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Question Number 23537 by math solver last updated on 01/Nov/17

Commented by math solver last updated on 01/Nov/17

q.7 ?

$$\mathrm{q}.\mathrm{7}\:? \\ $$

Commented by ajfour last updated on 01/Nov/17

yes 24 , a=2, h=2  i have nearly solved, but some  proof is still wanting..

$${yes}\:\mathrm{24}\:,\:{a}=\mathrm{2},\:{h}=\mathrm{2} \\ $$$${i}\:{have}\:{nearly}\:{solved},\:{but}\:{some} \\ $$$${proof}\:{is}\:{still}\:{wanting}.. \\ $$

Commented by math solver last updated on 01/Nov/17

sir it is not even in options !

$$\mathrm{sir}\:\mathrm{it}\:\mathrm{is}\:\mathrm{not}\:\mathrm{even}\:\mathrm{in}\:\mathrm{options}\:! \\ $$

Commented by math solver last updated on 01/Nov/17

the answer is 24.  but i don′t know how ?

$$\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{24}. \\ $$$$\mathrm{but}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{how}\:? \\ $$

Commented by math solver last updated on 01/Nov/17

did you get the required proof?

$$\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\mathrm{the}\:\mathrm{required}\:\mathrm{proof}? \\ $$

Commented by Rasheed.Sindhi last updated on 02/Nov/17

2a^2 +4ah=8a+4h  a^2 +2ah−4a=2h  a(a+2h−4)=2h  ((2h)/a)=a+2h−4  ((2h)/a)∈Z  a∣2   or   a∣h  a∣2⇒a=1,2  For a=1       2a^2 +4ah=8a+4h       ⇒2+4h=8+4h (impossible)       ⇒a=2      2a^2 +4ah=8a+4h        8+8h=16+4h⇒h=2  One solution is a=h=2  Hence sum of edges=24  For a∣h     Let h=ka for some integer k.      2a^2 +4ah=8a+4h      ⇒2a^2 +4a.ka=8a+4.ka      ⇒2a^2 +4ka^2 =8a+4ka      ⇒a+2ka=4+2k      ⇒2k(a−1)=4−a         ((4−a)/(2(a−1))) =k    k=1⇒4−a=2a−2⇒a=2    Again a=h=2 and sum of edges=24    k=2⇒4−a=4(a−1)⇒a=(8/5) ∉N     Continue

$$\mathrm{2a}^{\mathrm{2}} +\mathrm{4ah}=\mathrm{8a}+\mathrm{4h} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{2ah}−\mathrm{4a}=\mathrm{2h} \\ $$$$\mathrm{a}\left(\mathrm{a}+\mathrm{2h}−\mathrm{4}\right)=\mathrm{2h} \\ $$$$\frac{\mathrm{2h}}{\mathrm{a}}=\mathrm{a}+\mathrm{2h}−\mathrm{4} \\ $$$$\frac{\mathrm{2h}}{\mathrm{a}}\in\mathbb{Z} \\ $$$$\mathrm{a}\mid\mathrm{2}\:\:\:\mathrm{or}\:\:\:\mathrm{a}\mid\mathrm{h} \\ $$$$\mathrm{a}\mid\mathrm{2}\Rightarrow\mathrm{a}=\mathrm{1},\mathrm{2} \\ $$$$\mathrm{For}\:\mathrm{a}=\mathrm{1}\: \\ $$$$\:\:\:\:\mathrm{2a}^{\mathrm{2}} +\mathrm{4ah}=\mathrm{8a}+\mathrm{4h} \\ $$$$\:\:\:\:\:\Rightarrow\mathrm{2}+\mathrm{4h}=\mathrm{8}+\mathrm{4h}\:\left({impossible}\right) \\ $$$$\:\:\:\:\:\Rightarrow\mathrm{a}=\mathrm{2} \\ $$$$\:\:\:\:\mathrm{2a}^{\mathrm{2}} +\mathrm{4ah}=\mathrm{8a}+\mathrm{4h} \\ $$$$\:\:\:\:\:\:\mathrm{8}+\mathrm{8h}=\mathrm{16}+\mathrm{4h}\Rightarrow\mathrm{h}=\mathrm{2} \\ $$$$\mathrm{One}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{a}=\mathrm{h}=\mathrm{2} \\ $$$$\mathrm{Hence}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{edges}=\mathrm{24} \\ $$$$\mathrm{For}\:\mathrm{a}\mid\mathrm{h} \\ $$$$\:\:\:\mathrm{Let}\:\mathrm{h}=\mathrm{ka}\:\mathrm{for}\:\mathrm{some}\:\mathrm{integer}\:\mathrm{k}. \\ $$$$\:\:\:\:\mathrm{2a}^{\mathrm{2}} +\mathrm{4ah}=\mathrm{8a}+\mathrm{4h} \\ $$$$\:\:\:\:\Rightarrow\mathrm{2a}^{\mathrm{2}} +\mathrm{4a}.\mathrm{ka}=\mathrm{8a}+\mathrm{4}.\mathrm{ka} \\ $$$$\:\:\:\:\Rightarrow\mathrm{2a}^{\mathrm{2}} +\mathrm{4ka}^{\mathrm{2}} =\mathrm{8a}+\mathrm{4ka} \\ $$$$\:\:\:\:\Rightarrow\mathrm{a}+\mathrm{2ka}=\mathrm{4}+\mathrm{2k} \\ $$$$\:\:\:\:\Rightarrow\mathrm{2k}\left(\mathrm{a}−\mathrm{1}\right)=\mathrm{4}−\mathrm{a} \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{4}−\mathrm{a}}{\mathrm{2}\left(\mathrm{a}−\mathrm{1}\right)}\:=\mathrm{k} \\ $$$$\:\:\mathrm{k}=\mathrm{1}\Rightarrow\mathrm{4}−\mathrm{a}=\mathrm{2a}−\mathrm{2}\Rightarrow\mathrm{a}=\mathrm{2} \\ $$$$\:\:\mathrm{Again}\:\mathrm{a}=\mathrm{h}=\mathrm{2}\:\mathrm{and}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{edges}=\mathrm{24} \\ $$$$\:\:\mathrm{k}=\mathrm{2}\Rightarrow\mathrm{4}−\mathrm{a}=\mathrm{4}\left(\mathrm{a}−\mathrm{1}\right)\Rightarrow\mathrm{a}=\frac{\mathrm{8}}{\mathrm{5}}\:\notin\mathbb{N}\:\:\: \\ $$$$\mathrm{Continue} \\ $$

Commented by ajfour last updated on 01/Nov/17

yes, just now.

$${yes},\:{just}\:{now}. \\ $$

Commented by math solver last updated on 01/Nov/17

from here how you get a=h=2

$$\mathrm{from}\:\mathrm{here}\:\mathrm{how}\:\mathrm{you}\:\mathrm{get}\:\mathrm{a}=\mathrm{h}=\mathrm{2} \\ $$$$ \\ $$

Answered by ajfour last updated on 01/Nov/17

let the square base of cuboid be  of edge length x. Let the height  be y. Then,  sum of the length  of edges = 4x+4y+4x  and area of all the faces is   =x^2 +4xy+x^2   Given they are numerixally  equal ⇒         2x^2 +4xy = 8x+4y  or   x^2 +2xy = 4x+2y         x^2 −4x+4 − 2y+2xy= 4        (x−2)^2  +2y(x−1)=4  ⇒ (x−2)^2 +2y(x−2)+y^2 =4−2y+y^2    (x−2+y)^2 = (y−1)^2 +3  ⇒  (x−2+y)^2 −(y−1)^2 =3  (x−2+y+y−1)(x−2+y−y+1)=3  (x+2y−3)(x−1)=3×1 =1×3  As x, y are positive integers,  ⇒ both       x−1=1  and               x+2y−3=3   So         x=2  and   y=2      Sum of length of edges =24 .  otherwise if   x−1=3 ⇒ x=4  and  x+2y−3 = 1  with x=4  yields  y=0 ; this is true even  but not a proper cuboid.

$${let}\:{the}\:{square}\:{base}\:{of}\:{cuboid}\:{be} \\ $$$${of}\:{edge}\:{length}\:\boldsymbol{{x}}.\:{Let}\:{the}\:{height} \\ $$$${be}\:\boldsymbol{{y}}.\:{Then},\:\:{sum}\:{of}\:{the}\:{length} \\ $$$${of}\:{edges}\:=\:\mathrm{4}{x}+\mathrm{4}{y}+\mathrm{4}{x} \\ $$$${and}\:{area}\:{of}\:{all}\:{the}\:{faces}\:{is} \\ $$$$\:={x}^{\mathrm{2}} +\mathrm{4}{xy}+{x}^{\mathrm{2}} \\ $$$${Given}\:{they}\:{are}\:{numerixally} \\ $$$${equal}\:\Rightarrow \\ $$$$\:\:\:\:\:\:\:\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{xy}\:=\:\mathrm{8}{x}+\mathrm{4}{y} \\ $$$${or}\:\:\:{x}^{\mathrm{2}} +\mathrm{2}{xy}\:=\:\mathrm{4}{x}+\mathrm{2}{y} \\ $$$$\:\:\:\:\:\:\:{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}\:−\:\mathrm{2}{y}+\mathrm{2}{xy}=\:\mathrm{4} \\ $$$$\:\:\:\:\:\:\left({x}−\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{2}{y}\left({x}−\mathrm{1}\right)=\mathrm{4} \\ $$$$\Rightarrow\:\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}{y}\left({x}−\mathrm{2}\right)+{y}^{\mathrm{2}} =\mathrm{4}−\mathrm{2}{y}+{y}^{\mathrm{2}} \\ $$$$\:\left({x}−\mathrm{2}+{y}\right)^{\mathrm{2}} =\:\left({y}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3} \\ $$$$\Rightarrow\:\:\left({x}−\mathrm{2}+{y}\right)^{\mathrm{2}} −\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{3} \\ $$$$\left({x}−\mathrm{2}+{y}+{y}−\mathrm{1}\right)\left({x}−\mathrm{2}+{y}−{y}+\mathrm{1}\right)=\mathrm{3} \\ $$$$\left({x}+\mathrm{2}{y}−\mathrm{3}\right)\left({x}−\mathrm{1}\right)=\mathrm{3}×\mathrm{1}\:=\mathrm{1}×\mathrm{3} \\ $$$${As}\:{x},\:{y}\:{are}\:{positive}\:{integers}, \\ $$$$\Rightarrow\:{both}\:\:\:\:\:\:\:{x}−\mathrm{1}=\mathrm{1} \\ $$$${and}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}+\mathrm{2}{y}−\mathrm{3}=\mathrm{3} \\ $$$$\:{So}\:\:\:\:\:\:\:\:\:{x}=\mathrm{2}\:\:{and}\:\:\:{y}=\mathrm{2}\: \\ $$$$\:\:\:{Sum}\:{of}\:{length}\:{of}\:{edges}\:=\mathrm{24}\:. \\ $$$$\boldsymbol{{otherwise}}\:{if} \\ $$$$\:{x}−\mathrm{1}=\mathrm{3}\:\Rightarrow\:{x}=\mathrm{4} \\ $$$${and}\:\:{x}+\mathrm{2}{y}−\mathrm{3}\:=\:\mathrm{1}\:\:{with}\:{x}=\mathrm{4} \\ $$$${yields}\:\:{y}=\mathrm{0}\:;\:{this}\:{is}\:{true}\:{even} \\ $$$${but}\:{not}\:{a}\:{proper}\:{cuboid}. \\ $$

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