \rightarrow{l}.{d}=\mathrm{9} \\ $$$$\mathrm{19}^{\mathrm{2}} \rightarrow{l}.{d}=\mathrm{1} \\ $$$${cyclic}\:{of}\:\mathrm{9},\mathrm{1} \\ $$$${last}\:{digit}\:{of}\:\mathrm{19}^{\mathrm{1789}} \\ $$$$\mathrm{1789}/\mathrm{2}=\mathrm{894}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${last}\:{digit}\:{of}\:\mathrm{19}^{\mathrm{1789}} \:{is}=\mathrm{9} \\ $$$${so}\: \\ $$$${last}\:{digit}\:{of}\:\mathrm{15}^{\mathrm{1789}} +\mathrm{17}^{\mathrm{1789}} +\mathrm{19}^{\mathrm{1789}} \\ $$$${is}\left(\mathrm{5}+\mathrm{7}+\mathrm{9}=\mathrm{21}\:\:\right. \\ $$$${so}\:\boldsymbol{{last}}\:\boldsymbol{{digit}}\:\boldsymbol{{of}}\:\boldsymbol{{given}}\:\boldsymbol{{exoression}}=\mathrm{1} \\ $$
Commented by Rio Michael last updated on 15/Mar/20
thanks so much very clear
$$\mathrm{thanks}\:\mathrm{so}\:\mathrm{much}\:\mathrm{very}\:\mathrm{clear} \\ $$
Commented by mr W last updated on 15/Mar/20
what are the last two digits?
$${what}\:{are}\:{the}\:{last}\:\boldsymbol{{two}}\:{digits}? \\ $$
Commented by Rio Michael last updated on 15/Mar/20
from the calculation and my understanding it will be the sum of the last two digits of all three.
$$\mathrm{from}\:\mathrm{the}\:\mathrm{calculation}\:\mathrm{and}\:\mathrm{my}\:\mathrm{understanding} \\ $$$$\mathrm{it}\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{last}\:\mathrm{two}\:\mathrm{digits}\:\mathrm{of} \\ $$$$\mathrm{all}\:\mathrm{three}.\: \\ $$
Commented by mr W last updated on 15/Mar/20
that′s clear. but what are they?
$${that}'{s}\:{clear}.\:{but}\:{what}\:{are}\:{they}? \\ $$
Commented by Rio Michael last updated on 15/Mar/20
the last two digits of the number 1235432674 is 7 and 4 respectively this can be gotten when this number is divided by 100 and the remainder noted.
$$\mathrm{the}\:\mathrm{last}\:\mathrm{two}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{the}\:\mathrm{number}\:\mathrm{1235432674}\:\mathrm{is}\:\mathrm{7}\:\mathrm{and}\:\mathrm{4}\:\mathrm{respectively} \\ $$$$\mathrm{this}\:\mathrm{can}\:\mathrm{be}\:\mathrm{gotten}\:\mathrm{when}\:\mathrm{this}\:\mathrm{number}\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{100}\:\mathrm{and}\:\mathrm{the}\: \\ $$$$\mathrm{remainder}\:\mathrm{noted}. \\ $$$$\: \\ $$
Commented by mr W last updated on 15/Mar/20
that′s also clear. but can you get the last two digits from 15^(1789) and 17^(1789) and 19^(1789) respectively?
$${that}'{s}\:{also}\:{clear}.\:{but}\:{can}\:{you}\:{get}\:{the} \\ $$$${last}\:{two}\:{digits}\:{from}\:\mathrm{15}^{\mathrm{1789}} \:\:{and}\:\mathrm{17}^{\mathrm{1789}} \\ $$$${and}\:\mathrm{19}^{\mathrm{1789}} \:{respectively}? \\ $$
Commented by Rio Michael last updated on 15/Mar/20
that will be difficult to do
$$\mathrm{that}\:\mathrm{will}\:\mathrm{be}\:\mathrm{difficult}\:\mathrm{to}\:\mathrm{do} \\ $$
Commented by mr W last updated on 15/Mar/20
i got last two digits from 15^(1789) are 75 last two digits from 17^(1789) are 97 last two digits from 19^(1789) are 79 ⇒last two digits from 15^(1789) +17^(1789) +19^(1789) are 51.
$${i}\:{got} \\ $$$${last}\:{two}\:{digits}\:{from}\:\mathrm{15}^{\mathrm{1789}} \:{are}\:\mathrm{75} \\ $$$${last}\:{two}\:{digits}\:{from}\:\mathrm{17}^{\mathrm{1789}} \:{are}\:\mathrm{97} \\ $$$${last}\:{two}\:{digits}\:{from}\:\mathrm{19}^{\mathrm{1789}} \:{are}\:\mathrm{79} \\ $$$$\Rightarrow{last}\:{two}\:{digits}\:{from}\:\mathrm{15}^{\mathrm{1789}} +\mathrm{17}^{\mathrm{1789}} +\mathrm{19}^{\mathrm{1789}} \\ $$$${are}\:\mathrm{51}. \\ $$
Commented by Rio Michael last updated on 15/Mar/20
wow sir thats great work
$$\mathrm{wow}\:\mathrm{sir}\:\mathrm{thats}\:\mathrm{great}\:\mathrm{work} \\ $$
Commented by mr W last updated on 16/Mar/20
see Q84782
$${see}\:{Q}\mathrm{84782} \\ $$

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