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

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by Mo2men » Sun Jan 22, 2017 3:09 am
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??
Last edited by Mo2men on Sun Jan 22, 2017 3:47 am, edited 1 time in total.
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by GMATGuruNY » Sun Jan 22, 2017 3:44 am
Mo2men wrote: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.
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by Mo2men » Sun Jan 22, 2017 3:51 am
GMATGuruNY wrote: 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
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by GMATGuruNY » Sun Jan 22, 2017 3:53 am
Mo2men wrote: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.
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by Mo2men » Sun Jan 22, 2017 3:55 am
Thanks a lot for your help and prompt response :) :)
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by Matt@VeritasPrep » Wed Feb 01, 2017 5:35 pm
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.
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by Matt@VeritasPrep » Wed Feb 01, 2017 5:35 pm
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.
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