what is the remainder ?
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rossmj
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Generally these questions come down to prime factorization. This is a tough question the first thing to notice is that we will be adding 2 evens to 2 odds and thus will get an even number that will cancel to a large number divided by 29, which eliminates A, and C. Beyond that I would have to guess, I'm sure I'm missing a shortcut but within the time confines of the test I would take my chances.
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dwilliams05
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adding exponents is tricky but I believe that if the bases aer different yet the exponents ar the same, then you can simply add the bases together but will need to keep the exponenet the same.
in this question 13-14-15-16=58. being that you are dividing by 58, there will be no remainder (58^7/58 = 58^6). I say the answer is D.
in this question 13-14-15-16=58. being that you are dividing by 58, there will be no remainder (58^7/58 = 58^6). I say the answer is D.
thnx Ross n Williams
I think we cant just add the bases if exponents are same .
i.e., 2^3 + 3^3 is not eq to 5^3
But what i have figured out is , the result of 2^3 + 3^3 would definately be a multiple of 2+3 i.e 5
anyways in this case remainder is 0
Ans is D
I think we cant just add the bases if exponents are same .
i.e., 2^3 + 3^3 is not eq to 5^3
But what i have figured out is , the result of 2^3 + 3^3 would definately be a multiple of 2+3 i.e 5
anyways in this case remainder is 0
Ans is D
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iamcste
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dude, this worked for a^3+b^3 as a^3+b^3=(a+b)(a^2+...) hence a^3+b^3 was divisble by a+bmavesum wrote:
But what i have figured out is , the result of 2^3 + 3^3 would definately be a multiple of 2+3 i.e 5
anyways in this case remainder is 0
Ans is D
this wont work for other cases. Also, whats the source for the problem
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Ian Stewart
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It actually works for any odd power. For example:iamcste wrote:dude, this worked for a^3+b^3 as a^3+b^3=(a+b)(a^2+...) hence a^3+b^3 was divisble by a+bmavesum wrote:
But what i have figured out is , the result of 2^3 + 3^3 would definately be a multiple of 2+3 i.e 5
anyways in this case remainder is 0
Ans is D
this wont work for other cases. Also, whats the source for the problem
x^7 + y^7 = (x + y)(x^6 - (x^5)y + (x^4)(y^2) - (x^3)(y^3) + (x^2)(y^4) - x(y^5) + y^6)
That's not a factorization you'll ever need on the GMAT, but you can use it for this question:
13^7 + 14^7 + 15^7 + 16^7
= 13^7 + 16^7 + 14^7 + 15^7
= (13 + 16)*(a lot of terms) + (14 + 15)*(a lot of terms)
= 29*(a lot of terms plus a lot of terms)
So 13^7 + 14^7 + 15^7 + 16^7 is a multiple of 29, and since it's even (odd + even + odd + even = even), it's a multiple of 58, and the remainder will be zero when it's divided by 58.
It's not the kind of question you'll see on the GMAT, though.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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