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27 Sep 2024, 12:40
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
50% (03:25) correct
50% (01:20) wrong based on 2 sessions
History
Date
Time
Result
Not Attempted Yet
\(m = 10^{32} + 2\) When m is divided by 11, the remainder is r.
Quantity A
r
Quantity B
3
Quantity A
r
Quantity B
3
27 Feb 2025, 04:02
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OE
Actually dividing \(10^{32} + 2\) by 11 would be very time consuming, so it is worth trying to compare the quantities without actually doing the division. A good approach would be to compute the remainders when \(10^1 + 2, 10^2 + 2, 10^3 + 2, 10^4 + 2,\) etc., are divided by 11 to see if there is a pattern that can help you determine the remainder when \(10^{32} + 2\) is divided by 11. The following table shows the first few cases.
Note that the remainder is 1 when 10 is raised to an odd power, and the remainder is 3 when 10 is raised to an even power. This pattern suggests that since 32 is even, the remainder when \(10^{32} + 2\) is divided by 11 is 3.
To see that this is true, note that the integers 99 and 9,999 in the rows for n = 2 and n = 4, respectively, are multiples of 11. Tat is because they each consist of an even number of consecutive digits of 9. Also, these multiples of 11 are each 3 less than \(10^2 + 2 \) and \(10^4 + 2,\) respectively, so that is why the remainders are 3 when \(10^2 + 2 \) and \(10^4 + 2\) are divided by 11. Similarly, for n = 32, the integer with 32 consecutive digits of 9 is a multiple of 11 because 32 is even. Also, that multiple of 11 is 3 less than \(10^{32} + 2,\) so the remainder is 3 when \(10^{32} + 2\) is divided by 11. Thus the correct answer is Choice C.
Answer: C
GMAT-Club-Forum-82bymjob.png [ 46.28 KiB | Viewed 972 times ]
Actually dividing \(10^{32} + 2\) by 11 would be very time consuming, so it is worth trying to compare the quantities without actually doing the division. A good approach would be to compute the remainders when \(10^1 + 2, 10^2 + 2, 10^3 + 2, 10^4 + 2,\) etc., are divided by 11 to see if there is a pattern that can help you determine the remainder when \(10^{32} + 2\) is divided by 11. The following table shows the first few cases.
Note that the remainder is 1 when 10 is raised to an odd power, and the remainder is 3 when 10 is raised to an even power. This pattern suggests that since 32 is even, the remainder when \(10^{32} + 2\) is divided by 11 is 3.
To see that this is true, note that the integers 99 and 9,999 in the rows for n = 2 and n = 4, respectively, are multiples of 11. Tat is because they each consist of an even number of consecutive digits of 9. Also, these multiples of 11 are each 3 less than \(10^2 + 2 \) and \(10^4 + 2,\) respectively, so that is why the remainders are 3 when \(10^2 + 2 \) and \(10^4 + 2\) are divided by 11. Similarly, for n = 32, the integer with 32 consecutive digits of 9 is a multiple of 11 because 32 is even. Also, that multiple of 11 is 3 less than \(10^{32} + 2,\) so the remainder is 3 when \(10^{32} + 2\) is divided by 11. Thus the correct answer is Choice C.
Answer: C
Attachment:
GMAT-Club-Forum-82bymjob.png [ 46.28 KiB | Viewed 972 times ]
