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Algebra
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zorro
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Joined: 28 Apr 2008
Posts: 64
2 May 2008 12:30:13 IST
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Is the answer 23?
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2 May 2008 16:16:54 IST
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Given : a+11b=k(13) - (i)
a+13b=k'(11) - (ii)
and, (a,b) belongs to Natural no.'s
Proof : (a+11b)+2b=k'(11)
b=[k'(11)-k(13)]/2
Therefore, b(min.)=10 [because, b belongs to Natural no.)
Now, for a(min.)=46 eq. (i) & (ii) are satisfied
Hence, a+b(min.)=46+10=56
a+13b=k'(11) - (ii)
and, (a,b) belongs to Natural no.'s
Proof : (a+11b)+2b=k'(11)
b=[k'(11)-k(13)]/2
Therefore, b(min.)=10 [because, b belongs to Natural no.)
Now, for a(min.)=46 eq. (i) & (ii) are satisfied
Hence, a+b(min.)=46+10=56
5 Jun 2008 21:55:55 IST
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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..













