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Question Number 121732 by ZiYangLee last updated on 11/Nov/20

Given that a,b,c,d,e and f are positive  integers where a<b<c<d<e<f and  a+b+c+d+e+f=100, find the  largest value of e.  A.46   B.44    C.43    D.40   E.45

$$\mathrm{Given}\:\mathrm{that}\:{a},{b},{c},{d},{e}\:\mathrm{and}\:{f}\:\mathrm{are}\:\mathrm{positive} \\ $$ $$\mathrm{integers}\:\mathrm{where}\:{a}<{b}<{c}<{d}<{e}<{f}\:\mathrm{and} \\ $$ $${a}+{b}+{c}+{d}+{e}+{f}=\mathrm{100},\:\mathrm{find}\:\mathrm{the} \\ $$ $$\mathrm{largest}\:\mathrm{value}\:\mathrm{of}\:{e}. \\ $$ $${A}.\mathrm{46}\:\:\:{B}.\mathrm{44}\:\:\:\:{C}.\mathrm{43}\:\:\:\:{D}.\mathrm{40}\:\:\:{E}.\mathrm{45} \\ $$

Answered by bemath last updated on 11/Nov/20

e_(max)  when the value of a,b,c and   d must be minimum  ⇒ 1+2+3+4+e+f = 100   ⇒e + f = 90 , since e < f , it  follows that e_(max)  = 44 ∧ f_(min)  = 46.

$${e}_{{max}} \:{when}\:{the}\:{value}\:{of}\:{a},{b},{c}\:{and}\: \\ $$ $${d}\:{must}\:{be}\:{minimum} \\ $$ $$\Rightarrow\:\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+{e}+{f}\:=\:\mathrm{100}\: \\ $$ $$\Rightarrow{e}\:+\:{f}\:=\:\mathrm{90}\:,\:{since}\:{e}\:<\:{f}\:,\:{it} \\ $$ $${follows}\:{that}\:{e}_{{max}} \:=\:\mathrm{44}\:\wedge\:{f}_{{min}} \:=\:\mathrm{46}. \\ $$

Commented byZiYangLee last updated on 11/Nov/20

Yes thanks!

$$\mathrm{Yes}\:\mathrm{thanks}! \\ $$

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