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If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai
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14 Jul 2011, 01:45
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If \(x = (16^3+17^3+18^3+19^3)\), then x when divided by 70 leaves a remainder of: A. 0 B. 1 C. 69 D. 35 E. 42
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Re: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai
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14 Jul 2011, 02:21
AnkitK wrote: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remainder of :
OA. 0 We know: \(a^3 + b^3 = (a + b)*(a^2  ab + b^2)\) \(16^3 + 19^3 = (35)(16^2  16*19 + 19^2)\) \(17^3 + 18^3 = (35)(17^2  17*18 + 18^2)\) Therefore, \((16^3+17^3+18^3+19^3) = 35(16^2  16*19 + 19^2 + 17^2  17*18 + 18^2)\) \((16^2  16*19 + 19^2 + 17^2  17*18 + 18^2)\) is a multiple of 2 since we have (Even  Even + Odd + Odd  Even + Even). Hence the entire expression is Even. Therefore, x is divisible by 70. Why did I think of algebraic identities?  because exponents were 3 for all the terms and there seemed to be no +1, 1 symmetry. The numbers weren't close to a multiple of 70 and were too big. Most people don't know the squares/cubes of 16, 17 etc.
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Re: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai
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14 Jul 2011, 02:30
AnkitK wrote: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remainder of :
OA. 0 Hi Ankit, I dont think GMAT will ask this , but to answer your question , the best way to do this is a^+b^3 = (a+b)(a^2 ab +b^2) ( 16^3+19^3) + (17^3+18^3) (16+17)( 16^2 16*19+19^2) + (17+18)(17^217*18 +18^2) 35*( 16^2 16*19+19^2) (17^217*18 +18^2) . so we can re write the equation as 35^(even even+odd)(eveneven+odd) odd*odd = even = atleast one 2 35*2 (...........) (...........) when this equation is divided by 70 , we will have 0 as the remainder. hope this helps.



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Re: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai
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15 Jul 2011, 23:33
AnkitK wrote: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remainder of :
OA. 0 A good to know question IMO but still I would take it this way At first glance I read a symmetry here as I break the \(A^3 + B^3\)formula pair of 16, 19 and of 17,18 give me a total of 35 (something useful considering I am looking for divisibility by 70. Now my job is half done but I have still to check whether the remaining is even  to check for divisibility by 2). after opening the formula \(A^3 + B^3\) nd rearranging terms one would get Lets recall  \((A^3 + B^3) = (A+B)(A^2 + B^2  A*B)\).. so pair 1 gives \((16^2 + 19^2  16*19 ) (16+19) = 35 * (16^2 + 19^2  16*19 )\) pair 2 gives \((17^2 + 18^2  17*18) (17+18) = 35 * (17^2 + 18^2  17*18)\) now lets add them, so we have got now \((35) * (16^2 + 17^2 + 18^2 + 19^2  16*19  17*18) \) to get down finally we need to determine whether \((16^2 + 17^2 + 18^2 + 19^2  16*19  17*18)\) is divisible by 2  (Even + Odd + Even + Odd  Even  Even). You see we are done here (obviously remainder is big ZERO
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Re: If x = (16^3 + 17^3 + 18^3 + 19^3), then x divided by 70 leaves remai
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11 Sep 2018, 22:16
rahul16singh28 wrote: If \(x = (16^3 + 17^3 + 18^3 + 19^3)\), then x divided by 70 leaves remainder of
1. 0 2. 1 3. 69 4. 35 5. 42 Note: \(a^n+b^n\) is divisible by (a+b) when n is ODD. In line with the above, \(a^3+b^3\) is divisible by (a+b) a) \(17^3 + 18^3\) is divisible by 17+18=35 b) \(16^3 + 19^3\) is divisible by 16+19=35 c) Sum of unit digits of 17^3 + 18^3, 3+2=5, Sum of unit digits of 16^3 + 19^3, 6+9=15. So the given expression is an EVEN number Hence from (a),(b), and (c), we have \(x = (16^3 + 17^3 + 18^3 + 19^3)\)=\(17^3 + 18^3+16^3 + 19^3=35k+35p=35*2*y=70y\) So, the given expression when divided by 70 leaves a remainder 0. Ans. (A) Another approach: \(a^n+b^n+c^n+d^n\) is divisible by (a+b+c+d) when n is ODD , a, b, c, and d are in A.P.(Arithmetic progression) Here 16,17,18, and 19 are in AP. n=3, which is ODD’ Hence, \(x = (16^3 + 17^3 + 18^3 + 19^3)\) is divisible by (16+17+18+19=70) Therefore, remainder is zero.
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Re: If x = (16^3 + 17^3 + 18^3 + 19^3), then x divided by 70 leaves remai
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11 Sep 2018, 22:36
a^3+b^3=(a+b)(a^2+b^2ab) 16^3+19^3=35k 17^3+18^3=35k So x=35k +35k=70k 1) is the answer



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Re: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai
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12 Sep 2018, 03:43
VeritasKarishma wrote: AnkitK wrote: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remainder of :
OA. 0 We know: \(a^3 + b^3 = (a + b)*(a^2  ab + b^2)\) \(16^3 + 19^3 = (35)(16^2  16*19 + 19^2)\) \(17^3 + 18^3 = (35)(17^2  17*18 + 18^2)\) Therefore, \((16^3+17^3+18^3+19^3) = 35(16^2  16*19 + 19^2 + 17^2  17*18 + 18^2)\) \((16^2  16*19 + 19^2 + 17^2  17*18 + 18^2)\) is a multiple of 2 since we have (Even  Even + Odd + Odd  Even + Even). Hence the entire expression is Even. Therefore, x is divisible by 70. Why did I think of algebraic identities?  because exponents were 3 for all the terms and there seemed to be no +1, 1 symmetry. The numbers weren't close to a multiple of 70 and were too big. Most people don't know the squares/cubes of 16, 17 etc. Hi.... would the following logic work? 16^3 + 17^3 is iv by 33 18^3 + 19^3 is div by 37 therefore 16^3 + 17^3 + 18^3 + 19^3 is div by 33+37 or 70 regards



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Re: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai
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12 Sep 2018, 05:23
Mansoor50 wrote: VeritasKarishma wrote: AnkitK wrote: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remainder of :
OA. 0 We know: \(a^3 + b^3 = (a + b)*(a^2  ab + b^2)\) \(16^3 + 19^3 = (35)(16^2  16*19 + 19^2)\) \(17^3 + 18^3 = (35)(17^2  17*18 + 18^2)\) Therefore, \((16^3+17^3+18^3+19^3) = 35(16^2  16*19 + 19^2 + 17^2  17*18 + 18^2)\) \((16^2  16*19 + 19^2 + 17^2  17*18 + 18^2)\) is a multiple of 2 since we have (Even  Even + Odd + Odd  Even + Even). Hence the entire expression is Even. Therefore, x is divisible by 70. Why did I think of algebraic identities?  because exponents were 3 for all the terms and there seemed to be no +1, 1 symmetry. The numbers weren't close to a multiple of 70 and were too big. Most people don't know the squares/cubes of 16, 17 etc. Hi.... would the following logic work? 16^3 + 17^3 is iv by 33 18^3 + 19^3 is div by 37 therefore 16^3 + 17^3 + 18^3 + 19^3 is div by 33+37 or 70 regards Think about it: a is divisible by 3 (say a = 6) and b is divisible by 5 (say b = 5) Does this mean (a + b = 11) is divisible by 8? It is certainly not necessary.
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Re: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai
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12 Sep 2018, 06:04
VeritasKarishma wrote: Mansoor50 wrote: AnkitK wrote: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remainder of :
Hi....
would the following logic work?
16^3 + 17^3 is iv by 33 18^3 + 19^3 is div by 37
therefore
16^3 + 17^3 + 18^3 + 19^3 is div by 33+37 or 70
regards Think about it: a is divisible by 3 (say a = 6) and b is divisible by 5 (say b = 5) Does this mean (a + b = 11) is divisible by 8? It is certainly not necessary. Thanks!!! i should have tested the assumption myself....!!



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Re: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai
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12 Sep 2018, 09:42
This question can be solved by focusing on only the units digit of x and the answer choices. The units digit of the answer choices are all unique. So, by knowing only the units digit of x we can know the units digit of the right answer. This would NOT work if the units digit of the answer choices were all the same  in this case say 0, 10, 20, 30, and 40. But, of course, the GMAT is a test of your ability to think and simplify, not of your ability to do a bunch of arithmetic, so this is the type of thing you want to be on the lookout for. So in this case The units digit of 16 cubed is 6 The units digit of 17 cubed is 3 The units digit of 18 cubed is 2 The units digit of 19 cubed is 9 6 + 3 + 2 + 9 = 20. So, x has a units digit of 0. So, the remainder of x divided by 70 will also have a units digit of 0. A is the only answer choices that works. A is correct. Think before you math
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Re: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai
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