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If n is a 27digit positive integer, all of whose digits are [#permalink]
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23 Jul 2014, 09:48
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If n is a 27digit positive integer, all of whose digits are the same, which of the following must be true? I. n is divisible by 3 II. n is divisible by 9 III. n is divisible by 27 A. I only B. I and II only C. I and III only D. II and III only E. I, II and III Explanation given: Suppose n = 111,111,111,111,111,111,111,111,111. Then the digits of n sum up to 27 * 1, which is divisible by 9, so n is divisible by both 3 and 9 by the basic divisibility rules given in our Arithmetic book. (Any other 27digit integer with all identical digits is just a multiple of this first one, so we know that all the other values of n will divide by 3 and 9.) We might conclude that the divisibility rule "generalizes", and that any number whose digits sum to a multiple of 27 divides by 27. Be careful, though! This is exact the sort of natural, intuitive thinking that the testwriters like to exploit, so let's see if we can test a number. Suppose n = 111,111,111,111,111,111,111,111,111. Noticing that 111/3 = 37, we can divide n by 3 and obtain n/3 = 37,037,037,037,037,037,037,037,037. The sum of the digits of this number is 90, so n/3 divides by 9. (n/3)/9 = n/27, so n divides by 27. Since any 27digit number with one repeating digit is just a multiple of 111,111,111,111,111,111,111,111,111  i.e., a multiple of a multiple of 27  we conclude that n is divisible by 27. CONFUSION : DO WE HAVE TO WRITE OUT THE WHOLE NUMBER? IS THERE ANOTHER WAY TO SOLVE? ALSO WHY DO WE DIVIDE 111 by 3? AND THEN HOW DO WE GET TO n/3 = 37,037,037,037,037,037,037,037,037?
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Last edited by Bunuel on 23 Jul 2014, 09:55, edited 1 time in total.
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Re: If n is a 27digit positive integer, all of whose digits are [#permalink]
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24 Jul 2014, 00:19
sagnik2422 wrote: If n is a 27digit positive integer, all of whose digits are the same, which of the following must be true? I. n is divisible by 3 II. n is divisible by 9 III. n is divisible by 27 A. I only B. I and II only C. I and III only D. II and III only E. I, II and III Explanation given: Suppose n = 111,111,111,111,111,111,111,111,111. Then the digits of n sum up to 27 * 1, which is divisible by 9, so n is divisible by both 3 and 9 by the basic divisibility rules given in our Arithmetic book. (Any other 27digit integer with all identical digits is just a multiple of this first one, so we know that all the other values of n will divide by 3 and 9.) We might conclude that the divisibility rule "generalizes", and that any number whose digits sum to a multiple of 27 divides by 27. Be careful, though! This is exact the sort of natural, intuitive thinking that the testwriters like to exploit, so let's see if we can test a number. Suppose n = 111,111,111,111,111,111,111,111,111. Noticing that 111/3 = 37, we can divide n by 3 and obtain n/3 = 37,037,037,037,037,037,037,037,037. The sum of the digits of this number is 90, so n/3 divides by 9. (n/3)/9 = n/27, so n divides by 27. Since any 27digit number with one repeating digit is just a multiple of 111,111,111,111,111,111,111,111,111  i.e., a multiple of a multiple of 27  we conclude that n is divisible by 27. CONFUSION : DO WE HAVE TO WRITE OUT THE WHOLE NUMBER? IS THERE ANOTHER WAY TO SOLVE? ALSO WHY DO WE DIVIDE 111 by 3? AND THEN HOW DO WE GET TO n/3 = 37,037,037,037,037,037,037,037,037? Rule for divisibility by 3: The sum of the digits of the number should be multiple of 3 Rule for divisibility by 9: The sum of the nos digits of the number should be divisible by 9 or the number should be divisible by 3 two times. Rules for divisibility by 27: The sum of the digits should a multiple of 27 Consider no 11111111...27 times = The sum 27*1=27> divisbible by 3,9 and 27 consider number to be 222....27 times, then sum of the no. 27*2=54 divisibly by 3,9 and 27 So why so because when you sum the numbers either you can add the digits 27 times or multiply the digit *27.. Note that since 27 is divisble by 27,9 and 3 and thus the sum of the nos will be divisible by all the nos. Ans is E More on this: Refer to Number properties of GMAT CLUB Math Book mathnumbertheory88376.html
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Re: If n is a 27digit positive integer, all of whose digits are [#permalink]
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09 Aug 2014, 09:36
Wounded Tiger,
Do we have divisibility rule for 27 as "The sum of the digits should a multiple of 27"? Say we have 54 whose sum of digits is 9. Is 9 divisible by 27? Can you explain in detail? Thanks



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Re: If n is a 27digit positive integer, all of whose digits are [#permalink]
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09 Aug 2014, 10:31
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WoundedTiger wrote: sagnik2422 wrote: If n is a 27digit positive integer, all of whose digits are the same, which of the following must be true? I. n is divisible by 3 II. n is divisible by 9 III. n is divisible by 27 A. I only B. I and II only C. I and III only D. II and III only E. I, II and III Explanation given: Suppose n = 111,111,111,111,111,111,111,111,111. Then the digits of n sum up to 27 * 1, which is divisible by 9, so n is divisible by both 3 and 9 by the basic divisibility rules given in our Arithmetic book. (Any other 27digit integer with all identical digits is just a multiple of this first one, so we know that all the other values of n will divide by 3 and 9.) We might conclude that the divisibility rule "generalizes", and that any number whose digits sum to a multiple of 27 divides by 27. Be careful, though! This is exact the sort of natural, intuitive thinking that the testwriters like to exploit, so let's see if we can test a number. Suppose n = 111,111,111,111,111,111,111,111,111. Noticing that 111/3 = 37, we can divide n by 3 and obtain n/3 = 37,037,037,037,037,037,037,037,037. The sum of the digits of this number is 90, so n/3 divides by 9. (n/3)/9 = n/27, so n divides by 27. Since any 27digit number with one repeating digit is just a multiple of 111,111,111,111,111,111,111,111,111  i.e., a multiple of a multiple of 27  we conclude that n is divisible by 27. CONFUSION : DO WE HAVE TO WRITE OUT THE WHOLE NUMBER? IS THERE ANOTHER WAY TO SOLVE? ALSO WHY DO WE DIVIDE 111 by 3? AND THEN HOW DO WE GET TO n/3 = 37,037,037,037,037,037,037,037,037? Rule for divisibility by 3: The sum of the digits of the number should be multiple of 3 Rule for divisibility by 9: The sum of the nos digits of the number should be divisible by 9 or the number should be divisible by 3 two times. Rules for divisibility by 27: The sum of the digits should a multiple of 27 Consider no 11111111...27 times = The sum 27*1=27> divisbible by 3,9 and 27 consider number to be 222....27 times, then sum of the no. 27*2=54 divisibly by 3,9 and 27 So why so because when you sum the numbers either you can add the digits 27 times or multiply the digit *27.. Note that since 27 is divisble by 27,9 and 3 and thus the sum of the nos will be divisible by all the nos. Ans is E More on this: Refer to Number properties of GMAT CLUB Math Book mathnumbertheory88376.htmlDefinitely not right criteria for divisibility by 27.
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Re: If n is a 27digit positive integer, all of whose digits are [#permalink]
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09 Aug 2014, 10:34
sagnik2422 wrote: If n is a 27digit positive integer, all of whose digits are the same, which of the following must be true? I. n is divisible by 3 II. n is divisible by 9 III. n is divisible by 27 A. I only B. I and II only C. I and III only D. II and III only E. I, II and III Explanation given: Suppose n = 111,111,111,111,111,111,111,111,111. Then the digits of n sum up to 27 * 1, which is divisible by 9, so n is divisible by both 3 and 9 by the basic divisibility rules given in our Arithmetic book. (Any other 27digit integer with all identical digits is just a multiple of this first one, so we know that all the other values of n will divide by 3 and 9.) We might conclude that the divisibility rule "generalizes", and that any number whose digits sum to a multiple of 27 divides by 27. Be careful, though! This is exact the sort of natural, intuitive thinking that the testwriters like to exploit, so let's see if we can test a number. Suppose n = 111,111,111,111,111,111,111,111,111. Noticing that 111/3 = 37, we can divide n by 3 and obtain n/3 = 37,037,037,037,037,037,037,037,037. The sum of the digits of this number is 90, so n/3 divides by 9. (n/3)/9 = n/27, so n divides by 27. Since any 27digit number with one repeating digit is just a multiple of 111,111,111,111,111,111,111,111,111  i.e., a multiple of a multiple of 27  we conclude that n is divisible by 27. CONFUSION : DO WE HAVE TO WRITE OUT THE WHOLE NUMBER? IS THERE ANOTHER WAY TO SOLVE? ALSO WHY DO WE DIVIDE 111 by 3? AND THEN HOW DO WE GET TO n/3 = 37,037,037,037,037,037,037,037,037? I think the best solution is provided in spoiler.
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Re: If n is a 27digit positive integer, all of whose digits are [#permalink]
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10 Aug 2014, 21:19
sri30kanth wrote: Wounded Tiger,
Do we have divisibility rule for 27 as "The sum of the digits should a multiple of 27"? Say we have 54 whose sum of digits is 9. Is 9 divisible by 27? Can you explain in detail? Thanks Hmmm....I agree with smyarga... Look in general, we know a number is divisible by 3 if the sum of the digits of the nos is divisble by 3 For a number to divisible by 9, either the sum of the nos should be divisble by 9 or the number should be divisble by 3 two times Similarly, for a no to be divisible by 27, it should be divisble by 3 three times... There is no other method... In my approach initially, I just extended the rule for 3 and 9 to 27 and as it can be seen it does not work out...Just went with flow I guess... Like smyarga said, Best approach is the one given in the spoiler...
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Re: If n is a 27digit positive integer, all of whose digits are [#permalink]
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11 Aug 2014, 01:42
3 & 9 are multiples of 27; so divisibility applicable for 27, 3 & 9 would inherit from it Lets take n = 2 (Neither divisible by 3, nor 6 nor 27) Addition = 2*27 = 54 which is divisible by both 3,9 & 27 Answer = E
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Re: If n is a 27digit positive integer, all of whose digits are [#permalink]
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11 Aug 2014, 01:46
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sagnik2422 wrote: If n is a 27digit positive integer, all of whose digits are the same, which of the following must be true? I. n is divisible by 3 II. n is divisible by 9 III. n is divisible by 27 A. I only B. I and II only C. I and III only D. II and III only E. I, II and III Explanation given: Suppose n = 111,111,111,111,111,111,111,111,111. Then the digits of n sum up to 27 * 1, which is divisible by 9, so n is divisible by both 3 and 9 by the basic divisibility rules given in our Arithmetic book. (Any other 27digit integer with all identical digits is just a multiple of this first one, so we know that all the other values of n will divide by 3 and 9.) We might conclude that the divisibility rule "generalizes", and that any number whose digits sum to a multiple of 27 divides by 27. Be careful, though! This is exact the sort of natural, intuitive thinking that the testwriters like to exploit, so let's see if we can test a number. Suppose n = 111,111,111,111,111,111,111,111,111. Noticing that 111/3 = 37, we can divide n by 3 and obtain n/3 = 37,037,037,037,037,037,037,037,037. The sum of the digits of this number is 90, so n/3 divides by 9. (n/3)/9 = n/27, so n divides by 27. Since any 27digit number with one repeating digit is just a multiple of 111,111,111,111,111,111,111,111,111  i.e., a multiple of a multiple of 27  we conclude that n is divisible by 27. CONFUSION : DO WE HAVE TO WRITE OUT THE WHOLE NUMBER? IS THERE ANOTHER WAY TO SOLVE? ALSO WHY DO WE DIVIDE 111 by 3? AND THEN HOW DO WE GET TO n/3 = 37,037,037,037,037,037,037,037,037? Its not necessary to write the whole number. Though I have taken n=2 in my earlier post, its not necessary as well. We can also solve it by variable nnn...........n ......... 27 times Addition of digits = n+n+n+ ........... 27 times = 27n 27n is divisible by 27; so obviously by 3 & 9 Answer =E
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Re: If n is a 27digit positive integer, all of whose digits are [#permalink]
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If n is a 27digit positive integer, all of whose digits are [#permalink]
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12 Dec 2016, 04:16
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sagnik2422 wrote: If n is a 27digit positive integer, all of whose digits are the same, which of the following must be true? I. n is divisible by 3 II. n is divisible by 9 III. n is divisible by 27 A. I only B. I and II only C. I and III only D. II and III only E. I, II and III Explanation given: Suppose n = 111,111,111,111,111,111,111,111,111. Then the digits of n sum up to 27 * 1, which is divisible by 9, so n is divisible by both 3 and 9 by the basic divisibility rules given in our Arithmetic book. (Any other 27digit integer with all identical digits is just a multiple of this first one, so we know that all the other values of n will divide by 3 and 9.) We might conclude that the divisibility rule "generalizes", and that any number whose digits sum to a multiple of 27 divides by 27. Be careful, though! This is exact the sort of natural, intuitive thinking that the testwriters like to exploit, so let's see if we can test a number. Suppose n = 111,111,111,111,111,111,111,111,111. Noticing that 111/3 = 37, we can divide n by 3 and obtain n/3 = 37,037,037,037,037,037,037,037,037. The sum of the digits of this number is 90, so n/3 divides by 9. (n/3)/9 = n/27, so n divides by 27. Since any 27digit number with one repeating digit is just a multiple of 111,111,111,111,111,111,111,111,111  i.e., a multiple of a multiple of 27  we conclude that n is divisible by 27. CONFUSION : DO WE HAVE TO WRITE OUT THE WHOLE NUMBER? IS THERE ANOTHER WAY TO SOLVE? ALSO WHY DO WE DIVIDE 111 by 3? AND THEN HOW DO WE GET TO n/3 = 37,037,037,037,037,037,037,037,037? \(N=aaaaaaaaaaaaa…\). (27 times) \(a+a+a+a+a+ … (27 times) = 27*a\) Number is definitely divisible by 3 and 9. I and II are correct. With 27 things are a bit trickier. \(27*37=999\) which is \(10^3 – 1\) When we need to find remainder of a number in a form \(10^x – 1\) we need to take clusters of digits taken by \(x\) starting from the right and sum them up. In our case \(aaaaaa… 27\) times we split the number into the groups of \(aaa\) starting from the right. We’ll get: \(\frac{27}{3} *(aaa) = 9*aaa\) Because we have particular case where all digits are the same, our group of aaa is divisible by 3. We can put \(9*3x = 27x\) and our number is divisible by 27. Option III is also correct. Answer E In GENERAL divisibility by 3 and 9 does not automatically mean that number is divisible by 27. As example: 13779 (is divisible by 9 but not by 27). And we can't apply divisibility by 9 rule here. Although the sum of the diggits is 27 the number is not divisible by 27!



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Re: If n is a 27digit positive integer, all of whose digits are [#permalink]
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04 Apr 2017, 00:22
n= aaaaaaaaaa..................a (27 times) n=27a n= 3^3*a n is thus divisible by 3,9, and 27.
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Re: If n is a 27digit positive integer, all of whose digits are [#permalink]
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11 Apr 2017, 00:03
Here it was easy to determine that 27 divides this number but for other numbers it might become tricky. There is one rule that can be used (though it's little lengthy) Divisibility by 27: For example: We have to check the divisibility of 4567 by 27. (1) Multiply the last digit by 8 7*8=56 (2) Subtract 56 from the remaining number 45656= 400 (3) Repeat the process until you get a small number whose divisibility you know or can easily determine. 0*8=0 (4) 400= 40 (5) 40 is not divisible by 27. Therefore 4567 is also not divisible by 27. (169.148...) I hope it makes sense. But the solution given in spoiler is quicker.
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If n is a 27digit positive integer, all of whose digits are [#permalink]
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22 Apr 2017, 12:41
I started confusing myself on the algebra for 27.
So, by hand, I multiplied 27 by each of what would be identical digits, i.e. 27*1(27), 27*2 (54), 27*3 (81) ... 27*9 (243).
Just to be sure, I checked to ascertain that digits added up to 9 (hence divisible by 3 and 9).
By definition (27 times some #), each was a multiple of 27. Time taken, 1:15.
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If n is a 27digit positive integer, all of whose digits are
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