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# We define f(n) as the highest non-negative integer power of 7 that

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We define f(n) as the highest non-negative integer power of 7 that  [#permalink]

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01 Dec 2010, 00:30
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We define f(n) as the highest non-negative integer 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

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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 Nov 2017, 23:57
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3
soodia wrote:
can someone please explain the question?
I think I'm missing something but I don't know what!

"f(n) as the highest power of 7 that divides n." means highest power of 7 which can divide on N, when it says the highest power it could be anything

am I wrong?

We define f(n) as the highest non-negative integer 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

f(n) is the highest non-negative integer power of 7 that divides n. This means that f(n) is the highest value of x (where x is a non-negative integer) for which n/7^x is an integer. Or simply put f(n) is the value of the power of 7 in prime factorisation of n. For example:

If n = 14, then f(14) = 1 because the highest power of 7 that divides 14 is 1: 14 = 2*7^1;
If n = 15, then f(14) = 0 because the highest power of 7 that divides 14 is 0: 14 = 3*5*7^0;
...

You can notice that if n is NOT a multiple of 7, then f(n) = 0.

The question asks to find the value of k such that $$f(1)+f(2)+....+f(k)$$ is a positive multiple of 7. As discussed all values which are not multiples of 7 will give the value of f as 0. So, we should concentrate on the multiple of 7.

f(7) = 1;
f(14) = 1;
f(21) = 1;
f(28) = 1;
f(35) = 1;
f(42) = 1;
f(49) = 2 (because 49 = 7^2);

At this point the sum is $$f(1)+f(2)+....+f(49) = 1+1+1+1+1+1+2=8$$, which is not a multiple of 7.

f(56) = 1 (sum = 9);
f(63) = 1 (sum = 10);
f(70) = 1 (sum = 11);
f(77) = 1 (sum = 12);
f(84) = 1 (sum = 13);
f(91) = 1 (sum = 14 = multiple of 7).

So, basically we need thirteen multiples of 7, out of which twelve give the value of f as 1 and one gives the value of f as 2 (f(49) = 2):

$$f(1)+f(2)+....+f(91) = 1+1+1+1+1+1+2+1+1+1+1+1+1=14$$.

Hope it's clear.
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Re: We define f(n) as the highest non-negative integer power of 7 that  [#permalink]

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01 Dec 2010, 11:06
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5

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

##### General Discussion
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Re: We define f(n) as the highest non-negative integer power of 7 that  [#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.
Kudos?
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Re: We define f(n) as the highest non-negative integer power of 7 that  [#permalink]

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Updated on: 02 Dec 2010, 04:16
2
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 $$49+6*7=91$$

Originally posted by medanova on 01 Dec 2010, 14:15.
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 non-negative integer power of 7 that  [#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 non-negative integer power of 7 that  [#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 non-negative integer power of 7 that  [#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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Posts: 55
Re: We define f(n) as the highest non-negative integer power of 7 that  [#permalink]

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02 Nov 2017, 11:04
can someone please explain the question?
I think I'm missing something but I don't know what!

"f(n) as the highest power of 7 that divides n." means highest power of 7 which can divide on N, when it says the highest power it could be anything

am I wrong?
Manager
Joined: 30 Apr 2017
Posts: 55
Re: We define f(n) as the highest non-negative integer power of 7 that  [#permalink]

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03 Nov 2017, 14:58
Bunuel wrote:
soodia wrote:
can someone please explain the question?
I think I'm missing something but I don't know what!

"f(n) as the highest power of 7 that divides n." means highest power of 7 which can divide on N, when it says the highest power it could be anything

am I wrong?

We define f(n) as the highest non-negative integer 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

f(n) is the highest non-negative integer power of 7 that divides n. This means that f(n) is the highest value of x (where x is a non-negative integer) for which n/7^x is an integer. Or simply put f(n) is the value of the power of 7 in prime factorisation of n. For example:

If n = 14, then f(14) = 1 because the highest power of 7 that divides 14 is 1: 14 = 2*7^1;
If n = 15, then f(14) = 0 because the highest power of 7 that divides 14 is 0: 14 = 3*5*7^0;
...

You can notice that if n is NOT a multiple of 7, then f(n) = 0.

The question asks to find the value of k such that $$f(1)+f(2)+....+f(k)$$ is a positive multiple of 7. As discussed all values which are not multiples of 7 will give the value of f as 0. So, we should concentrate on the multiple of 7.

f(7) = 1;
f(14) = 1;
f(21) = 1;
f(28) = 1;
f(35) = 1;
f(42) = 1;
f(49) = 2 (because 49 = 7^2);

At this point the sum is $$f(1)+f(2)+....+f(49) = 1+1+1+1+1+1+2=8$$, which is not a multiple of 7.

f(56) = 1 (sum = 9);
f(63) = 1 (sum = 10);
f(70) = 1 (sum = 11);
f(77) = 1 (sum = 12);
f(84) = 1 (sum = 13);
f(91) = 1 (sum = 14 = multiple of 7).

So, basically we need thirteen multiples of 7, out of which twelve give the value of f as 1 and one gives the value of f as 2 (f(49) = 2):

$$f(1)+f(2)+....+f(91) = 1+1+1+1+1+1+2+1+1+1+1+1+1=14$$.

Hope it's clear.

wow!
I don't know how should I say thank you Bunuel
it was the amazing explanation
I am really thank you
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We define f(n) as the highest non-negative integer power of 7 that  [#permalink]

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29 Nov 2018, 13:36
shrouded1 wrote:
We define f(n) as the highest non-negative integer 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

Interesting question. Not sure if there is a better way, but I solved by brute force.

Any value of n that is not a multiple of 7 will give a value of 0 since 7 is prime.

I just started counting up

f(7) = 1
f(14) = 1
f(21) = 1
f(28) = 1
f(35) = 1
f(42) = 1
f(49) = 2 (shoot our sum went from 6 to 8 so need to keep going)
f(56) = 1
f(63) = 1
f(70) = 1
f(77) = 1
f(84) = 1
f(91) = 1 (Our sum is now 14 and divisible by 7)

We define f(n) as the highest non-negative integer power of 7 that   [#permalink] 29 Nov 2018, 13:36
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