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Which of the following is divisible by the first nine positive integer
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04 Feb 2015, 07:50
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Re: Which of the following is divisible by the first nine positive integer
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07 Feb 2015, 08:57
Bunuel wrote: Which of the following is divisible by the first nine positive integer multiples of 17?
(i) 14,280 (ii) 85,689 (iii) 368,420
A. none B. iii only C. i and ii only D. ii and iii only E. i, ii, and iii
Kudos for a correct solution. lets not think of 17 first.. the number has to be divisible by 8 and also 9.. lets see each number.. 1)14280...number even and last three digits div by 8.. but sum of digits of number 15, so not div by 9...NO. 2)85689... number is not even.. no need to look any further... NO 3)368420..sum of digits 23, so not div by 9.. also 420 not div by 8.. NO ans A none
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Re: Which of the following is divisible by the first nine positive integer
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07 Feb 2015, 09:51
Bunuel wrote: Which of the following is divisible by the first nine positive integer multiples of 17?
(i) 14,280 (ii) 85,689 (iii) 368,420
A. none B. iii only C. i and ii only D. ii and iii only E. i, ii, and iii
Kudos for a correct solution. Seem's like a tricky question but I hope that I have been able to crack it. Here's my solution: I think that the first 9 positive integers multiples of 17; which are (17)*1 + (17)*2...... + (17)*9 can be written as 17*(1 + 2..... + 9). This value can be simplified to 17*(45). Therefore the options (i), (ii) and (iii) need to be divisible by all the prime factors of (17)*(45) which can be broken down into (17)*(5)*(3)*(3). Using the factors 9 and 5. we can deduce (using divisibility rules) that option (ii) is not divisible by 5, and option's (i) and (iii) are not divisible by 9. Thus it seems like none of the answer choices are divisible by the the first 9 positive integer multiples of 17. I think the answer is A. Please consider giving me kudos, if you felt this post was helpful. Thanks.



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Re: Which of the following is divisible by the first nine positive integer
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07 Feb 2015, 18:53
kdatt1991 wrote: Bunuel wrote: Which of the following is divisible by the first nine positive integer multiples of 17?
(i) 14,280 (ii) 85,689 (iii) 368,420
A. none B. iii only C. i and ii only D. ii and iii only E. i, ii, and iii
Kudos for a correct solution. Seem's like a tricky question but I hope that I have been able to crack it. Here's my solution: I think that the first 9 positive integers multiples of 17; which are (17)*1 + (17)*2...... + (17)*9 can be written as 17*(1 + 2..... + 9). This value can be simplified to 17*(45). Therefore the options (i), (ii) and (iii) need to be divisible by all the prime factors of (17)*(45) which can be broken down into (17)*(5)*(3)*(3). Using the factors 9 and 5. we can deduce (using divisibility rules) that option (ii) is not divisible by 5, and option's (i) and (iii) are not divisible by 9. Thus it seems like none of the answer choices are divisible by the the first 9 positive integer multiples of 17. I think the answer is A. Please consider giving me kudos, if you felt this post was helpful. Thanks. hi kdatt, although the ans is correct, you have misunderstood the question because in some other values your answer would have been wrong.. the Q asks to check div by each of the first nine multiple integers of 17.. this means the value should be div by 17*1,17*2,17*3.....and 17*9.. it translates into the value to be div by each of 5,7,8,9 and 17.... but if we look at 17*45.. it means only by 5,9 and 17... i hope it is clear...
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Re: Which of the following is divisible by the first nine positive integer
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09 Feb 2015, 02:44
Here we go 
17*1 17*2 17*3 17*4 17*5 17*6 17*7 17*8 17*9
So we need to find out...
Does the number in the options contain the following prime numbers with their powers?
17 * (2^7) * (3^4) * (5) * (7)
here's the easy way.. pick any anumber from above prime numbers and try to divide the numbers in the given options.
I picked 9.
Option 1  not divisible by 9 OUT Option 2  Yes divisible by 9 > but we need to accomodate 3 to the power of 4> again divide by 9> not divisible > out Option 3  not divisible by 9 OUT
Hence option A is correct



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Re: Which of the following is divisible by the first nine positive integer
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09 Feb 2015, 03:07
DesiGmat wrote: Here we go 
17*1 17*2 17*3 17*4 17*5 17*6 17*7 17*8 17*9
So we need to find out...
Does the number in the options contain the following prime numbers with their powers?
17 * (2^7) * (3^4) * (5) * (7)
here's the easy way.. pick any anumber from above prime numbers and try to divide the numbers in the given options.
I picked 9.
Option 1  not divisible by 9 OUT Option 2  Yes divisible by 9 > but we need to accomodate 3 to the power of 4> again divide by 9> not divisible > out Option 3  not divisible by 9 OUT
Hence option A is correct hi desigmat, the Q does not ask if the number is div by product of first nine multiples of 17 but asks if the number is div by first nine multiples of 17..... so we have to check the number separately for each of the multiples... so option 2 is out because it is not even...
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Re: Which of the following is divisible by the first nine positive integer
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09 Feb 2015, 03:17
chetan2u wrote: DesiGmat wrote: Here we go 
17*1 17*2 17*3 17*4 17*5 17*6 17*7 17*8 17*9
So we need to find out...
Does the number in the options contain the following prime numbers with their powers?
17 * (2^7) * (3^4) * (5) * (7)
here's the easy way.. pick any anumber from above prime numbers and try to divide the numbers in the given options.
I picked 9.
Option 1  not divisible by 9 OUT Option 2  Yes divisible by 9 > but we need to accomodate 3 to the power of 4> again divide by 9> not divisible > out Option 3  not divisible by 9 OUT
Hence option A is correct hi desigmat, the Q does not ask if the number is div by product of first nine multiples of 17 but asks if the number is div by first nine multiples of 17..... so we have to check the number separately for each of the multiples... so option 2 is out because it is not even... hi chetan2u, That is what I listed... 17*2 will be the first multiple of 17 and hence forth. A number to be divisible by all the nine positive integer multiples of 17 has to contain minimum of 2^7, 3^4, 5, 7 and 17. Can you tell me the first number/lowest number that will be divisible by all the nine positive integer multiples of 17?



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Re: Which of the following is divisible by the first nine positive integer
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09 Feb 2015, 04:16
Bunuel wrote: Which of the following is divisible by the first nine positive integer multiples of 17?
(i) 14,280 (ii) 85,689 (iii) 368,420
A. none B. iii only C. i and ii only D. ii and iii only E. i, ii, and iii
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:Solution: A. For a number to be divisible by the first nine positive integer multiples of 17, it must be divisible by 2 * 17, 3 * 17, 4 * 17 … 9 * 17, or by 17 and every integer from 1 through 9. (ii) is not even, so it must not be divisible by 2 and is therefore out. Neither (i) nor (iii) have digits that sum to 9, so neither result can be divisible by 9 and none of the results is divisible by all of the first nine positive integers; (A)
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Re: Which of the following is divisible by the first nine positive integer
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09 Feb 2015, 05:03
chetan2u wrote: kdatt1991 wrote: Bunuel wrote: Which of the following is divisible by the first nine positive integer multiples of 17?
(i) 14,280 (ii) 85,689 (iii) 368,420
A. none B. iii only C. i and ii only D. ii and iii only E. i, ii, and iii
Kudos for a correct solution. Seem's like a tricky question but I hope that I have been able to crack it. Here's my solution: I think that the first 9 positive integers multiples of 17; which are (17)*1 + (17)*2...... + (17)*9 can be written as 17*(1 + 2..... + 9). This value can be simplified to 17*(45). Therefore the options (i), (ii) and (iii) need to be divisible by all the prime factors of (17)*(45) which can be broken down into (17)*(5)*(3)*(3). Using the factors 9 and 5. we can deduce (using divisibility rules) that option (ii) is not divisible by 5, and option's (i) and (iii) are not divisible by 9. Thus it seems like none of the answer choices are divisible by the the first 9 positive integer multiples of 17. I think the answer is A. Please consider giving me kudos, if you felt this post was helpful. Thanks. hi kdatt, although the ans is correct, you have misunderstood the question because in some other values your answer would have been wrong.. the Q asks to check div by each of the first nine multiple integers of 17.. this means the value should be div by 17*1,17*2,17*3.....and 17*9.. it translates into the value to be div by each of 5,7,8,9 and 17.... but if we look at 17*45.. it means only by 5,9 and 17... i hope it is clear... Hi chetan2u, Thanks for pointing that out to me. Seems like I need to practise on reading the questions more carefully!



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Re: Which of the following is divisible by the first nine positive integer
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09 Feb 2015, 05:20
DesiGmat wrote: chetan2u wrote: DesiGmat wrote: Here we go 
17*1 17*2 17*3 17*4 17*5 17*6 17*7 17*8 17*9
So we need to find out...
Does the number in the options contain the following prime numbers with their powers?
17 * (2^7) * (3^4) * (5) * (7)
here's the easy way.. pick any anumber from above prime numbers and try to divide the numbers in the given options.
I picked 9.
Option 1  not divisible by 9 OUT Option 2  Yes divisible by 9 > but we need to accomodate 3 to the power of 4> again divide by 9> not divisible > out Option 3  not divisible by 9 OUT
Hence option A is correct hi desigmat, the Q does not ask if the number is div by product of first nine multiples of 17 but asks if the number is div by first nine multiples of 17..... so we have to check the number separately for each of the multiples... so option 2 is out because it is not even... hi chetan2u, That is what I listed... 17*2 will be the first multiple of 17 and hence forth. A number to be divisible by all the nine positive integer multiples of 17 has to contain minimum of 2^7, 3^4, 5, 7 and 17. Can you tell me the first number/lowest number that will be divisible by all the nine positive integer multiples of 17? hi desigmat, why 2^7.. highest power of 2 in first nine integers is 8, which is 2^3... highest power of 3 is 3^2 or 9.... u cant have 3^4..
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1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
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Re: Which of the following is divisible by the first nine positive integer
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