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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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27% (01:50) correct
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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 ?

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

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.

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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