If in a six-digit integer n, f(k) represents the value of the

Question

If in a six-digit integer  n ,  f(k)  represents the value of the  k_{th}  digit from the leftmost digit, is  n  is divisible by 7? (For example,  f(4)  corresponds to 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(3) = f(4) = f(5) = f(6)

M15-29

4test1 · Accepted Answer

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.

atish · Answer

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.

aditya8062 · Answer

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 !!

Bunuel · Answer

123,456
1 - HUNDRED THOUSANDS
2 - TEN THOUSANDS
3 - THOUSANDS
 4 - HUNDREDS 
5 - TENS
6 - UNITS

23a2012 · Answer

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?

Bunuel · Answer

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.

Bunuel · Answer

Official Solution:

If in a six-digit integer  n ,  f(k)  represents the value of the  k_{th}  digit from the leftmost digit, is  n  is divisible by 7? (For example,  f(4)  corresponds to the value of the hundreds digit of  n .) 
 
(1)  f(1) = f(4), f(2) = f(5), f(3) = f(6) 
 
The above condition implies that the last three digits of  n  are the same as the first three digits of  n . Therefore,  n = abc,abc = abc*1000 + abc = abc*(1000 + 1) = abc*1001 . Since 1001 is divisible by 7 (1,001 = 143*7), then  n = abc*1001  must also be divisible by 7. Sufficient.
 
(2)  f(1) = f(2) = f(3) = f(4) = f(5) = f(6) 
 
The above condition implies that all digits of  n  are the same. Therefore,  n = aaa,aaa = aaa*1000 + aaa = aaa*(1000 + 1) = aaa*1001 . Since 1001 is divisible by 7 (1,001 = 143*7), then  n = aaa*1001  must also be divisible by 7. Sufficient.

Answer: D

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If in a six-digit integer n, f(k) represents the value of the

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