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The threedigit positive integer n can be written as ABC, in which [#permalink]
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11 Nov 2014, 08:09
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Re: The threedigit positive integer n can be written as ABC, in which [#permalink]
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11 Nov 2014, 13:31
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I found this quite challenging to conceptualize so perhaps this is off:
The threedigit positive integer n can be written as ABC, in which A, B, and C stand for the unknown digits of n. What is the remainder when n is divided by 37?
(1) A + \frac{B}{10} + \frac{C}{100} = B + \frac{C}{10} + \frac{A}{100}
(2) A + \frac{B}{10} + \frac{C}{100} = C + \frac{A}{10} + \frac{B}{100}
1). This can be rewritten as 100 A + 10 B + C = 100 B + 10 C + A.
Simplify for a value: 99A = 90 B + 9C. I almost gave up there 11A = 10B + C. No value can be negative. All are integers... can A=B=C? There 111, 222, 333, etc. With 111 being 37 * 3 (90+21) then 111, 222, 333 must all be multiples of 37, so remainder is 0
Can we have anything else? Lets make A 2. 22 = 10+2. So we can have 212 as a number. Different remainder. A is insufficient.
2) rewrite to 100A + 10V + C = 100C + 10A + B > 90A = 9B +99C... 10A = B+11C.
382 works, 473, 564. These give different remainders, insufficient.
Together... hold on.
11a = 10b + c and 10a = b + 11c. Can't be negative, both have been disproven, and these say conflicting things. Must be E



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Re: The threedigit positive integer n can be written as ABC, in which [#permalink]
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11 Nov 2014, 14:55
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Ground rules: 0<= A,B,C <=9  to form a 3 digit number.
1) Rearranging: 11A = 10B + C  Applying the ground rule; A = (10B+C)/11 => 10B+C has to be multiple of 11 starting from 0 to 99. 0 isn't going to give a 3 digit number (A=0 makes it 2 digit). So, 10B+C=11,22....99 and A = 1,2,....9. So the numbers that fit this condition are 111,222,333......999. All these numbers are divisible by 37. So the remainder = 0. A gives an unambiguous answer
2) Rearranging: 11C = 10A + B  Same as above; C=1,2,....9 and 10A+B = 11,22...99 and the numbers that fit this condition are again 111,222,333.....999. Same as above and B too gives an unambiguous answer.
Answer should be (D). OA?



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Re: The threedigit positive integer n can be written as ABC, in which [#permalink]
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11 Nov 2014, 15:22
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Hi there,
(1) implies ABC=BCA therefore A=B=C A is at least 1 so ABC is 111, 222, etc. 111/37 = 3 so the remainder is zero. Same will be true with 222, etc. as they are multiples of 111.
(2) same reasoning as above
Solution: D



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Re: The threedigit positive integer n can be written as ABC, in which [#permalink]
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11 Nov 2014, 18:14
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n = ABC = 100A +10B + C
From statement 1 : A + B/10 + C/100 = B + C/10 + A /100 or, 100A + 10B+C = 100B + 10C + A
From this we can say A=B=C , therefore the possible numbers are 111, 222, 333, 444... , which are all divisible by 37.
From statement 2, we can deduce the same.
Therefore the answer has to be D) , since each of the statements is sufficient.



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Re: The threedigit positive integer n can be written as ABC, in which [#permalink]
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11 Nov 2014, 19:54
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From either 1 or 2, we can deduce that
A=B=C
111 = 37*3 so all multiples of 111 (222,333, 444,...,999) will be divisible by 37.
Thus, each statement alone is sufficient. Answer should be D.



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Re: The threedigit positive integer n can be written as ABC, in which [#permalink]
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12 Nov 2014, 03:45
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Bunuel wrote: Tough and Tricky questions: Remainders. The threedigit positive integer \(n\) can be written as \(ABC\), in which \(A\), \(B\), and \(C\) stand for the unknown digits of \(n\). What is the remainder when \(n\) is divided by 37? (1) \(A + \frac{B}{10} + \frac{C}{100} = B + \frac{C}{10} + \frac{A}{100}\) (2) \(A + \frac{B}{10} + \frac{C}{100} = C + \frac{A}{10} + \frac{B}{100}\) Kudos for a correct solution. Official Solution: The question stem tells us that the positive integer \(n\) has three unknown digits: \(A\), \(B\), and \(C\), in that order. In other words, \(n\) can be written as \(ABC\). Note that in this context, \(ABC\) does not represent the product of the variables \(A\), \(B\), and \(C\), but rather a threedigit integer with unknown digit values. It is important to note that since \(A\), \(B\), and \(C\) stand for digits, their values are restricted to the ten digits 0 through 9. Moreover, \(A\) cannot equal 0, since we know that \(n\) is a "threedigit" integer and therefore must be at least 100. We are asked for the remainder after \(n\) is divided by 37. We could rephrase this question in a variety of ways, but none of them are particularly better than simply leaving the question as is. Statement (1): SUFFICIENT. We can translate this statement to a decimal representation, which will be easier to understand. The left side of the equation, in words, is "\(A\) units plus \(B\) tenths plus \(C\) hundredths." We can write this in shorthand: \(A.BC\) (that is, "A point BC"). After performing the same translation to the right side of the equation, we can see that we get the following: \(A.BC = B.CA\) Since \(A\), \(B\), and \(C\) stand for digits, we can match up the decimal representations and observe that \(A = B\) and \(B = C\). Thus, all the digits are the same. This means that we can write \(n\) as \(AAA\), which is simply \(111 \times A\). Now, 111 factors into \(3 \times 37\), so \(n = 3 \times 37 \times A\). Thus, \(n\) is a multiple of 37, and the remainder after division by 37 is zero. Statement (2): SUFFICIENT. Again, we can match up the decimal representations of the given equation and find that all the digits are the same. The logic from that point forward is identical to that shown above. Answer: D
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Re: The threedigit positive integer n can be written as ABC, in which [#permalink]
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11 Jan 2016, 23:36
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. The threedigit positive integer n can be written as ABC , in which A , B , and C stand for the unknown digits of n . What is the remainder when n is divided by 37? When you modify the original condition, (1) A+B10 +C100 =B+C10 +A100 Multiply 100 to the both equations. It becomes 100A+10B+C=100B+10C+A > 99A=90B+9C, 11A=10B+C. (2) A+B10 +C100 =C+A10 +B100 Multiply 100 to the both equations. It becomes 100A+10B+C=100C+10A+B > 99C=90A+9B, 11C=10A+B. In the original condition, there are 3 variables, which should match with the number of equations. So, you need 3 more equations. For 1) 1 equation, for 2) 1 equation, which is likely to make E the answer as 1 more equation is needed. When 1) & 2), in 1)=2), n=111,222,333,.......,999. In 111=37*3+0, 0 is the remainder when n is divided by 37, which is unique and sufficient. Therefore, the answer is D. When 1)=2), it is about 95% that D is the answer.
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The threedigit positive integer n can be written as ABC, in which [#permalink]
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20 Feb 2016, 08:28
Bunuel wrote: Tough and Tricky questions: Remainders. The threedigit positive integer \(n\) can be written as \(ABC\), in which \(A\), \(B\), and \(C\) stand for the unknown digits of \(n\). What is the remainder when \(n\) is divided by 37? (1) \(A + \frac{B}{10} + \frac{C}{100} = B + \frac{C}{10} + \frac{A}{100}\) (2) \(A + \frac{B}{10} + \frac{C}{100} = C + \frac{A}{10} + \frac{B}{100}\) Kudos for a correct solution.To solve this math in a CAT u need to stick a couple of math rules to your head otherwise u cant in 3 minutes. FIRSTLY \(234 = 100(2) + 10(3) + 4\) \(200 + 20 + 4 = 234\) This works for any number such that the four digit number \(ABDE = 1000(A) + 100(B) + 10(D) + E\) SECONDLY, the 37 rule: 37 is can divide 111, 222, 333, 444, 555, ..., 999. Now SOLUTION to the question above (1) if rearranged, simply states that \(ABC = BCA\) (i.e. for the 3digit integer, when u place the first digit as the last and the second as the first it must equal the original number in value) only if the three digits are equal will that condition hold. So ABC must equal 111 or 222 or 333 or 444 or 555 or 666 or 777 or 888 or 999. We dont care which of them it is. Each leaves a remainder 0 when divided by 37. Sufficient! For 2, same condition in 1. sufficient. D. Each.



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Re: The threedigit positive integer n can be written as ABC, in which [#permalink]
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