Reminder addition and subtraction........Expert needed

[Mo2men]

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

A. 1
B. 2
C. 3
D. 5
E. 7

OA:A

Source: Veritas

I solved the problem by this way but the answer is not presented here

(10^49 + 2)/11= 10^49/11 +2/11

Studying each term:
10/11 = reminder is 10
10^2/11= reminder is 1
10^3/11= reminder is 10

so 10^49/11 == reminder is 10
and 2/11 =reminder is 2

so total reminder is 10+2= 12 but the reminder can not be greater than the divisor.

Is adding reminders incorrect?


Any comment please??

[GMATGuruNY]

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

A. 1
B. 2
C. 3
D. 5
E. 7

Generally, remainders repeat in a CYCLE.
(10¹ + 2)/11 = 12/11 = 1 R1.
(10² + 2)/11 = 102/11 = 9 R3.
(10³ + 2)/11 = 1002/11 = 91 R1.
(10� + 2)/11 = 10002/11 = 909 R3.
Here, the remainders ALTERNATE between 1 and 3.
More specifically:
Every ODD exponent yields a remainder of 1.
Every EVEN exponent yields a remainder of 3.
Since 10�� + 2 includes an odd exponent, it will yield a remainder of 1.

The correct answer is A.

[Mo2men]

Generally, remainders repeat in a CYCLE.
(10¹ + 2)/11 = 12/11 = 1 R1.
(10² + 2)/11 = 102/11 = 9 R3.
(10³ + 2)/11 = 1002/11 = 91 R1.
(10� + 2)/11 = 10002/11 = 909 R3.
Here, the remainders ALTERNATE between 1 and 3.
More specifically:
Every ODD exponent yields a remainder of 1.
Every EVEN exponent yields a remainder of 3.
Since 10�� + 2 includes an odd exponent, it will yield a remainder of 1.

The correct answer is A.

Can I subtract or add reminders? what is wrong with my answer above?

Thanks in advance for your help

[GMATGuruNY]

I solved the problem by this way but the answer is not presented here

(10^49 + 2)/11= 10^49/11 +2/11

Studying each term:
10/11 = reminder is 10
10^2/11= reminder is 1
10^3/11= reminder is 10

so 10^49/11 == reminder is 10
and 2/11 =reminder is 2

so total reminder is 10+2= 12 but the reminder can not be greater than the divisor.

10��/11 + 2/11 --> R10 + R2 = R12.
Since the resulting remainder (12) cannot be greater than the divisor (11), we divide once more by 11:
12/11 = 1 R1.
Thus, dividing 10�� + 2 by 11 will yield a remainder of 1.

[Mo2men]

Thanks a lot for your help and prompt response

[Matt@VeritasPrep]

Here's a way that generalizes to more of these types of problems:

When 10*10 is divided by 11, the remainder is 1. So if we think of everything in terms of 10², we've got:

10�� =
(10²)²� * 10 =
1²� * 10

Now we'll add the 2, giving us

1 * 10 + 2

or 12. 12 divided by 11 has remainder 1, so we're done.

[Matt@VeritasPrep]

The trick to the approach above is finding a friendly remainder that allows you to greatly reduce the exponent that you're working with. As soon as we find a power of 10 that leaves remainder 1 or -1 when divided by 11, we're in business, and the remainder is a lot easier to tease out.