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If K is a positive integer, and 5^k is the factor of the [#permalink ]
07 Sep 2006, 09:59

If K is a positive integer, and 5^k is the factor of the product of the numbers from 99 to 199, what is the value of k?
1). When divided by 2, the remainder is 1
2). When divided by 3, the remainder is 2

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OK I am totally out today..dunno..maybe its lack of sleep...
I will go with E on this
lets see..
stem says 5^K is factor of 199!-99!...
anyways briefly looking at the factorial...
we have 100 which is a factor of 5^2, 105 has one 5 is one, 110 is another, 115 is another 125 has 3 factors., 120 has 2 5's...
and so..so 5^k is some value..
1) tells us that K is odd...thats its..it can 5^5, 5^3 etc...we dont know Insuff
3) tell us that K is a prime number , could be 5, 7...11..we dont know..Insuff
together...K can be 5, 7, 11 any of these values...we just dont know...Insuff

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E

Product of 99 to 199 contains

Numbers divisible by 5 = (195-100)/5 + 1 = 20

Numbers divisible by 25 = 3

Number divisible by 125 = 1

Total = 20+3+1 = 24

so k can be from 1 to 24.

St1: k could be 1,3,5,7, 9,11.....: INSUFF

St2: k could be 2,5,11,17....: INSUFF

Together:

k could be 5,11,17,23: INSUFF

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DAHIYA , I HAVE NOTHING TO SAY EXCEPT

WOW

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Q : If 5^k is a factor of P = 99 x 100 x 101 x ... x 199, what is K?
S1: k is odd...
P = 99 x 100 x 101 x ... x 199
P = 5^20 x 25 x 30 x 35 x (21 x 22 x 23 x 24 x... 39) x 99 x 101 x ... x 199)
= 5^24 x 6 x 7 x (remaining)..
= 5^24 x M
Where M is some number ...
Depending upon M we can have multiple values of K.
not sufficient.
S2: k divided by 3 has remainder 2.
k can be (3n+2) where n is an integer i.e .k = {5,8, 11, 15, 20 ..}
Not sufficient.
S1 & S2:
k is odd and of the form 3n+2
k = {5, 11, 17, 23...}
Not sufficient.
Answer: E

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Re: Good remainder problem [#permalink ]
09 Sep 2006, 22:24

yezz wrote:

If K is a positive integer, and 5^k is the factor of the product of the numbers from 99 to 199, what is the value of k? 1). When divided by 2, the remainder is 1 2). When divided by 3, the remainder is 2

Product of numbers from 99-199 can be expressed as

199!/98!

Highest power of 5 in 199! is

199/5 = 39

39/5 = 7

7/5= = 1

Total = 39+7+1 = 47

Highest power of 5 in 98! is

98/5 = 19

19/5 = 3

total = 19+3 = 22

Difference = 25 (not sure how psd gets 24....am i missing something???)

Anyway k can have a max value of 25.

From 1:

k is odd { 1,3,5,7,9,11 }

From 2:

k = 3j+2 { 2,5,8,11...}

Clearly there are values like 5, 11 satisfying both 1 and 2....E

Re: Good remainder problem
[#permalink ]
09 Sep 2006, 22:24