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
K = 123456789101112....
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- Brent@GMATPrepNow
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- logitech
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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
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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"DON'T LET ANYONE STEAL YOUR DREAM!"
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Everything looks good right up until the end.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
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
- logitech
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NICE! So it is almost like 129 divided by 6 - which gives us the remainder 3.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
129 = 6K + R
3L = 6K + R
3L - R = 6K
AHA!!
3(L-1) = 6K
(L-1) =2K
Got it! Thank you.
LGTCH
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Hi Logitech,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
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...
I firmly believe that one can succeed at almost anything for which one has unlimited enthusiasm.
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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!
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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Cramya in the house. Master explained everything in details.nitinrai wrote:Thanks Cramya.. for GR8 explaination
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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"DON'T LET ANYONE STEAL YOUR DREAM!"