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Manager  Joined: 05 Oct 2008
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If in a six-digit integer N , F(k) is the value of  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 42% (02:26) correct 58% (02:48) wrong based on 198 sessions

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If in a six-digit integer $$N$$, $$F(k)$$ is the value of the $$k-th$$ digit, is $$N$$ divisible by 7 (For example, $$F(4)$$ is the value of the hundreds digit of $$N$$)?

(1) $$F(1) = F(4), F(2) = F(5), F(3) = F(6)$$

(2) $$F(1) = F(2) = ... = F(6)$$

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Re: If in a six-digit integer N , F(k) is the value of  [#permalink]

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5
7
From the question stem, F(k) gives us the k-th digit in the k-th place. Since F(4) is the hundredths digit, we are counting the digits from the left for the 6-digit number N.
So F(1) gives us the 1st digit, F(2) the 2nd digit and so on.

Stmt 1.
Let $$F(1) = x, F(2) = y, F(3) = z$$, where x,y,z are any of the digits from 0 to 9.
So the first 3 digits are x,y,z.
From the statement, the next 3 digits are the same as these. So the number is the form xyzxyz (e.g. 123123 or 375375)
Now a number like 123123 can be written as $$123*1000 + 123$$
$$123123 = 123*1000 + 123 = 1001*123$$
So $$xyzxyz = 1001*xyz$$

The key thing here is that 1001 is divisible by 7. So that means the other side of the above equation is also divisible by 7 i.e. the number xyzxyz is divisible by 7.
So if the first 3 digits are the same as the last 3, it is divisible by 7.

SUFF

Stmt 2.
This says that all the digits are the same e.g. (111111,444444,55555 etc). This is just a special case of the same principle in Stmt 1. Here too, the first 3 digits are equal to last 3 digits. (in addition, all are equal)

SUFF

So D.

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

This can lead into a nice test for checking whether a number is divisible by 7. Take the digits of the number from the right three at a time and take the alternating sum. If that sum is divisible, then the number is divisible too.

E.g. Is 12348763 divisible by 7?
Take numbers in groups of 3 from the right: 12, 348, 763
Alternately add and subtract each group $$= 12-348+763 = 427$$, which is divisible by 7.
So 12348763 is divisible by 7.
General Discussion
Intern  Joined: 31 Oct 2009
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Re: If in a six-digit integer N , F(k) is the value of  [#permalink]

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The answer is D that both are sufficient, but I don't really know why...
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Re: If in a six-digit integer N , F(k) is the value of  [#permalink]

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3
1
My Solution is same as above, however I would never base my answers on a single set of sample numbers i.e. 123. Instead we can work with the number xyzxyz
this can be written as z+1000z+10y+10,000y+100x+100,000x
=1001z+1001*10y+1001*100x
=1001(z+10y+100x)

Now we can lead to the same conclusion that 1001 is divisible by 7.
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Re: If in a six-digit integer N , F(k) is the value of  [#permalink]

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Thats a nice work..

+1.

4test1 wrote:
From the question stem, F(k) gives us the k-th digit in the k-th place. Since F(4) is the hundredths digit, we are counting the digits from the left for the 6-digit number N.
So F(1) gives us the 1st digit, F(2) the 2nd digit and so on.

Stmt 1.
Let $$F(1) = x, F(2) = y, F(3) = z$$, where x,y,z are any of the digits from 0 to 9.
So the first 3 digits are x,y,z.
From the statement, the next 3 digits are the same as these. So the number is the form xyzxyz (e.g. 123123 or 375375)
Now a number like 123123 can be written as $$123*1000 + 123$$
$$123123 = 123*1000 + 123 = 1001*123$$
So $$xyzxyz = 1001*xyz$$

The key thing here is that 1001 is divisible by 7. So that means the other side of the above equation is also divisible by 7 i.e. the number xyzxyz is divisible by 7.
So if the first 3 digits are the same as the last 3, it is divisible by 7.

SUFF

Stmt 2.
This says that all the digits are the same e.g. (111111,444444,55555 etc). This is just a special case of the same principle in Stmt 1. Here too, the first 3 digits are equal to last 3 digits. (in addition, all are equal)

SUFF

So D.

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

This can lead into a nice test for checking whether a number is divisible by 7. Take the digits of the number from the right three at a time and take the alternating sum. If that sum is divisible, then the number is divisible too.

E.g. Is 12348763 divisible by 7?
Take numbers in groups of 3 from the right: 12, 348, 763
Alternately add and subtract each group $$= 12-348+763 = 427$$, which is divisible by 7.
So 12348763 is divisible by 7.

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Manager  Joined: 05 Oct 2008
Posts: 216
Re: If in a six-digit integer N , F(k) is the value of  [#permalink]

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Beware..This rule doesn't really work with all numbers. Could be a false check.

For eg: 111111

111+111= 222

222 is not divisible by 7, but 111111 is!
Retired Moderator Joined: 05 Sep 2010
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Re: If in a six-digit integer N , F(k) is the value of  [#permalink]

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if F(k) gives us the k-th digit in the k-th place then how F(4) gives hundredth place? F (4) should give thousandth place !!
Math Expert V
Joined: 02 Sep 2009
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Re: If in a six-digit integer N , F(k) is the value of  [#permalink]

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1
1
if F(k) gives us the k-th digit in the k-th place then how F(4) gives hundredth place? F (4) should give thousandth place !!

123,456
1 - HUNDRED THOUSANDS
2 - TEN THOUSANDS
3 - THOUSANDS
4 - HUNDREDS
5 - TENS
6 - UNITS
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If in a six-digit integer N , F(k) is the value of  [#permalink]

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Dear Bunuel, I answered the above question by using this way: Anumber is divisible by 7 if the difference between its units digit multiplied by 2 and the rest of the number is a multiple of 7 so I tried by using the number 523523
52352-3*2=52346
5234-6*2=5222
522-2*2=518
51-8*2=35
3-5*2=7
I found this way in Kaplan book but I need to be sure that I unberstand the way correctly and whether it is the same the way that I found in Gmat club Math book where in this book they said that take the last digit and double it and subtract it from the rest of the number, if the answer is divisible by 7 , then the number is divisible by 7. Does the last digit mean units digit?
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Math Expert V
Joined: 02 Sep 2009
Posts: 58464
Re: If in a six-digit integer N , F(k) is the value of  [#permalink]

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23a2012 wrote:
Dear Bunuel, I answered the above question by using this way: Anumber is divisible by 7 if the difference between its units digit multiplied by 2 and the rest of the number is a multiple of 7 so I tried by using the number 523523
52352-3*2=52346
5234-6*2=5222
522-2*2=518
51-8*2=35
3-5*2=7
I found this way in Kaplan book but I need to be sure that I unberstand the way correctly and whether it is the same the way that I found in Gmat club Math book where in this book they said that take the last digit and double it and subtract it from the rest of the number, if the answer is divisible by 7 , then the number is divisible by 7. Does the last digit mean units digit?

Yes, its units digit.

For example, let's check whether 1519 is divisible by 7: 151-2*9=133. Since 133 is divisible by 7 (133=7*19), then so is 1,519.
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Re: If in a six-digit integer N , F(k) is the value of  [#permalink]

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is N divisible by 7?

Statement 1: F(1) = F(4), F(2) = F(5), F(3) = F(6)

Taking any example 123,123 is divisible by 7 = yes
146,146 is divisble by 7 = yes

clearly sufficient

Statement 2:

F(1) = F(2) = F(3)........= F(6)

Clearly sufficient

Therefore D is the answer
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Re: If in a six-digit integer N , F(k) is the value of  [#permalink]

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yup, this question frequently appears in test prep.

st: abcabc = abc * 1001 is divisible 7
st 2 is true
Non-Human User Joined: 09 Sep 2013
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Re: If in a six-digit integer N , F(k) is the value of  [#permalink]

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_________________ Re: If in a six-digit integer N , F(k) is the value of   [#permalink] 27 Feb 2019, 12:34
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