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# If N is a positive three-digit number that is greater than 2

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If N is a positive three-digit number that is greater than 2  [#permalink]

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09 Dec 2013, 03:54
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27% (02:19) correct 73% (02:13) wrong based on 650 sessions

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If N is a positive three-digit 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 three-digit number that is greater than 2  [#permalink]

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09 Dec 2013, 06:06
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If N is a positive three-digit 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.

Hope it's clear.
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Re: If N is a positive three-digit number that is greater than 2  [#permalink]

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09 Dec 2013, 06:37
Bunuel wrote:
If N is a positive three-digit 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.

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 three-digit number that is greater than 2  [#permalink]

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23 Sep 2014, 17:01
Bunuel wrote:
If N is a positive three-digit 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.

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 three-digit number that is greater than 2  [#permalink]

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24 Sep 2014, 02:25
parry wrote:
Bunuel wrote:
If N is a positive three-digit 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.

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 = a5c, 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 three-digit number that is greater than 2  [#permalink]

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25 Sep 2014, 05:10
Bunuel wrote:
parry wrote:
Bunuel wrote:
If N is a positive three-digit 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.

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 = a5c, 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 three-digit number that is greater than 2  [#permalink]

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09 Mar 2015, 03:02
(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.

Hope it's clear.[/quote]

Hey Bunuel,
Cant 551 be other number in statement A?
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Re: If N is a positive three-digit number that is greater than 2  [#permalink]

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09 Mar 2015, 03:14
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.

Hope it's clear.

Hey Bunuel,
Cant 551 be other number in statement A?

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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29 Jun 2015, 04:40
3
(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.

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Re: If N is a positive three-digit number that is greater than 2  [#permalink]

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26 Jan 2017, 09:10
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 three-digit number that is greater than 2  [#permalink]

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27 Jan 2017, 00:59
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 three-digit number that is greater than 2  [#permalink]

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31 Jan 2018, 11:07
xhimi wrote:
If N is a positive three-digit 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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If N is a positive three-digit number that is greater than 2  [#permalink]

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17 Aug 2018, 17:35
xhimi wrote:
If N is a positive three-digit 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.

Official Solution (Credit: Manhattan Prep)

(1) SUFFICIENT: If 5 is a digit of N, then N is a multiple of 5. All multiples of 5 end with a units digit of either 5 or 0. However, the units digit of N cannot be 0, since 0 is not a factor of any number. Therefore, the units digit of N must also be 5. Therefore N has the form _55, with only the hundreds digit left to consider.

Consider the possible cases for the hundreds digit:
It can’t be 1, since N > 200.
It can’t be an even number, because N is odd (it ends in 5) and thus doesn’t have any even factors.
If it were 3, then N would be 355—but that doesn’t work, since 355 is not a multiple of 3. (Check: 3 + 5 + 5 = 13, which is not divisible by 3.)
If it were 5, then N would be 555. This is a possible value.
If it were 7, then N would be 755—but that doesn’t work, since 755 is not a multiple of 7. (Check: 700 + 55 = 755. 700 is divisible by 7 but 55 is not, so the whole thing is not.)
If it were 9, then N would be 955—but that doesn’t work, since 955 is not a multiple of 9. (Check: 9 + 5 + 5 = 19, which is not divisible by 9.)

The only possible value for N is 555, so statement 1 is sufficient.

(2) INSUFFICIENT: If the units digit of N is 5, then 5 must be a factor of N. All integers ending in 5 are multiples of 5, though, so this fact doesn’t narrow the possibilities any further.

N is a multiple of 5 and any number is a multiple of 1. Using only these digits, try to formulate two numbers that satisfy the statement. Both 515 and 555 satisfy statement (2) so it is not sufficient to answer the question.

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