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Celestine Preetham's Avatar
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2 May 2008 12:09:09 IST
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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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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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pramod's Avatar

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5 Jun 2008 23:30:49 IST
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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.....


 

pramod's Avatar

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6 Jun 2008 00:45:11 IST
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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.....

Haresh's Avatar

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6 Jun 2008 05:37:17 IST
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Is the answer 28.....If correct i will give u the answer
IITAIEEE's Avatar

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29 Jan 2009 14:25:47 IST
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Kya khicri paka raha hai, please tell the correct if u are no baby.
IITAIEEE's Avatar

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29 Jan 2009 14:26:36 IST
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kya khicri hai
Celestine Preetham's Avatar

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29 Jan 2009 14:40:51 IST
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yes 28 is the ans pramod is right


Scorching goIITian

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29 Jan 2009 15:27:36 IST
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 Since a+11b is divisible by 13, so a-2b is also divisible by 13, hence 6a-12b is also divisible by 13

This implies that 6a+b is divisible by 13. ........(1)

Similarly, a+13b is div. by 11 implies that a+2b is div. by 11 which means that 6a+12b is div. by 11

Hence, 6a+b is divisible by 11. .............(2)

Now, (11,13) = 1

Thus, from (1) and (2), we have that 11*13 = 143 divides 6a+b.

So, 6a+b = 143k for some natural no. k.

Thus, 6a+6b = 143k+5b=144k+6b-(k+b)

This shows that 6 divides k+b and hence, k+b >=6 (both are nat. nos.)

So, 6a+6b = 143k + 5b = 138k+ 5(k+b) >=(138*1) + (5*6) = 168

thus, a+b>=168/6 =28

Hence, minimum value of (a+b) = 28

 

 


Scorching goIITian

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31 Jan 2009 14:36:11 IST
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Guys, I think I deserve some reps for this solution, don't I?
sarang's Avatar

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31 Jan 2009 20:39:28 IST
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24

Scorching goIITian

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31 Jan 2009 21:42:34 IST
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Hey sarang, the answer has been found. It is 28. See the solutions above. It is a question from RMO (probably 2004).

Cool goIITian

Joined: 24 Jan 2009
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1 Feb 2009 22:48:23 IST
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is it 28 when a=23 and b = 5


Cool goIITian

Joined: 24 Jan 2009
Posts: 59
1 Feb 2009 22:49:28 IST
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is it 28 when a=23 and b=5




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