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The-sum-of-four-consecutive-2-digit-numbers-when-divided-by-10-becomes-a-perfect-square-Which-of-the-following-can-possibly-be-one-of-these-four-numbers-a-21-b-25-c-41-d-67-e-73-please-show-work




Question Number 19009 by chux last updated on 03/Aug/17
The sum of four consecutive  2−digit numbers when divided by  10 becomes a perfect square.Which  of the following can possibly be  one of these four numbers?  (a)21(b)25(c)41(d)67(e)73          please show workings
Thesumoffourconsecutive2digitnumberswhendividedby10becomesaperfectsquare.Whichofthefollowingcanpossiblybeoneofthesefournumbers?(a)21(b)25(c)41(d)67(e)73pleaseshowworkings
Answered by ajfour last updated on 03/Aug/17
sum=10n^2   a=((5n^2 )/2)−(3/2)    ,   b=((5n^2 )/2)−(1/2)  c=((5n^2 )/2)+(1/2)    ,   d=((5n^2 )/2)+(3/2)  For n=3 ,  a=21 , so (a).
sum=10n2a=5n2232,b=5n2212c=5n22+12,d=5n22+32Forn=3,a=21,so(a).
Commented by chux last updated on 03/Aug/17
please how did you come about  the 10n^2  and other representation
pleasehowdidyoucomeaboutthe10n2andotherrepresentation
Commented by ajfour last updated on 03/Aug/17
((sum)/(10))=n^2   (the perfect square)
sum10=n2(theperfectsquare)
Commented by chux last updated on 03/Aug/17
ok..... its clear.
ok..itsclear.
Commented by chernoaguero@gmail.com last updated on 03/Aug/17
sir adfour why is ±(3/2) and ±(1/2) applicable
siradfourwhyis±32and±12applicable

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