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# We define f(n) as the highest power of 7 that divides n. Wha

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We define f(n) as the highest power of 7 that divides n. Wha [#permalink]

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01 Dec 2010, 00:30
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We define f(n) as the highest power of 7 that divides n. What is the minimum value of k such that $$f(1)+f(2)+....+f(k)$$ is a positive multiple of 7 ?

(A) 7
(B) 49
(C) 56
(D) 91
(E) 98
[Reveal] Spoiler: OA

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Re: We define f(n) as the highest power of 7 that divides n. Wha [#permalink]

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01 Dec 2010, 02:03
i go with B i.e.49.

As 7 is a prime #, 7 can devide multiples of 7 only but can not devide the multiples of any other #

for n = 1 thru 6 the highest power of 7 should be 0 as 7^0 = 1 that evenly devides 1 thru 6

for n = 7 the highest power of 7 should be 1 as 7^1 = 7

again for n = 8 thru 13 the highest power of 7 shuld be 0 as 7^0 = 1

hence, every n that is multiple of 7, can be devided by 7

hence in f(0)+f(1)....f(k) , K has to be a multiple of 7

However it is also given that f(0)+f(1)....f(k) is a +ve multiple of 7
k when = 7 gives the SUM = 1 that is NOT a mupltiple of 7
k when 7*2=14 gives the SUM = 2 that is NOT a mupltiple of 7
.
.
.
when k = 7*6 give the SUM = 6 that is NOT a multiple of 7
hence K should be atleast 7^2 = 49 gives the SUM = 8
and for K>49 (precisely when K = 70) we can arrive for the SUM (precisely the SUM = 14)that is a multiple of 7

min(K)=49

ASNWER "B"

WHat is the OA.

Regards,
Murali.
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Re: We define f(n) as the highest power of 7 that divides n. Wha [#permalink]

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01 Dec 2010, 11:06
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All function f(x) values will be divisible by power of 7 as 0 except for those where x is divisible by 7.
Thus,

f(1)+........ +f(7) gives sum as 1
f(8)+.........+f(14) gives sum as 1+1=2
...till ... + f(42)............ gives total sum as 6

however at f(49), 49 is 7 power 2 to give value as 2

total = 6(1)+ 2 = 8..........
going by this way 1 keeps on adding at places 56, 63, 70, 77, 84 and 91

at f(91) we reach across total sum as 6 +2 + 6 =14 which is divisible by 7

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Re: We define f(n) as the highest power of 7 that divides n. Wha [#permalink]

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01 Dec 2010, 14:15
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shrouded1 wrote:
We define $$f(n)$$ as the highest power of 7 that divides n. What is the minimum value of k such that $$f(1)+f(2)+....+f(k)$$ is a positive multiple of 7 ?

(a) 7
(b) 49
(c) 56
(d) 91
(e) 98

Everything from $$f(1)$$ until $$f(6)$$ ist $$0$$. $$f(7)$$ is the first that equals $$1$$.
Therefore multiples of $$7$$ will all equal $$1$$ until we reach $$7^2=49$$ which eqquals $$2$$ . For easier understanding:
7-->1
14-->1
21-->1
28-->1
35-->1
42-->1
49-->2 and the sum of all so far is 8 $$(1*6+2)$$
Meaning we need 6 more multiples of 7 to reach a sum that will be divisible by 7, in the case 14 $$(14-8=6)$$
6 more multiples of 7 means that the number we are looking for is $$49+6*7=91$$

Last edited by medanova on 02 Dec 2010, 04:16, edited 1 time in total.

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Re: We define f(n) as the highest power of 7 that divides n. Wha [#permalink]

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02 Dec 2010, 00:07

F(n) will be 0 for all numbers that are not multiples of 7.

So the sum till k=7 is 1, k=14 is 2 and so on ...
When you get to k=49, the sum will jump from 6 to 8, as 49 is divisible by 7^2

So we need another 7*6 or 42 numbers to get to 8+6=14 on the sum .... So answer is 49+42=91
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Re: We define f(n) as the highest power of 7 that divides n. Wha [#permalink]

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02 Dec 2010, 04:10
medanova wrote:
shrouded1 wrote:
We define $$f(n)$$ as the highest power of 7 that divides n. What is the minimum value of k such that $$f(1)+f(2)+....+f(k)$$ is a positive multiple of 7 ?

(a) 7
(b) 49
(c) 56
(d) 91
(e) 98

Everything from $$f(1)$$ until $$f(6)$$ ist $$0$$. $$f(7)$$ is the first that equals $$1$$.
Therefore multiples of $$7$$ will all equal $$1$$ until we reach $$7^2=49$$ which eqquals $$2$$ . For easier understanding:
7-->1
14-->1
21-->1
28-->1
35-->1
42-->1
49-->2 and the sum of all so far is 8 $$(4*6+2)$$
Meaning we need 6 more multiples of 7 to reach a sum that will be divisible by 7, in the case 14 $$(14-8=6)$$
6 more multiples of 7 means that the number we are looking for is $$49+6*7=91$$

I am not good in functions...
Can somebody please guide me through this problem step by step?

Also, some kind of material for function would be of great help...

thanks..
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Re: We define f(n) as the highest power of 7 that divides n. Wha [#permalink]

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02 Dec 2010, 04:11
Is functions a topic for GMAT?
i thought GMAT never asks questions from Funvtions
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Re: We define f(n) as the highest power of 7 that divides n. Wha [#permalink]

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02 Dec 2010, 10:04
Quote:
Is functions a topic for GMAT?
i thought GMAT never asks questions from Funvtions

Yes. Functions are included in the official syllabus of GMAT.

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Re: We define f(n) as the highest power of 7 that divides n. Wha [#permalink]

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Re: We define f(n) as the highest power of 7 that divides n. Wha [#permalink]

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Re: We define f(n) as the highest power of 7 that divides n. Wha   [#permalink] 15 Aug 2017, 04:57
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