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Question Number 195194 by Abdullahrussell last updated on 26/Jul/23

Answered by Frix last updated on 26/Jul/23

a     b     c     d  2     3    15   10  2     4    12    6  2     6    12    4  2    10   15    3  3     2    10   15  3     3     6     6  3     6     6     3  3    15   10    2  4     2     6    12  4     4     4     4  4    12    6     2  6     2     4    12  6     3     3     6  6     6     3     3  6    12    4     2  10   2     3    15  10  15    3     2  12   4     2     6  12   6     2     4  15   3     2    10  15  10    2     3

abcd231510241262612421015332101533663663315102426124444412626241263366633612421023151015321242612624153210151023

Answered by deleteduser1 last updated on 27/Jul/23

WLOG,let a be the maximum  ac=a+b+c+d≤4a⇒c≤4;c=1⇒b+d=−1(absurd)  c=2⇒b+d=a−2 and 2a=bd  b and d satisfy x^2 −(a−2)x+2a=0  b,d=(((a−2)+_− (√(a^2 −12a+4)))/2)∈Z^+ ⇒a^2 −12a+4=p^2   (a−6)^2 −32=p^2 ⇒(a−6−p)(a−6+p)=32=1×32  =2×16=3×8... (upto permutation,negativefactors)  Z^+  solutions:(a,p)=(12,2),(15,7)  ⇒b,d=((10+_− 2)/2)=6 or 4     OR ((13+_− 7)/2)=10 or 3  c=3⇒b+d=2a−3 and 3a=bd  ⇒b,d=(((2a−3)+_− (√(4a^2 −24a+9)))/2)  ⇒(2a−6)^2 −27=p^2 ⇒(2a−6−p)(2a−6+p)=27  ⇒a=6,p=3⇒b,d=((9+_− 3)/2)  =6  or 3  For c=4⇒b+d=3a−4,bd=4a  ⇒b,d=(((3a−4)+_− (√(9a^2 −40a+16=(3a−((20)/3))^2 −(((16)/3))^2 )))/2),  ⇒(3a−((20)/3)−p)(3a−((20)/3)+p)=((256)/9)⇒a=4,p=0  ⇒b,d=4,4  ⇒(a,b,c,d)=(12,6,2,4),(15,10,2,3),(6,6,3,3),  (4,4,4,4) upto permutations of b and d.

WLOG,letabethemaximumac=a+b+c+d4ac4;c=1b+d=1(absurd)c=2b+d=a2and2a=bdbanddsatisfyx2(a2)x+2a=0b,d=(a2)+a212a+42Z+a212a+4=p2(a6)232=p2(a6p)(a6+p)=32=1×32=2×16=3×8...(uptopermutation,negativefactors)Z+solutions:(a,p)=(12,2),(15,7)b,d=10+22=6or4OR13+72=10or3c=3b+d=2a3and3a=bdb,d=(2a3)+4a224a+92(2a6)227=p2(2a6p)(2a6+p)=27a=6,p=3b,d=9+32=6or3Forc=4b+d=3a4,bd=4ab,d=(3a4)+9a240a+16=(3a203)2(163)22,(3a203p)(3a203+p)=2569a=4,p=0b,d=4,4(a,b,c,d)=(12,6,2,4),(15,10,2,3),(6,6,3,3),(4,4,4,4)uptopermutationsofbandd.

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