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K = 123456789101112....

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by Brent@GMATPrepNow » Sat Jan 10, 2009 7:18 am
Source: Beat The GMAT Practice Questions

The integer K is created by writing the digits from 1 to 29 inclusive, so that K=1234567891011...272829
What is the remainder when K is divided by 6?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 5
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Re: K = 123456789101112....

by logitech » Sat Jan 10, 2009 7:30 am
Interesting question Brent..

Sum of the digits of this crazy integer will be

1....9 = 45
10.....19 = 45 + 10x1
20.....29= 45 + 20 x1

=135+30 = 165

So the number is actually divisible by 3

But we are asked to find the remainder when the number is divided by 6.

So we need to check both 3 and 2 divisibility.

The number end with an odd number, and every odd number that is divisible by 3 gives us 1 as a remainder when divided by 2

3
9
15
,...

So I would go with 1
LGTCH
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Re: K = 123456789101112....

by Brent@GMATPrepNow » Sat Jan 10, 2009 7:48 am
logitech wrote:Interesting question Brent..

Sum of the digits of this crazy integer will be

1....9 = 45
10.....19 = 45 + 10x1
20.....29= 45 + 20 x1

=135+30 = 165

So the number is actually divisible by 3

But we are asked to find the remainder when the number is divided by 6.

So we need to check both 3 and 2 divisibility.

The number end with an odd number, and every odd number that is divisible by 3 gives us 1 as a remainder when divided by 2

3
9
15
,...

So I would go with 1
Everything looks good right up until the end.
We know that K is divisible by 3, but since it is odd it cannot be divisible by 6.
So, what must we do to K so that it becomes even AND remains divisible by 3? We must add/subtract multiples of 3 from K. If we add (or subtract) 3 to K, the result is even AND it remains divisible by 3.
If adding 3 gives us a number that is even and divisible by 3, then the current remainder must be 3
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Re: K = 123456789101112....

by logitech » Sat Jan 10, 2009 8:00 am
Brent Hanneson wrote:
Everything looks good right up until the end.
We know that K is divisible by 3, but since it is odd it cannot be divisible by 6.
So, what must we do to K so that it becomes even AND remains divisible by 3? We must add/subtract multiples of 3 from K. If we add (or subtract) 3 to K, the result is even AND it remains divisible by 3.
If adding 3 gives us a number that is even and divisible by 3, then the current remainder must be 3
NICE! So it is almost like 129 divided by 6 - which gives us the remainder 3.

129 = 6K + R

3L = 6K + R


3L - R = 6K

AHA!!

3(L-1) = 6K


(L-1) =2K

Got it! Thank you.
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by woo » Sat Jan 10, 2009 9:52 pm
Hey, I don't get the explanations at all..

Why would you want to know the sum of all the digits in the first place? what is the logic behind?

Can either of you GMAT math geniuses answer my question?
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Re: K = 123456789101112....

by nitinrai » Sat Jan 10, 2009 10:11 pm
logitech wrote:Interesting question Brent..

Sum of the digits of this crazy integer will be

1....9 = 45
10.....19 = 45 + 10x1
20.....29= 45 + 20 x1

=135+30 = 165

So the number is actually divisible by 3

But we are asked to find the remainder when the number is divided by 6.

So we need to check both 3 and 2 divisibility.

The number end with an odd number, and every odd number that is divisible by 3 gives us 1 as a remainder when divided by 2

3
9
15
,...

So I would go with 1
Hi Logitech,

I got the solution but I have one query here for ur ur explanation. i.e.
There is formula for sum of natural number which is (n(n+1))/2 so in this case n = 29 .. hence the sum is 435..which is divisible by 3... how come u get 165.. and what is this method used...
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by cramya » Sat Jan 10, 2009 10:22 pm
Nitin/Woo,
n(n+1)/2 will work if u r adding n natural numbers

Here there is a number k which is 1,2,3........29 listed side by side

When the sum of digits of a number is divisible by 3 then the number is divisible by 3. Hence we are finding the sum of digits of the number which is 165

1,....9 -> (1+2.....+9)sum is 45

10-19

From 10 to 19 there is 10 1's in the tens digit from 10-19 so sum = 10
Again we have 1 to 9(units digit) in the 10....19 sos sum = 45

20-29
ten 2's in the tens digit= 20

Again 1-9 in the units digit = 45

Sum=165 divisible by 3


What Brent is saying is this number k (ODD) is divisible by 3 but not by 2

If u add 1 to this number the number will be divisible by 2 since odd+odd = EVEN BUT IT WILL NOT BE DIVISIBLE BY 3

So u have to add 3 to the number again k(odd) + 3(odd) = even. This number will be divisible by 3 and 2. So the original number must leave a remainder of 3 when divided by 6



Easier example that u can related to big number k above:

Take k=9

Whats the remainder when k is divided by 6? Answer is 3

9 was divisible by 3 and not by 2. If we added 1 to it then its divisible by 2 but not by 3

We have to add 3 (9+3=12) to be divisible by 3 and 2 therefore by 6 so the original number 9 must have left a remainder of 3 when divided by 6

Hope this helps!
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by nitinrai » Sat Jan 10, 2009 11:32 pm
Thanks Cramya.. for GR8 explaination
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by logitech » Sun Jan 11, 2009 1:09 am
nitinrai wrote:Thanks Cramya.. for GR8 explaination
Cramya in the house. Master explained everything in details.
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