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Re: When positive integer x is divided by 7 the quotient is q and the rema
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04 Aug 2018, 07:09
Official Explanation from Veritasprep:In quotient remainder problems, number picking is often a valid technique to show patterns and answer the question. For this problem, however, it is not very helpful as the question is really testing a tricky property of division with integers. From statement one, you learn that when x is divided by 5 the quotient is q and the remainder is 1. In the question stem, you learned that when x is divided by 7 the quotient is also q and the remainder is also 1. Most students understand that if you divide a number by two different divisors and get the same remainder, you must be at the LCM (lowest common multiple) of those two numbers plus the remainder (and then at the same location down the number line to infinity). For instance if you divide a number by 5 and 7 and get a remainder of 1, then that can happen at 35+1, 70+1, 105+1, etc. However, if you read carefully you see that the quotient is the same in both operations, a puzzling result. Clearly if you divide 36 by 7 it will give you a different quotient then when you divide 36 by 5. The only way this can happen is if x is equal to 1. Remember that when a smaller integer is divided by a larger integer, the quotient is always 0 and the remainder is the dividend itself. Here when 1 is divided by 7 the quotient is 0 and the remainder is 1 and when 1 is divided by 5 the quotient is also 0 and the remainder is 1. Statement 1 is sufficient as x must be 1 and the remainder when x is divided by 10 is also 1. Statement 2 is clearly not sufficient by itself so the answer is A. This statement is here for people who miss that the quotient is the same and think that x could be any multiple of 35+1. By knowing that x is less than 50 it would lock in the value at 36. However, from the discussion above it is clear that this would be incorrect.
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Re: When positive integer x is divided by 7 the quotient is q and the rema
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07 Feb 2019, 01:58
Bunuel wrote: When positive integer x is divided by 7 the quotient is q and the remainder is 1. What is the remainder when x is divided by 10?
(1) When x is divided by 5 the quotient is q and the remainder is 1
(2) x is less than 50 I have calculated the problem in the following way. My answer is C. What's wrong in my way? Pls explain it. According to the question: x=7q+1 so, x can be 8, 15,22, 29,36, 43................. Statement 1: x=5q+1; so x can be 6,11,16,21,26,31,36...................... so x=35q+36 x can be 36, 71,141. so when x is divided by 10, reminder will be 6 or 1. Statement : 2 , x<50, which is insufficient. Statement (1) + (2) x=36, so reminder is 6 . Hence answer is c.



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Re: When positive integer x is divided by 7 the quotient is q and the rema
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07 Feb 2019, 02:25
Rashed12 wrote: Bunuel wrote: When positive integer x is divided by 7 the quotient is q and the remainder is 1. What is the remainder when x is divided by 10?
(1) When x is divided by 5 the quotient is q and the remainder is 1
(2) x is less than 50 I have calculated the problem in the following way. My answer is C. What's wrong in my way? Pls explain it. According to the question: x=7q+1 so, x can be 8, 15,22, 29,36, 43................. Statement 1: x=5q+1; so x can be 6,11,16,21,26,31,36...................... so x=35q+36 x can be 36, 71,141. so when x is divided by 10, reminder will be 6 or 1. Statement : 2 , x<50, which is insufficient. Statement (1) + (2) x=36, so reminder is 6 . Hence answer is c. Rashed12for Statement 1, q could have been 0 as well which when put in x=7q+1 so, x can be 1, 8, 15,22, 29,36, 43................. Statement 1: x=5q+1; so x can be 1,6,11,16,21,26,31,36...................... Giving the remainder as 1, thereby giving you an unique value for the remainder.
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Re: When positive integer x is divided by 7 the quotient is q and the rema
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07 Feb 2019, 02:25



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