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For positive integers k and n, the kpower remainder of n is defined
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23 Apr 2015, 04:04
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For positive integers k and n, the kpower remainder of n is defined as r in the following equation: \(n = k^w + r\), where w is the largest integer such that r is not negative. For instance, the 3power remainder of 13 is 4, since 13 = 3^2 + 4. In terms of k and w, what is the largest possible value of r that satisfies the given conditions? A. \((k – 1)k^w – 1\) B. \(k^w – 1\) C. \((k + 1)k^w – 1\) D. \(k^{(w+1)} – 1\) E. \((k + 1)k^{(w+1)} – 1\) Kudos for a correct solution.
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Re: For positive integers k and n, the kpower remainder of n is defined
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25 Apr 2015, 04:57
I found the variable substitution method to be easier than the pure algebraic method , assuming my line of reasoning is correct ! Given equation ==> n = \(k^w\) + r Try a few values to get the hang of what is given in the question stem. 3 = \(3^1\) + 0 4 = \(3^1\) + 1 5 = \(3^1\) + 2 6 = \(3^1\) + 3 7 = \(3^1\) + 4 8 = \(3^1\) + 5 .. Here 5 is the largest value of r . ==> r = 5 , k = 3 , w = 19 = \(3^2\) + 0 Substituting the values of k = 3 and w= 1 in the answer choices, we should get the value of r = 5 (A.) (k – 1)k^w – 1 = (31) \(3^1\)  1 = 2 * 3  1 = 5 (Bingo !) (B.) k^w – 1 = \(3^1\)  1 = 2 (Oops !) (C.) (k + 1)k^w – 1 = (3 + 1) \(3^1\)  1 = 4 * 3  1 = 11 (Oops !) (D.) k^(w+1) – 1 = \(3^2\)  1 = 8 (Oops !) (E.) (k + 1)k^(w+1) – 1 = 4 * \(3^2\)  1 = 36  1 = 35 (Oops !) Correct Answer should be (A)P.S. Beware ! if you are using the powers of 2, we will get stuck between options A and B because k= 2 and k1 will always be 1. Therefore, (k – 1)k^w – 1 will be equal to k^w – 1 .




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Re: For positive integers k and n, the kpower remainder of n is defined
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23 Apr 2015, 09:30
n = k^w + r, where w is the largest integer such that r is not negative This means k^w <= n, and k^(w+1) > n So n lies between k^w and k^(w+1)
Now, r = n  k^w, which means r is the distance between n and k^w This distance is maximised at the highest possible value of n, which can be just below k^(w+1) So n = k^(w+1)  1, as n is an integer
Therefore, highest value of r = n  k^w = k^(w+1)  1  k^w = (k  1)k^w  1
A is the correct choice here.



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Re: For positive integers k and n, the kpower remainder of n is defined
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24 Apr 2015, 10:03
\(n\,=\,k^w\,+\,r\)
\(3\)power remainder of \(27\) is \(0\), since \(27\) = \(3^3\) + \(0\)
\(3\)power remainder of \(80\) is \(53\), since \(80\) = \(3^3\) + \(53\) ; here remainder \(r\) is largest \(r\,=\,(811)\,\,27\) ; \(k^w\,=\,27\,\,and\,\,k^{w+1}\,=\,81\) \(r\,=\,(k^{w+1}1)\,\,k^w\) \(r\,=\,(k^w*k)1\,\,k^w\) \(r\,=\,k^w(k1)\,\,1\)
Answer A



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Re: For positive integers k and n, the kpower remainder of n is defined
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24 Apr 2015, 16:09
sudh wrote: \(n\,=\,k^w\,+\,r\)
\(3\)power remainder of \(27\) is \(0\), since \(27\) = \(3^3\) + \(0\)
\(3\)power remainder of \(80\) is \(53\), since \(80\) = \(3^3\) + \(53\) ; here remainder \(r\) is largest \(r\,=\,(811)\,\,27\) ; \(k^w\,=\,27\,\,and\,\,k^{w+1}\,=\,81\) \(r\,=\,(k^{w+1}1)\,\,k^w\) \(r\,=\,(k^w*k)1\,\,k^w\) \(r\,=\,k^w(k1)\,\,1\)
Answer A Where did you get 80 from? How did you know to pick 80? I am not understanding this question and how we are picking values for k and w.



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Re: For positive integers k and n, the kpower remainder of n is defined
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24 Apr 2015, 20:58
peachfuzz wrote: sudh wrote: \(n\,=\,k^w\,+\,r\)
\(3\)power remainder of \(27\) is \(0\), since \(27\) = \(3^3\) + \(0\)
\(3\)power remainder of \(80\) is \(53\), since \(80\) = \(3^3\) + \(53\) ; here remainder \(r\) is largest \(r\,=\,(811)\,\,27\) ; \(k^w\,=\,27\,\,and\,\,k^{w+1}\,=\,81\) \(r\,=\,(k^{w+1}1)\,\,k^w\) \(r\,=\,(k^w*k)1\,\,k^w\) \(r\,=\,k^w(k1)\,\,1\)
Answer A Where did you get 80 from? How did you know to pick 80? I am not understanding this question and how we are picking values for k and w. Normal method of division: \(N_{dividend}\) = \(K_{divisor}\)*\(W_{quotient}\) + \(R\); \(0\leq remainder\,<\,divisor\) i.e. \(27\) = \(3_{divisor}\)*\(9_{quotient}\) + \(0\) if we want to maximize the remainder with the same quotient and divisor, then the dividend should be 29, i.e. \(29\) = \(3_{divisor}\)*\(9_{quotient}\) + \(2\); (Note 30/3 gives a remainder 0) Generalizing the above 27 = 3*9 = K*W 29 = 3*(9+1)  1 = K*(W+1)  1 so maximum remainder can be obtained by R = 29  27 =K*(W+1)  1  K*W = K1 In the problem above quotient \(W\) is expressed in terms of power, so \(27\) = \(3^3\)+\(0\) = \(K^W\)+\(R\) ; R = 0 To maximize the remainder increase the quotient by one and minus one from the result, \(80\) = \(3^3\)+\(53\) = \(K^W\)+\(R\) ; R = 53; 81/3 (\(\frac{3^4}{3}\)) gives remainder of 0 27 = \(3^3\) = \(K^W\) 80 = \(3^{3+1}1\) = (\(K^{W+1}\)1) Remainder 53 = 80  27 = (\(K^{W+1}\)1)  \(K^W\) = \((K1)K^W  1\)



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Re: For positive integers k and n, the kpower remainder of n is defined
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24 Apr 2015, 21:06
Hi peachfuzz peachfuzz wrote: I am not understanding this question and how we are picking values for k and w. We don't need to pick any values of k and w in order to solve this question. The equation says \(n = k^w + r\), and then asks us to find the largest value of r in terms of k and w. So we can treat k and w as constant and vary the value of n in order to find the highest r. This is what I have posted in reply earlier. Picking values of k and w isn't needed because our final answer for r needs to be in terms of k and w anyway. For any value of k and w, we can find the highest r. When you take k and w as 3 and 3, in order to get the highest r, n will have to be k^(w+1)  1 = 3^4  1 = 80, that's all. Hope that clears your doubt.



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Re: For positive integers k and n, the kpower remainder of n is defined
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25 Apr 2015, 09:19
Thanks guys, Makes all the sense now



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Re: For positive integers k and n, the kpower remainder of n is defined
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27 Apr 2015, 02:10
Bunuel wrote: For positive integers k and n, the kpower remainder of n is defined as r in the following equation: n = k^w + r, where w is the largest integer such that r is not negative. For instance, the 3power remainder of 13 is 4, since 13 = 3^2 + 4. In terms of k and w, what is the largest possible value of r that satisfies the given conditions?
A. (k – 1)k^w – 1 B. k^w – 1 C. (k + 1)k^w – 1 D. k^(w+1) – 1 E. (k + 1)k^(w+1) – 1
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION:We are told that n = k^w + r, with a number of conditions on the possible values of the variables (k and n are positive integers, w is an integer, and r is nonnegative). An example is given to us: 13 = 3^2 + 4. The question is this: for a given k and w, what is the largest possible value of r? Let’s keep with the example. The given k and w are 3 and 2, respectively, in the expression 3^2. The question becomes “how big can r get?” At first, it might seem that there’s no cap on the size of r, but if you consider n = 30, for instance, you can write it using a larger power of 3: 30 = 3^2 + 21 30 = 3^3 + 3 So the first equation doesn’t fit the conditions (w has to be the largest integer such that r is not negative). The tipping point is the next power of 3, namely 3^3 = 27. 27 itself would be written as 3^3 + 0, with r = 0, so the number that gives the largest r for k = 3 and w = 2 must be 26: 26 = 3^2 + 17, giving r = 17. At this point, we could take a numbertesting approach: which answer choice equals 17 when k = 3 and w = 2? After a little computation, we’d find that the answer is (A). We can also take a more algebraic approach. The maximum r is going to come when n is the integer just below the next power of k above k^w, in other words when n equals k^(w+1) – 1. Plug this expression for n into the equation: k^(w+1) – 1 = k^w + r Now split k^(w+1) into k*k^w: k*k^w – 1 = k^w + r Finally, subtract k^w from both sides: (k – 1)(k^w) – 1 = r The left side matches the expression in choice (A). The correct answer is A.
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Re: For positive integers k and n, the kpower remainder of n is defined
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28 Apr 2015, 20:38
alt approach,Bunuel plz let me know if it flawed
in order to get max = r either put (n=8 , r=5,k=3,w=1) or (n=26, r=17, k=3,w=2) or (n=24, r=19,k=5,w=1) only A will give the desired result
I am taking first case (n=8 , r=5,k=3,w=1) , you can take any case A. (k – 1)k^w – 1 2.31=5 yes B. k^w – 1 31 = 2 NO C. (k + 1)k^w – 1 4.31 no D. k^(w+1) – 1 3^21 no E. (k + 1)k^(w+1) – 1 4.3^21 no



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Re: For positive integers k and n, the kpower remainder of n is defined
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19 Feb 2016, 10:37
n=kw+rn=kw+r 33power remainder of 2727 is 00, since 2727 = 3333 + 00 33power remainder of 8080 is 5353, since 8080 = 3333 + 5353 ; here remainder rr is largest r=(81−1)−27r=(81−1)−27 ; kw=27andkw+1=81kw=27andkw+1=81 r=(kw+1−1)−kwr=(kw+1−1)−kw r=(kw∗k)−1−kwr=(kw∗k)−1−kw r=kw(k−1)−1
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Re: For positive integers k and n, the kpower remainder of n is defined
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12 Jun 2019, 04:46
Let the number be n= k^w+r For r to be greatest and non negative is possible if we add 1 to the number and the new number becomes n'= k^(w+1) ie the r is zero .
Hence k^w+r+1= k^(w+1) On further calculation we get the Answer A.




Re: For positive integers k and n, the kpower remainder of n is defined
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