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If N is a positive threedigit number that is greater than 2 [#permalink]
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09 Dec 2013, 04:54
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If N is a positive threedigit number that is greater than 200, and each digit of N is a factor of N itself, what is the value of N? (1) The tens digit of N is 5. (2) The units digit of N is 5.
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Re: If N is a positive threedigit number that is greater than 2 [#permalink]
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If N is a positive threedigit number that is greater than 200, and each digit of N is a factor of N itself, what is the value of N?Given: N = abc, where a>1 and each of a, b, and c is a factor of N. (1) The tens digit of N is 5. N = a5c. This implies that 5 is a factor of N, which means that the units digit of N ( c) must be 5. c cannot be 0 since we are told that each digit of N is a factor of N itself and 0 is not a factor of any number. Thus we have that N = a55. Because N is odd, then a to be a factor of N, must be odd too. N can be: 355; 555; 755; 955. Only 555 satisfies all the conditions: each digit of N>200 is a factor of N itself . Sufficient. (2) The units digit of N is 5. N = ab5 > more than one values of N are possible: 315, 515, 555. Not sufficient. Answer: A. Hope it's clear.
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Re: If N is a positive threedigit number that is greater than 2 [#permalink]
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09 Dec 2013, 07:37
Bunuel wrote: If N is a positive threedigit number that is greater than 200, and each digit of N is a factor of N itself, what is the value of N?
Given: N = abc, where a>1 and each of a, b, and c is a factor of N.
(1) The tens digit of N is 5.
N = a5c. This implies that 5 is a factor of N, which means that the units digit of N (c) must be 5. c cannot be 0 since we are told that each digit of N is a factor of N itself and 0 is not a factor of any number. Thus we have that N = a55. Because N is odd, then a to be a factor of N, must be odd too. N can be: 355; 555; 755; 955.
Only 555 satisfies all the conditions: each digit of N>200 is a factor of N itself . Sufficient. (2) The units digit of N is 5.
N = ab5 > more than one values of N are possible: 315, 515, 555. Not sufficient.
Answer: A.
Hope it's clear. Excellent question and excellent explanation Bunuel. Took me 5 minutes to understand your explanation. I marked E while solving this question.
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Re: If N is a positive threedigit number that is greater than 2 [#permalink]
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23 Sep 2014, 18:01
Bunuel wrote: If N is a positive threedigit number that is greater than 200, and each digit of N is a factor of N itself, what is the value of N?
Given: N = abc, where a>1 and each of a, b, and c is a factor of N.
(1) The tens digit of N is 5.
N = a5c. This implies that 5 is a factor of N, which means that the units digit of N (c) must be 5. c cannot be 0 since we are told that each digit of N is a factor of N itself and 0 is not a factor of any number. Thus we have that N = a55. Because N is odd, then a to be a factor of N, must be odd too. N can be: 355; 555; 755; 955.
Only 555 satisfies all the conditions: each digit of N>200 is a factor of N itself . Sufficient. (2) The units digit of N is 5.
N = ab5 > more than one values of N are possible: 315, 515, 555. Not sufficient.
Answer: A.
Hope it's clear. As done in option (2), why not choose 551 in option (1), as 1 is multiple of 5?



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Re: If N is a positive threedigit number that is greater than 2 [#permalink]
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24 Sep 2014, 03:25
parry wrote: Bunuel wrote: If N is a positive threedigit number that is greater than 200, and each digit of N is a factor of N itself, what is the value of N?
Given: N = abc, where a>1 and each of a, b, and c is a factor of N.
(1) The tens digit of N is 5.
N = a5c. This implies that 5 is a factor of N, which means that the units digit of N (c) must be 5. c cannot be 0 since we are told that each digit of N is a factor of N itself and 0 is not a factor of any number. Thus we have that N = a55. Because N is odd, then a to be a factor of N, must be odd too. N can be: 355; 555; 755; 955.
Only 555 satisfies all the conditions: each digit of N>200 is a factor of N itself . Sufficient. (2) The units digit of N is 5.
N = ab5 > more than one values of N are possible: 315, 515, 555. Not sufficient.
Answer: A.
Hope it's clear. As done in option (2), why not choose 551 in option (1), as 1 is multiple of 5? I guess you meant that 1 is a factor of 5. Anyway, since N = a 5c, then 5 must a factor of N, which means that the units digit of N ( c) must be 5, so it cannot be 1: 5 is not a factor of 551. Hope it's clear.
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Re: If N is a positive threedigit number that is greater than 2 [#permalink]
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25 Sep 2014, 06:10
Bunuel wrote: parry wrote: Bunuel wrote: If N is a positive threedigit number that is greater than 200, and each digit of N is a factor of N itself, what is the value of N?
Given: N = abc, where a>1 and each of a, b, and c is a factor of N.
(1) The tens digit of N is 5.
N = a5c. This implies that 5 is a factor of N, which means that the units digit of N (c) must be 5. c cannot be 0 since we are told that each digit of N is a factor of N itself and 0 is not a factor of any number. Thus we have that N = a55. Because N is odd, then a to be a factor of N, must be odd too. N can be: 355; 555; 755; 955.
Only 555 satisfies all the conditions: each digit of N>200 is a factor of N itself . Sufficient. (2) The units digit of N is 5.
N = ab5 > more than one values of N are possible: 315, 515, 555. Not sufficient.
Answer: A.
Hope it's clear. As done in option (2), why not choose 551 in option (1), as 1 is multiple of 5? I guess you meant that 1 is a factor of 5. Anyway, since N = a 5c, then 5 must a factor of N, which means that the units digit of N ( c) must be 5, so it cannot be 1: 5 is not a factor of 551. Hope it's clear. I really appreciate your early reply.



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Re: If N is a positive threedigit number that is greater than 2 [#permalink]
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09 Mar 2015, 04:02
(1) The tens digit of N is 5. N = a5c. This implies that 5 is a factor of N, which means that the units digit of N ( c) must be 5. c cannot be 0 since we are told that each digit of N is a factor of N itself and 0 is not a factor of any number. Thus we have that N = a55. Because N is odd, then a to be a factor of N, must be odd too. N can be: 355; 555; 755; 955. Only 555 satisfies all the conditions: each digit of N>200 is a factor of N itself . Sufficient. (2) The units digit of N is 5. N = ab5 > more than one values of N are possible: 315, 515, 555. Not sufficient. Answer: A. Hope it's clear.[/quote] Hey Bunuel, Cant 551 be other number in statement A? Please explain.



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Re: If N is a positive threedigit number that is greater than 2 [#permalink]
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09 Mar 2015, 04:14
ssriva2 wrote: (1) The tens digit of N is 5.
N = a5c. This implies that 5 is a factor of N, which means that the units digit of N (c) must be 5. c cannot be 0 since we are told that each digit of N is a factor of N itself and 0 is not a factor of any number. Thus we have that N = a55. Because N is odd, then a to be a factor of N, must be odd too. N can be: 355; 555; 755; 955.
Only 555 satisfies all the conditions: each digit of N>200 is a factor of N itself . Sufficient. (2) The units digit of N is 5.
N = ab5 > more than one values of N are possible: 315, 515, 555. Not sufficient.
Answer: A.
Hope it's clear. Hey Bunuel, Cant 551 be other number in statement A? Please explain.[/quote] The stem says that each digit of N is a factor of N itself, thus N cannot be 551 because 5 is not a factor of 551. By the way, this doubt is addressed in my previous post HERE. So, please read the whole thread before posting a question.
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Data sufficiency question [#permalink]
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29 Jun 2015, 05:40
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(1) The tens digit of N is 5.We know that all of the digits should be factors of N, hence this statement effectively tells us that last digit is also 5 (any number divisible by 5 should end with 5 or 0, as 0 can't be a factor of any number, we are left with 5). Now, let's plug numbers! First digit should be >1 (question stem) and a can't be even (as our number ends with 5, hence it's not divisible by 2): 355 sum of digits = 13, not divisible by 3. 555  works 755  not divisible by 7 (just use 7*100 = 700 + 7*7 = 49 = 749, then 7*8=756). 955  sum = 19, not divisible by 9. Sufficient.(2) The units digit of N is 5.First, notice that we already have this info from A, this gives us that xy can't be even. Let's plug numbers: xy5  where x and y are digits, x>1. 315, sum of digits 9, hence divisible by 3 and obviously by 1 and 5. Works. 515, divisible by 5's and 1's. Works. We already have 2 solutions. Not sufficient.Answer:
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Re: If N is a positive threedigit number that is greater than 2 [#permalink]
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26 Jan 2017, 10:10
I answered E but didn't realize all the numbers don't have to be different Never considered that it could be same number.
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Re: If N is a positive threedigit number that is greater than 2 [#permalink]
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27 Jan 2017, 01:59
Thanks Bunuel .. I missed the fact "0 is not a factor of any number". So took 550 and 555 as possibility and answered C...



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If N is a positive threedigit number that is greater than 2 [#permalink]
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31 Jan 2018, 12:07
xhimi wrote: If N is a positive threedigit number that is greater than 200, and each digit of N is a factor of N itself, what is the value of N?
(1) The tens digit of N is 5. (2) The units digit of N is 5. I solved it this way.... Statement I: Let the number be \(abc\)... as per the statement I, the number is a5c... So, basically 5 is a factor of \(a5c\). But for \(a5c\) to be a factor of 5, the last digit should be 0 or 5. We cannot have 0 as per the given question as each digit is a factor of the Number itself. So, \(c = 5\). Using same logic we can identify \(a = 5\). So, Sufficient. Statement II: From this the number can be written as \(ab5\).. So, 5 is a factor of \(ab5\). But in this case we can have b = 3/5/7/9 because each of these numbers MAY divide any number ending with 5. So, Insufficient.
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