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A positive integer X is a six digit number of the form
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08 Aug 2018, 21:46
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A positive integer X is a six digit number of the form ababbb where a and b are distinct digits. What is the value of a? (1) X is divisible by 9. (2) X is divisible by each integer from 1 to 5. Posted from my mobile device
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Re: A positive integer X is a six digit number of the form
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08 Aug 2018, 22:29
amanvermagmat wrote: A positive integer X is a six digit number of the form ababbb where a and b are distinct digits. What is the value of a?
(1) X is divisible by 9.
(2) X is divisible by each integer from 1 to 5.
Posted from my mobile device The answer shall be C... Sent from my SMJ210F using GMAT Club Forum mobile app



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A positive integer X is a six digit number of the form
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09 Aug 2018, 00:31
St1: Insufficient as it gives no clue about the number other than divisible by 9 St2: Only clue that 120 is a factor of the number and last digit or b is 0. Hence the number is a0a000. But no conclusion can be drawn from St2. Insufficient.
Together St1 & St2, you know that sum of a+a should be divisible by 9 and a=9 only solution that satisfies the equation.
Answer is C



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Re: A positive integer X is a six digit number of the form
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09 Aug 2018, 01:51
amanvermagmat wrote: A positive integer X is a six digit number of the form ababbb where a and b are distinct digits. What is the value of a?
(1) X is divisible by 9.
(2) X is divisible by each integer from 1 to 5.
Posted from my mobile device Question stem: b=? St1: X is divisible by 9 When sum of all the digits of an integer is divisible by 9, then the integer is divisible by 9. 2a+4b=9k, where k>1 (\(a\neq0\)) a) a=1, b=4 b) a=9, b=0 So many (a.b) pairs possible. Insufficient. St2: X is divisible by each integer from 1 to 5. a) when the unit digit of any number is zero, then that number is divisible by at least 1,2 ,and 5. b) when the last two digits of a number is divisible by 4, then that number is divisible by 4. Since, here the last two digits are 'b'. So '00' is divisible by 4. c) when the sum of all the digits of an integer is divisible by 3, then the integer is divisible by 3. Now a can be 3 or 6 or 9. Or, six digit number is divisible by LCM(1,2,3,4,5)=120. So, unit digit has to be 0. Or, b=0. Insufficient. Combining, the only possibility of (a.b) is (9,0). Ans. (C)
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Re: A positive integer X is a six digit number of the form
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09 Aug 2018, 02:26
PKN wrote: amanvermagmat wrote: A positive integer X is a six digit number of the form ababbb where a and b are distinct digits. What is the value of a?
(1) X is divisible by 9.
(2) X is divisible by each integer from 1 to 5.
Posted from my mobile device Question stem: b=? St1: X is divisible by 9 When sum of all the digits of an integer is divisible by 9, then the integer is divisible by 9. 2a+4b=9k, where k>1 (\(a\neq0\)) a) a=1, b=4 b) a=9, b=0 So many (a.b) pairs possible. Insufficient. St2: X is divisible by each integer from 1 to 5. a) when the unit digit of any number is zero, then that number is divisible by at least 1,2 ,and 5. b) when the last two digits of a number is divisible by 4, then that number is divisible by 4. Since, here the last two digits are 'b'. So '00' is divisible by 4. c) when the sum of all the digits of an integer is divisible by 3, then the integer is divisible by 3. Now a can be 3 or 6 or 9. Or, six digit number is divisible by LCM(1,2,3,4,5)=120. So, unit digit has to be 0. Or, b=0. Insufficient. Combining, the only possibility of (a.b) is (9,0). Ans. (C) Hi PKN can you please elaborate on statement one, the highlihted part. i didnt get the logic behind it 2a+4b=9k is it some formula for checking divisibility St1: X is divisible by 9 When sum of all the digits of an integer is divisible by 9, then the integer is divisible by 9. 2a+4b=9k,where k>1 (\(a\neq0\)) a) a=1, b=4 b) a=9, b=0



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A positive integer X is a six digit number of the form
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09 Aug 2018, 02:38
dave13 wrote: PKN wrote: amanvermagmat wrote: A positive integer X is a six digit number of the form ababbb where a and b are distinct digits. What is the value of a?
(1) X is divisible by 9.
(2) X is divisible by each integer from 1 to 5.
Posted from my mobile device Question stem: b=? St1: X is divisible by 9 When sum of all the digits of an integer is divisible by 9, then the integer is divisible by 9. 2a+4b=9k, where k>1 (\(a\neq0\)) a) a=1, b=4 b) a=9, b=0 So many (a.b) pairs possible. Insufficient. St2: X is divisible by each integer from 1 to 5. a) when the unit digit of any number is zero, then that number is divisible by at least 1,2 ,and 5. b) when the last two digits of a number is divisible by 4, then that number is divisible by 4. Since, here the last two digits are 'b'. So '00' is divisible by 4. c) when the sum of all the digits of an integer is divisible by 3, then the integer is divisible by 3. Now a can be 3 or 6 or 9. Or, six digit number is divisible by LCM(1,2,3,4,5)=120. So, unit digit has to be 0. Or, b=0. Insufficient. Combining, the only possibility of (a.b) is (9,0). Ans. (C) Hi PKN can you please elaborate on statement one, the highlihted part. i didnt get the logic behind it 2a+4b=9k is it some formula for checking divisibility St1: X is divisible by 9 When sum of all the digits of an integer is divisible by 9, then the integer is divisible by 9. 2a+4b=9k,where k>1 (\(a\neq0\)) a) a=1, b=4 b) a=9, b=0 Hi dave13, Question maker has given us a 6digit integer Given 6digit ababbb. St1 holds when sum of all the digits of the given integer is divisible by 9. What is the sum of the digits? Isn't it a+b+a+b+b+b=2a+4b If I say 'x' is divisible by 'y', then 'x' is a multiple of 'y'. Hope you agree. Here, 2a+4b has to be divisible by 9. So, (2a+4b) has to be a multiple of 9. Since we don't know the value of multiplying factor, I have assigned it as k. So, 2a+4b=multiple of 9=9*k Hope it clarifies your query. Waiting for further queries(if any).
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Re: A positive integer X is a six digit number of the form
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17 Aug 2018, 06:48
my approach: (1)divisible by 9: 2*a+4*b=9m, m is an integer a b m 1 4 2 2 8 4 insufficient (2)consider 2,3,4,5 since there're 2 and 5, b must be 0; 3 zeros at the end also means it's divisible by 4 divisible by 3: 2*a divisible by 3 means a is a multiple of 3. a can be 3,6,9; insufficient 1+2: only 9 ensures x is divisble by 9. so a=9




Re: A positive integer X is a six digit number of the form &nbs
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17 Aug 2018, 06:48






