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X is a threedigit positive integer in which each digit is either 1 or
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27 Apr 2015, 04:38
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X is a threedigit positive integer in which each digit is either 1 or 2. Y has the same digits as X, but in reverse order. What is the remainder when X is divided by 3? (1) The hundreds digit of XY is 6. (2) The tens digit of XY is 4. Kudos for a correct solution.
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Re: X is a threedigit positive integer in which each digit is either 1 or
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04 May 2015, 04:53
Bunuel wrote: X is a threedigit positive integer in which each digit is either 1 or 2. Y has the same digits as X, but in reverse order. What is the remainder when X is divided by 3?
(1) The hundreds digit of XY is 6. (2) The tens digit of XY is 4.
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION:You are told that X is a threedigit integer in which each digit is either 1 or 2. There are 8 possibilities for X (8 = 2×2×2), but rather than list out this many possibilities, you might just write something like this: X = abc product of a, b, and c, but rather the threedigit number formed from the digits in that order (a = hundreds, b = tens, c = units). Likewise, you can now write Y = cba. You need to find the remainder when X is divided by 3. Since divisibility by 3 depends on the sum of the digits, you really just need to find a + b + c, or more precisely, whether this sum is itself a multiple of 3, one more than a multiple of 3, or two more than a multiple of 3. Statement 1: SUFFICIENT. To express the hundreds digit of XY in terms of a, b, and c, you need to write X and Y in a more formal algebraic way. A threedigit number is 100 times its hundreds digit, plus 10 times its tens digit, plus its units digit. So X = 100a + 10b + c. Likewise, Y = 100c + 10b + a. We now have to multiply these two expressions together. The efficient way to do so is to think of the possible coefficients (10,000; 1,000; 100; 10; and 1) and then match the terms that will create those coefficients. (100a+ 10b + c)(100c + 10b + a) =10,000ac + 1,000(ab + bc) + 100(a^2 + b^2 + c^2) + 10(ab + bc) + ac. So you are told that the hundreds digit is 6. Before you match that to the expression above, you have to think about carried digits—could the expressions in the units or the tens place cause a digit to be carried? The answer is no: even if all the digits were 2’s (the maximum), the tens product would only be 8, with no carrying. So we can now say that 6 = a^2 + b^2 + c^2. Since each variable can only be 1 or 2, what are the possible values of the digits? By testing numbers, you can quickly see that exactly one of the digits must be 2; the other two digits must be 1. Thus, the sum of the digits of X is 2 + 1 + 1 = 4, so the remainder after division by 3 is 1. Statement 2: INSUFFICIENT. Using the same work from Statement 1, and checking that you don’t have to worry about carried digits, you get ab + bc = 4. Factor the left side: b(a + c) = 4. Given the possible digit values of 1 and 2, there are two possible solutions. One solution is b = 2 and a + c = 2, or a = 1 and b = 1. The other solution is b = 1 and a + c = 4, or a = 2 and c = 2. That means that the number is either 121 or 212. If the number is 121, the remainder is 1. If the number is 212, the remainder is 2. The correct answer is A.
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X is a threedigit positive integer in which each digit is either 1 or
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Updated on: 27 Apr 2015, 08:06
100x+10y+z  first number, 100x + 10y + x  second number, x, y and z  either 1 or 2 product will be quite a sum but the parts which matter are these ones: 100(x^2+y^2+z^2) and 10(xy+yz) #1  x^2+y^2 +z^2 = 6 #2  xy+yz = y(x+z) = 4 #1 gives us 3 choices for (x,y,z): (1,1,2), (1,2,1),(2,1,1) #2 gives us these choices for (x,y,z): (2,1,2),(1,2,1)
#1 gives us 3 numbers, all of them have the sum of digits equal to 4 which lets us explicitly answer the question about the remainder (1) #2 doesn't
A
Originally posted by Zhenek on 27 Apr 2015, 07:54.
Last edited by Zhenek on 27 Apr 2015, 08:06, edited 1 time in total.




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Re: X is a threedigit positive integer in which each digit is either 1 or
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27 Apr 2015, 07:30
Bunuel wrote: X is a threedigit positive integer in which each digit is either 1 or 2. Y has the same digits as X, but in reverse order. What is the remainder when X is divided by 3?
(1) The hundreds digit of XY is 6. (2) The tens digit of XY is 4.
Kudos for a correct solution. From statement 1:If hundredth digit of product XY is 6, then: Number pairs can be (211, 112) or (121, 121). Of all three different numbers from the pairs, anyone when divided by 3 gives remainder as 1. Hence it doesn't really matters which one of the three is X and which one Y. Statement 1 is sufficient. From statement 2:If tens digit of product XY is 4, then: Number pairs can be (121, 121) or (212, 212). If X=121, remainder when X is divided by 3 is 1 but if X=212, remainder would be 2. Hence statement 2 is not sufficient. IMO answer should be A.
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X is a threedigit positive integer in which each digit is either 1 or
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27 Apr 2015, 07:57
Bunuel wrote: X is a threedigit positive integer in which each digit is either 1 or 2. Y has the same digits as X, but in reverse order. What is the remainder when X is divided by 3?
(1) The hundreds digit of XY is 6. (2) The tens digit of XY is 4.
Kudos for a correct solution. We know that we have two numbers \(abc\) and \(cba\) and from statements we have information about tens and hundreds. In this case (when only \(1\) and \(2\) can be numbers) we know that tens will be equal to \(ab + bc\) and hundreds will be equal to \(a^2+b^2+c^2\) 1)\(a^2+b^2+c^2 = 6\) This possible only in variants when two of numbers are equal to \(1\) and other number is equal to \(2\) \(1^2+1^2+2^2 = 6\) etc. So \(abc\) can be \(112\), \(211\) or \(121\) All this numbers when divided by \(3\) give us remainder \(1\) Sufficient 2) \(ab + bc = 4\) This possible when \(ab\) and \(bc\) equal to \(2\). So we have two variants: \(a\) and \(c\) equal to \(2\) and \(b\) equal to \(1\) or \(a\) and \(c\) equal to \(1\) and \(b\) equal to \(2\) \(212\) and \(121\). When divided by \(3\) first number gives remainder \(2\) and second number gives remainder \(1\) Insufficient. Answer is A
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Re: X is a threedigit positive integer in which each digit is either 1 or
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05 May 2015, 12:48
Harley1980 wrote: Bunuel wrote: X is a threedigit positive integer in which each digit is either 1 or 2. Y has the same digits as X, but in reverse order. What is the remainder when X is divided by 3?
(1) The hundreds digit of XY is 6. (2) The tens digit of XY is 4.
Kudos for a correct solution. We know that we have two numbers \(abc\) and \(cba\) and from statements we have information about tens and hundreds. In this case (when only \(1\) and \(2\) can be numbers) we know that tens will be equal to \(ab + bc\) and hundreds will be equal to \(a^2+b^2+c^2\) 1)\(a^2+b^2+c^2 = 6\) This possible only in variants when two of numbers are equal to \(1\) and other number is equal to \(2\) \(1^2+1^2+2^2 = 6\) etc. So \(abc\) can be \(112\), \(211\) or \(121\) All this numbers when divided by \(3\) give us remainder \(1\) Sufficient 2) \(ab + bc = 4\) This possible when \(ab\) and \(bc\) equal to \(2\). So we have two variants: \(a\) and \(c\) equal to \(2\) and \(b\) equal to \(1\) or \(a\) and \(c\) equal to \(1\) and \(b\) equal to \(2\) \(212\) and \(121\). When divided by \(3\) first number gives remainder \(2\) and second number gives remainder \(1\) Insufficient. Answer is A Dear Harley I have never seen this rule before even i went through the Manhattan Books  maybe I have overlooked it. However, is this rule with the tens and hundreds a general rule for that type of question?: "We know that we have two numbers abc and cba and from statements we have information about tens and hundreds. In this case (when only 1 and 2 can be numbers) we know that tens will be equal to ab+bc and hundreds will be equal to a2+b2+c2" How does it behave if the question tells you that each digit can be 1, 2 or 3? Thank you
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X is a threedigit positive integer in which each digit is either 1 or
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Updated on: 06 May 2015, 20:10
reto wrote: Harley1980 wrote: Bunuel wrote: X is a threedigit positive integer in which each digit is either 1 or 2. Y has the same digits as X, but in reverse order. What is the remainder when X is divided by 3?
(1) The hundreds digit of XY is 6. (2) The tens digit of XY is 4.
Kudos for a correct solution. We know that we have two numbers \(abc\) and \(cba\) and from statements we have information about tens and hundreds. In this case (when only \(1\) and \(2\) can be numbers) we know that tens will be equal to \(ab + bc\) and hundreds will be equal to \(a^2+b^2+c^2\) 1)\(a^2+b^2+c^2 = 6\) This possible only in variants when two of numbers are equal to \(1\) and other number is equal to \(2\) \(1^2+1^2+2^2 = 6\) etc. So \(abc\) can be \(112\), \(211\) or \(121\) All this numbers when divided by \(3\) give us remainder \(1\) Sufficient 2) \(ab + bc = 4\) This possible when \(ab\) and \(bc\) equal to \(2\). So we have two variants: \(a\) and \(c\) equal to \(2\) and \(b\) equal to \(1\) or \(a\) and \(c\) equal to \(1\) and \(b\) equal to \(2\) \(212\) and \(121\). When divided by \(3\) first number gives remainder \(2\) and second number gives remainder \(1\) Insufficient. Answer is A Dear Harley I have never seen this rule before even i went through the Manhattan Books  maybe I have overlooked it. However, is this rule with the tens and hundreds a general rule for that type of question?: "We know that we have two numbers abc and cba and from statements we have information about tens and hundreds. In this case (when only 1 and 2 can be numbers) we know that tens will be equal to ab+bc and hundreds will be equal to a2+b2+c2" How does it behave if the question tells you that each digit can be 1, 2 or 3? Thank you Hi reto, This rule is just based on how we multiply, there is nothing more to it. X = abc = a*100 + b*10 + c Y = cba = c*100 + b*10 + a XY = (a*100 + b*10 + c)*(c*100 + b*10 + a) =>XY = 10000*ac + 1000*ab + 100a^2 + 1000bc + 100b^2 + 10ab + 100c^2 + 10*bc + ac => XY = 10000*ac + 1000*(ab + bc) + 100*(a^2 + b^2 + c^2) + 10*(ab + bc) + ac If it can be proved that ac, (ab+bc) and (a^2 + b^2 + c^2) are single digits then the unit digit will be ac, 10th place will be (ab+bc) and 100th place will be (a^2 + b^2 + c^2) let's take the example of 112 and 211 112 * 211 = 23632 unit digit of 112 * 211 is ac = 1*2 = 2 < 10 10th digit of 112 * 211 is ab + bc = 1+2 = 3 < 10 100th digit 0f 112 * 211 is a^2 + b^2 + c^2 = 1 + 1 + 4 = 6 < 10 Now let's take the example 223 and 322 223*322 = 71806 ac = 2*3 = 6 < 10, so unit digit of 223*322 is 6 ab + bc = 4+6 = 10, so 10th digit of 223*322 is 0 not 10 a^2 + b^2 + c^2 = 4 + 4 + 9 = 17 > 10, so 100th digit of 223*322 is 7 + 1(carry over from the 10th digit) = 8 and not 17 Hope this helps. Do let me know if you have more questions.
Originally posted by PrepTap on 06 May 2015, 00:01.
Last edited by PrepTap on 06 May 2015, 20:10, edited 1 time in total.



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Re: X is a threedigit positive integer in which each digit is either 1 or
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06 May 2015, 14:19
reto wrote: Harley1980 wrote: Bunuel wrote: X is a threedigit positive integer in which each digit is either 1 or 2. Y has the same digits as X, but in reverse order. What is the remainder when X is divided by 3?
(1) The hundreds digit of XY is 6. (2) The tens digit of XY is 4.
Kudos for a correct solution. We know that we have two numbers \(abc\) and \(cba\) and from statements we have information about tens and hundreds. In this case (when only \(1\) and \(2\) can be numbers) we know that tens will be equal to \(ab + bc\) and hundreds will be equal to \(a^2+b^2+c^2\) 1)\(a^2+b^2+c^2 = 6\) This possible only in variants when two of numbers are equal to \(1\) and other number is equal to \(2\) \(1^2+1^2+2^2 = 6\) etc. So \(abc\) can be \(112\), \(211\) or \(121\) All this numbers when divided by \(3\) give us remainder \(1\) Sufficient 2) \(ab + bc = 4\) This possible when \(ab\) and \(bc\) equal to \(2\). So we have two variants: \(a\) and \(c\) equal to \(2\) and \(b\) equal to \(1\) or \(a\) and \(c\) equal to \(1\) and \(b\) equal to \(2\) \(212\) and \(121\). When divided by \(3\) first number gives remainder \(2\) and second number gives remainder \(1\) Insufficient. Answer is A Dear Harley I have never seen this rule before even i went through the Manhattan Books  maybe I have overlooked it. However, is this rule with the tens and hundreds a general rule for that type of question?: "We know that we have two numbers abc and cba and from statements we have information about tens and hundreds. In this case (when only 1 and 2 can be numbers) we know that tens will be equal to ab+bc and hundreds will be equal to a2+b2+c2" How does it behave if the question tells you that each digit can be 1, 2 or 3? Thank you Hello retoI don't think that you overlooked it. Books give you foundation, but you should invent new ways to use this knowledge (or look for this ways on forums About this task: I make one example with numbers and look for the pattern. it's hard to write, so I make screenshot; I hope it make sense. If not, don't hesitate to ask I will try to explain it in words.
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Re: X is a threedigit positive integer in which each digit is either 1 or
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14 Jan 2018, 06:39
Even if you dont take the equation 100a+10b+c, statement 1 can be solved just by using variables.
let X = abc , then Y = cba Now X*Y = abc*cba abc* cba  a^2 ab ac ab b^2 bc X ac bc c^2 X X  This will give a number that looks like> (ac) (ab+bc) (a^2+b^2+c^2) (ba+bc) (ac)Now back to statement 1. It says that hundreds digit of the number we got is 6. That means a^2+b^2+c^2 = 6. Since only 2 or 1 can be used, the only combination possible is where any one of the a,b or c is 4 and the remaining two are 1. so the sum of digits will always be 4. therefore the remainder will be 1. Hence sufficient.
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Re: X is a threedigit positive integer in which each digit is either 1 or
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