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If in a sixdigit integer N , F(k) is the value of
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02 Dec 2009, 11:52
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If in a sixdigit integer \(N\), \(F(k)\) is the value of the \(kth\) 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) \(F(1) = F(2) = ... = F(6)\) M1529
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Re: If in a sixdigit integer N , F(k) is the value of
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03 Dec 2009, 23:23
From the question stem, F(k) gives us the kth digit in the kth place. Since F(4) is the hundredths digit, we are counting the digits from the left for the 6digit 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 \(= 12348+763 = 427\), which is divisible by 7. So 12348763 is divisible by 7.




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Re: If in a sixdigit integer N , F(k) is the value of
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02 Dec 2009, 14:24
The answer is D that both are sufficient, but I don't really know why...



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Re: If in a sixdigit integer N , F(k) is the value of
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04 Dec 2009, 13:12
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 sixdigit integer N , F(k) is the value of
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05 Dec 2009, 11:52
Thats a nice work.. +1. 4test1 wrote: From the question stem, F(k) gives us the kth digit in the kth place. Since F(4) is the hundredths digit, we are counting the digits from the left for the 6digit 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 \(= 12348+763 = 427\), which is divisible by 7. So 12348763 is divisible by 7.



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Re: If in a sixdigit integer N , F(k) is the value of
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24 May 2010, 00:32
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!



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Re: If in a sixdigit integer N , F(k) is the value of
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07 Nov 2014, 05:17
if F(k) gives us the kth digit in the kth place then how F(4) gives hundredth place? F (4) should give thousandth place !!



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Re: If in a sixdigit integer N , F(k) is the value of
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07 Nov 2014, 05:28
aditya8062 wrote: if F(k) gives us the kth digit in the kth 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  HUNDREDS5  TENS 6  UNITS
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If in a sixdigit integer N , F(k) is the value of
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27 Jan 2015, 13:32
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 523523*2=52346 52346*2=5222 5222*2=518 518*2=35 35*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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Re: If in a sixdigit integer N , F(k) is the value of
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28 Jan 2015, 00:52
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 523523*2=52346 52346*2=5222 5222*2=518 518*2=35 35*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: 1512*9=133. Since 133 is divisible by 7 (133=7*19), then so is 1,519.
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Re: If in a sixdigit integer N , F(k) is the value of
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24 Feb 2016, 15:42
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 sixdigit integer N , F(k) is the value of
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29 Sep 2017, 09:42
yup, this question frequently appears in test prep.
st: abcabc = abc * 1001 is divisible 7 st 2 is true



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Re: If in a sixdigit integer N , F(k) is the value of
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28 Mar 2020, 03:18
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Re: If in a sixdigit integer N , F(k) is the value of
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