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

Question

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

AVRonaldo · Accepted Answer

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 .

PrepTap · Answer

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.

sudh · Answer

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

peachfuzz · Answer

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.

sudh · Answer

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

PrepTap · Answer

Hi peachfuzz

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.

peachfuzz · Answer

Thanks guys, :-D

Makes all the sense now

Bunuel · Answer

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.

vipulgoel · Answer

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

Lorenzo3688 · Answer

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

msali1983 · Answer

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.

suu007 · Answer

equation is n=k^w+r . In order to solve it quickly , take any constant value of k and w . 
It can be 2^2 or 3^3 or ....
Let's take 2^2 for our understanding.
r will be minimum when value is 0. So the equation can be written as -
4=2^2+0
r will be maximum when value is= 2^2 -1= 3. So the equation can be written as-
7=2^2+3
so, 
maximum value (3)= 7 - 2^2
 = (2^3 -1) -2^2
 r= k^(w+1) -1 -k^w
 r= (k–1)k^w–1.....(A)

naA2025 · Answer

n=k^w+r
k^w<n<k^(w+1)
0<n-k^w<k^(w+1)-k^w
0<r<(k-1)k^w
Therefore, r (max)=(k-1)k^w-1

MeghaYadav01 · Answer

Can a question like this actually come in the exam

bumpbot · Answer

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

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