What is the remainder when 10^49 + 2 is divided by 11?

the property of 11 as divisor is (sum of odd digits -sum of even digits)=0 starting from rightmost digit..
this means for 10^even number, (sum of odd digits -sum of even digits) =1-0=1 so 1 is the remainder...
and for 10^odd number, (sum of odd digits -sum of even digits) =0-1=-1 so 10 is the remainder...
therefore 10^49 will have a remainder of 10.. so ans =12/11 ie rem=1 ans A

Hi quitdreaming,

You have to be very careful with your generalizations. If you're "off", even a little bit, then you'll likely get whatever question you're working on wrong.

With this calculation, you ARE correct:

100/11 = 9r1

But with this one, your generalization is incorrect:

1,000/11 = 90r10 (NOT remainder 1)

Knowing this, what would you do differently to solve this problem?

GMAT assassins aren't born, they're made,
Rich

VERITAS PREP OFFICIAL SOLUTION:

With exponent-based problems and huge numbers, it's often helpful to try to establish a pattern using small numbers. On this problem, while 1049+2 is a massive number that you'd never want to try to perform calculations with, you can start by using smaller numbers to get a feel for what it would look like:

10^2+2=102, which when divided by 11 produces a remainder of 3 (as 9 * 11 = 99, leaving 3 left over)

10^3+2=1002, which when divided by 11 produces a remainder of 1 (as 9 * 110 = 990, and when you add one more 11 to that you get to 1001, leaving one left)

10^4+2=10002, which when divided by 11 produces a remainder of 3 (as you can get to 9999 as a multiple of 11, which would leave 3 left over)

10^5+2=100002, which when divided by 11 produces a remainder of 1 (as you can get to 99990 and then add 11 more, bringing you to 100001 leaving one left over).

By this point, you should see that the pattern will repeat, meaning that when 10 has an even exponent the remainder is 3 and when it has an odd exponent the remainder is 1. Therefore, since 10 has an odd exponent in the problem, the remainder will be 1.

what is the problem in solving indivitually like this..

10^49/11 + 2/11
from first part..we get 1 (odd places are 0 and even places are 9, as per cyclicity)
from second part, we get 2

1+2=3 =rem..
kindly correct me (we have used this line of method in example 1 also in math remainders theory..so wats the difference)

Thanku in advance

hi shreygupta3192,
your approach is correct and can be done the way you have solved...
however you have to be careful while finding remainder..
when you divide 10^n by 11, the remainder will be 1 or -1, that is 10..... it will depend on n..
when you divide 10 by 11.. remainder is 10 and not 1... you add odd and even from right most digit and then subtract even from odd..
in case of 10... 0-1=-1or 10.... similarily 10^49 will give a remainder of 10..
overall remainder=10+2=12, which will give remainder as 1 when divided by 11..
hope it is clear..

chetan

Without using cyclity:

\(10^{49} + 2\)
\(=(10^2)^{24}*10 + 2\)
\(=100^{24}*10 + 2\)

100/11 leaves remainder 1

\(=1^{24}*10+2\)
\(=12\)

12/11 leaves remainder 1
\(=1\)

Time invested on this question was worth.

NOTE FOR SELF.

Sometimes try finding the pattern and it will be easier to find the correct answer.

Would the following approach be correct?

(10)^49= (11-1)^49

Thus the remainer of (11-1)^49 will be -1

and (-1+2)/11 implies a remainer of 1