Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 14 Sep 2014, 20:35

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# K^4 is divisible by 32. What is the remainder when K is

Author Message
TAGS:
Intern
Joined: 02 Aug 2004
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 0

K^4 is divisible by 32. What is the remainder when K is [#permalink]  12 May 2005, 07:48
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
K^4 is divisible by 32. What is the remainder when K is divided by 32?
1) The remainder is 3 when K is divided by 4
2) The remainder is 2 when K is divided by 3
Manager
Joined: 22 Apr 2005
Posts: 129
Location: Los Angeles
Followers: 1

Kudos [?]: 5 [0], given: 0

Nightmare.
I'd pick E and give up.
First one doesn't make sense at all.

K^4 is divisable by 32=2^5
(1) K = 4m + 3

K^4 =(4m + 3)^4 = (4m)^4 + 4 * (4m)^3 * 3 + 6 * (4m)^2 * 3^2 + 4 * (4m) * 3^3 + 3^4

First two are divisable by 32 => 4*(4m)*3^3 + 3^4 = 3^3 (4*(4m) + 3)) - divisable by 32 => 4*(4m) + 3 is divisable by 32 => 16m = 29 (mod 32)

You just can't find an integer n so that 16m = 32n + 29 => 16(m-2n) = 29, 16 and 29 have no common divisors at all.

this does not hold. You can't find a number so that k^4 = 0 (mod 2^5) and k = 3 (mod 2^2)

Out.
VP
Joined: 13 Jun 2004
Posts: 1128
Location: London, UK
Schools: Tuck'08
Followers: 6

Kudos [?]: 24 [0], given: 0

Tyr wrote:
Nightmare

I definitely agree...it's not the first time that I see this exact problem but I still can not solve it ...shame on me
Manager
Joined: 05 May 2005
Posts: 92
Location: Kyiv, Ukraine
Followers: 1

Kudos [?]: 0 [0], given: 0

is this really going to appear in the test? my friend wrote a very complicated solution which for sure cannot take 2-3 minutes. his answer was also E. if you are interested i can post it.
Manager
Joined: 27 Jul 2004
Posts: 85
Followers: 1

Kudos [?]: 0 [0], given: 0

E
Note:- The first one does not make sense as there is no value of K that can fulfill the given condition as well as option 1 condition.
Intern
Joined: 06 Apr 2004
Posts: 42
Followers: 0

Kudos [?]: 0 [0], given: 0

1st statement indicates that K is odd but our problem states that
((k/2)^4)*(1/2). That is, K must be even to have this statement. So, statement contradicts with the problem. I guess it can't be sufficient. It's something different

2nd statement is insufficient. When k=8 (it also satisfies original statement) the remainder is 1. When k=20, the remainder is 5.

Combining statements won't suffice either. 1st one is weird

Intern
Joined: 30 Apr 2005
Posts: 11
Followers: 0

Kudos [?]: 0 [0], given: 0

answer is B :D <-- this is smilly not d :d [#permalink]  17 May 2005, 23:35
this question is bit diffy not one of the most diffy questions those who r targetting 700+ can xpect even harder questions in last five Qs
anyways coming back to the solution from statment B K can b 5,8 and so on as 5 do not fit in orginal statment so it is 8 then 8*8*8*8 leaves remainder as 0, then it cant b 11, 14.... (x*3+2 where x can b from 1...32)
32 may work but i havent checked till tht value.
_________________

i believe that all the problem solving calculations in Quantitative section have simple calculations. No Long Mahcine requiring calculations r required :D

Manager
Joined: 05 May 2005
Posts: 92
Location: Kyiv, Ukraine
Followers: 1

Kudos [?]: 0 [0], given: 0

here it is. i pasted this from an email, so the power sign "^" was gone. i tried to fix it, but if you find any more inconsistences like that, please let me know.

---------------

If you are still interested, you can refer to K. Rosen's "Discrete Mathematics," the part dealing with modular arithmetic.

The original statement
K^4 = 0 mod 32 (K^4 is divisible by 32)
can be reduced to
K = 0 mod 4 (K is divisible by 4)

Let's look at this closely.
K^4 = n * 32
K^4 = n * 2^5
K^4 = 2n * 2^4
K = rt4(2n) * 2

Since K is an integer, rt4(2n) must also be an integer. This is the case
only when n is divisible by 8, let n = 8m.
K = rt4(2*8m) * 2 = rt4(16m) * 2 = 2 * rt4(m) * 2 = 4 * rt4(m)

In this expression rt4(m) may be any integer. In other words, K = 4x. This is sufficient and necessary condition on K. The statements K=4x and K^4 is divisible by 32 are equivalent.

Let us proceed. Original statement in its new form K=4x contradicts (1).
Therefore we should not consider (1). Let us look into (2). Taken alone, (2) suggests the number set K = 2+3y: 2, 5, 8, 11, 14, etc.

Let us combine (2) with our original condition of K=4x:
K = 4x
K = 2+3y

Since 3 and 4 are mutually prime, by some theorem (maybe it's Chinese
Remainder Theorem, i'm not sure), the solution to this equation set is a
number range with a period of 3*4 = 12. The first number satisfying both
conditions is 8. All the numbers can be represented as K = 8 + 12z, where z
is some integer.

They are 8, 20, 32, 44, etc.

These numbers have different remainders by 32, for example 8 has 8, 20 has
20, and 32 has 0. It means that (2) is insufficient.

(1) and (2) taken together contradict the (0) condition because (1)

The answer is (E). Neither (1), nor (2) is sufficient.
Similar topics Replies Last post
Similar
Topics:
5 Remainder Question-K^4 6 12 Aug 2009, 20:19
K^4 is divisible by 32. What is the remainder when K is 4 05 Oct 2007, 16:11
When k^4 is divided by 32, the remainder is 0. Which of the 9 09 Jun 2007, 21:06
K^4 is divisible by 32. What is the remainder when K is 4 05 May 2006, 23:20
K is a positive integer. K^4 can be divided by 32. What can 1 26 Nov 2005, 06:35
Display posts from previous: Sort by