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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: find min a + b such that a+11b and a+13b are multiples of 13 and 11 resp. a,b>0
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celestine (76)

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sorry kirpal
ur reasoning that b min is 10 is wrong

IM NO BABY
I HAVE THE SEVENTH SENSE
NONSENSE!!!!!!!!!!!
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kirpal_06391 (0)

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Acc. to u reasoning that b min is 10 is wrong. Could u pls. explain me how.

Kirpal Singh
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celestine (76)

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well bmin is not 10


It is actually ....................... Youwant me to tell u ???


ne way   Kirpal


could u please explain in detail how u r telling bmin is 10 . In that process u may find ur error


IM NO BABY
I HAVE THE SEVENTH SENSE
NONSENSE!!!!!!!!!!!
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hsbhatt (5020)

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My attempt:




 


a+11b = 13  




 


a+13b = 11




 


Eliminating b we get 2a = 169  - 121  




 


Also 2b = 11 - 13




 


Hence 2(a+b) = 156 - 110


1 and 2, and  11 - 13   0 and  and  are of the same parity and hence the least difference is 2


give up!


Time wounds all heels
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kislay (1118)

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 no value of a+B exists as such.....


2a = 169  - 121  .....................................1


2b = 11 - 13 ..................................................2


a+b = 78 - 55 .................................................3


since a and b are natural no.s(as mentioned by author of this ques)...from eqn 1


    169  121 


=>  (121/169) 


=> (1.18) .....................................4


 


also from eqn.2


11  13 


 


=>  (13/11) 


=>  (1.39) ..................................5


 


thus from eqn (4) and (5) no such value of a and b exists..


 


B.Tech CSE, ISMU
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pramod6990 (955)

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my try.....


a + 11b=13m


a + 13b=11n


solving we have a= (169m-121n)/2


and b=(11n-13m)/2


and a+b =(156m-110n)/2...........1)


now as a>0 and b>0


we have.....   11/13> m/n>121/169


or 0.85 > m/n > 0.72


we take a favourable case of 0.75to get m and n as close to each other as possible so as to get the minimum value of the expression of 1)


due to which m/n=3/4.....m=3c and n=4c.......for integral values of a+b , c shud be even and 2 being the min evn number we have m=6 and n=8


using this we have min(a+b) as (936-880)/2 = 56/2 = 28....


so the minimum value of the expression (a+b) = 28.....


 


"Logic is the systematic way of reaching the wrong conclusion with confidence" lol.....
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pramod6990 (955)

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justification for using 0.75 as m/n ratio....


to produce min value of a+b......(153m-110n)/2 shud be minimum rite....nd 0.72<m/n<0.85......tell me the 2 evn integers such that they are in the ratio that differs frm 0.72 by the minimum and also the difference btween the two is minimum....=2 the minimum difference between 2 evn numbers(unequal)....u will say numbers like 6 and 8......(ratio 0.75) 10 and 8( ratio 0.8 but not closer to 0.72 than 0.75) u may say 10 and 12.....( ratio =0.833 but not as close to 0.72 as 0.75) in the same way u cud take a number of couples.....but u will find that the minimum difference btween 0.72 and the m/n ratio will only be achieved in the case of 6 and 8....if u go one step down taking 6 and 4...ull land up in truble as ratio is 0.66 which is out of the eqn...


so clearly the two integers m and n are 6 and 8 inserting which we get the min value of (a+b) as 28...


tell me whether its rite or wrong.....


"Logic is the systematic way of reaching the wrong conclusion with confidence" lol.....
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hpudipeddi (77)

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Is the answer 28.....If correct i will give u the answer

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