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09 Dec 2013, 04:54
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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.
(1) The tens digit of N is 5.
(2) The units digit of N is 5.
09 Dec 2013, 07:06
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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.
Answer: A.
Hope it's clear.
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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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:
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):
555 - works
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:
