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![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 Jan 2007 13:48:57 IST
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Let a be the 81-didit number all digits of which are equal to 1. Then the number a is 1) divisible by 9 but not divisible by 27 2) divisible by 27 but not divisible by 81 3) divisible by 81 but not divisible by 243 4) divisible by 243
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You never know what is enough till you know what is more than enough.
Titun |
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11 Jan 2007 14:11:00 IST
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If There Will Be No Repurcussions :
My Answer Will Be (C)
I Doubt It Is Correct....But I Applied All The Concepts I Knew...
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11 Jan 2007 16:10:35 IST
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 The correct answer is the option(3).As 9,27,81,243 are some powers of 3 Therefore for a no. to be divisible by them the sum of the digits should be divisible by them.
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11 Jan 2007 16:28:47 IST
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Asmita, I don't agree with your solution. Because, 54, 108 are divisible by 27 but the sum of the digits in either case is not divisible by 27. So, can u provide another solution?
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You never know what is enough till you know what is more than enough.
Titun |
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11 Jan 2007 16:40:30 IST
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a = 111111...............81 times a = (1/9)9999999.............81 tmes a = (10^81 - 1)/ 9 a = (10^81 -1)/ (10 -1) a is the sum of G.P whose common ratio is 10 consisting of 81 terms hence a is divisible by 9 the correct option is a) i may be wrong
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11 Jan 2007 17:01:01 IST
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I completely agre with Rahul and would like to give the same solution as given by him.
i.e. a is divisible by 9 and not by 27
the CORRECT option hence is (a)
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11 Jan 2007 17:09:13 IST
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it is obviously divisible by 81 as the no. consists of 9 sets of nine 1s which is divisible by nine.it yields(division)123456790 this repeated 9 times sum is also divisible by 9.
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11 Jan 2007 20:22:34 IST
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Here we can write the no. 'a' as follows
a = 111111...............81 times a = (1/9)9999999.............81 times a = (10^81 - 1)/ 9 a = (10^81 -1)/ 9
a = (1027x3 - 13)/9
a = (1027 - 1)(1054 + 1027 + 1)/9
a = (109 - 1)(1018 + 109 + 1)(1054 + 1027 + 1) / 9
a = 9 * 111111111*(1018 + 109 + 1)(1054 + 1027 + 1) / 9
a = 111111111*(1018 + 109 + 1)(1054 + 1027 + 1) .......(1)
now any no. is divisible by 3 if sum of the digits is 3 or integral multiple of 3.
Further for divisibility by 9 the sum of digits should be 9.
Thus, 111111111 = 9 * m = 32(m) ........(2)
1018 + 109 + 1 = 3*n ( However, this no.is not divisible by 9) .....(3)
1054 + 1027 + 1 = 3*p ( However, this no.is not divisible by 9) .....(4)
Where m,n and p are +ve integers not divisible by 3
Combining (1), (2), (3) and (4) we obtain
a = 34 (mnp) = 81mnp
so a is divisible by 81 but not by 243 = 35
so option (3) is correct
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11 Jan 2007 20:38:09 IST
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I just mean to say that just calculate the sum of the digits upto a 2 digit no. if the no. by which u r dividing is a 2 digit no.This can be generalised for 1 digit, 3 digits & ... no.
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12 Jan 2007 20:20:42 IST
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I am thankful to Edison. The correct option is 3
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You never know what is enough till you know what is more than enough.
Titun |
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12 Jan 2007 20:56:56 IST
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good question as well as good answer too.
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