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For positive integers k and n, the k-power remainder of n is defined

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For positive integers k and n, the k-power remainder of n is defined  [#permalink]

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New post 23 Apr 2015, 04:04
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For positive integers k and n, the k-power 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 3-power 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 k-power remainder of n is defined  [#permalink]

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New post 25 Apr 2015, 04:57
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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 = 1
9 = \(3^2\) + 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 = (3-1) \(3^1\) - 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 k-1 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 k-power remainder of n is defined  [#permalink]

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New post 23 Apr 2015, 09:30
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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 k-power remainder of n is defined  [#permalink]

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New post 24 Apr 2015, 10:03
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\(n\,=\,k^w\,+\,r\)

\(3\)-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
\(r\,=\,(81-1)\,-\,27\) ; \(k^w\,=\,27\,\,and\,\,k^{w+1}\,=\,81\)
\(r\,=\,(k^{w+1}-1)\,-\,k^w\)
\(r\,=\,(k^w*k)-1\,-\,k^w\)
\(r\,=\,k^w(k-1)\,-\,1\)

Answer A
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Re: For positive integers k and n, the k-power remainder of n is defined  [#permalink]

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New post 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
\(r\,=\,(81-1)\,-\,27\) ; \(k^w\,=\,27\,\,and\,\,k^{w+1}\,=\,81\)
\(r\,=\,(k^{w+1}-1)\,-\,k^w\)
\(r\,=\,(k^w*k)-1\,-\,k^w\)
\(r\,=\,k^w(k-1)\,-\,1\)

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 k-power remainder of n is defined  [#permalink]

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New post 24 Apr 2015, 20:58
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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
\(r\,=\,(81-1)\,-\,27\) ; \(k^w\,=\,27\,\,and\,\,k^{w+1}\,=\,81\)
\(r\,=\,(k^{w+1}-1)\,-\,k^w\)
\(r\,=\,(k^w*k)-1\,-\,k^w\)
\(r\,=\,k^w(k-1)\,-\,1\)

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 = K-1

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) - \(K^W\)
= \((K-1)K^W - 1\)
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Re: For positive integers k and n, the k-power remainder of n is defined  [#permalink]

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New post 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 k-power remainder of n is defined  [#permalink]

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New post 25 Apr 2015, 09:19
Thanks guys, :-D

Makes all the sense now
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Re: For positive integers k and n, the k-power remainder of n is defined  [#permalink]

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New post 27 Apr 2015, 02:10
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Bunuel wrote:
For positive integers k and n, the k-power 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 3-power 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 non-negative). 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 number-testing 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 k-power remainder of n is defined  [#permalink]

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New post 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 2.3-1=5 yes
B. k^w – 1 3-1 = 2 NO
C. (k + 1)k^w – 1 4.3-1 no
D. k^(w+1) – 1 3^2-1 no
E. (k + 1)k^(w+1) – 1 4.3^2-1 no
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Re: For positive integers k and n, the k-power remainder of n is defined  [#permalink]

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New post 19 Feb 2016, 10:37
n=kw+rn=kw+r

33-power remainder of 2727 is 00, since 2727 = 3333 + 00

33-power 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 k-power remainder of n is defined  [#permalink]

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New post 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.
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Re: For positive integers k and n, the k-power remainder of n is defined   [#permalink] 12 Jun 2019, 04:46
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