A decade is defined as a complete set of consecutive nonnegative integ

a decade = 10 years

by the definition above a decade starts with a nonnegative integer (0) that have identical digits in identical places, except for their units digits
so first decade contains the years - 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
second decade contains the years - 10, 11, 12, 13,1 4, 15, 17, 18, 19

we have to find next decade which has prime numbers contain the same set of units digits as the second decade
prime numbers can be expressed as (6n+1) or (6n-1)
prime numbers in the second decade

(6n-1)-type
11 = (2-1) = 6*2 - 1
17 = (8-1) = 6*3 - 1
(6n+1)-type
13 = (2+1) = 6*2 + 1
19 = (8+1) = 6*3 + 1
6n-1 will yield 1 or 3 in unit's place if n = 2, 3, .. 7, 8, ... 12, 13, ... 17, 18, ... 22, 23, ...
6n+1 will yield 7 or 9 in unit's place if n = 2, 3, .. 7, 8, ... 12, 13, ... 17, 18, ... 22, 23, ...

with multiples of 7 and 8 we have 41, 43, 47, 49 (not a prime)
with multiples of 12 and 13 we have 71, 73, 77 (not prime), 79
with multiples of 17 and 18 we have 101, 103, 107, 109; matched the second decade
so the decade {100, ... 109 } is (10 - 0 + 1) 11th decade; first decade has no tens' digit

Answer E

MANHATTAN GMAT OFFICIAL SOLUTION:

This problem is all about reading. “A complete set of consecutive nonnegative integers…” should conjure in your mind the numbers {0, 1, 2, 3, …} – and you’re going to take a consecutive subset of those numbers. Here are the key words: “… that have identical digits in identical places, except for their units digits…” So the units digits of the numbers in a “decade” can differ, but all the other digits are the same. For example, the integers 50 through 59 would form a decade; 150 through 159 would form a different decade.

The first decade, we are told, consists of the smallest integers that meet the criteria. The smallest nonnegative integers are 0, 1, 2, 3, … and in fact, the integers 0 through 9 meet the criteria (they have identical digits in identical places, except for their units digits – and since these numbers only consist of units digits, they have no digits in common, but they’re still part of the same decade). So the first decade is {0, 1, 2, 3, …, 9}. The second decade is {10, 11, 12, 13, …, 19}, and so on.

The prime numbers in the second decade are 11, 13, 17, and 19. So you must find a decade in which the primes are xxx1, xxx3, xxx7, and xxx9 (where xxx represents the unknown identical digits in the decade). Search backwards from the answer choices, noting that since the first decade has no tens digit and the second decade has 1 as the tens digit, the “ordinal” number (first, second, third, etc.) of the decade is one more than the tens digit. That is to say, the fifth decade is {40, 41, … 49}.

(A) cannot be right, because 49 is not prime.
(B) cannot be right, because 63 and 69 are not prime.
(C) cannot be right, because 77 is not prime.
(D) cannot be right, because 81 is not prime.

Hence, the answer must be (E): 101, 103, 107, and 109 are all prime. If you really want to check, look for divisibility by primes up to the square root of the number in question—and since they’re all less than 121, which is 11^2, you can just check divisibility by 2, 3, 5, and 7. Numbers ending in 1, 3, 7, and 9 are not divisible by either 2 or 5, so you only really have to check 3 and 7.

The correct answer is E.

is there any way to solve this problem within 2 mins?..understanding the question itself took me like 4 mins!...

Yes, the question has very unconventional language for something very simple. The concept involved is very basic. It does take some time to figure out what the question is asking and there is no other method to do that except read the question one statement at a time and try to figure out what it is saying. What helps is subtle clues such as the use of the term "decade" - perhaps every "decade" has 10 numbers.

Irrespective of the popular opinion I understood the question within time, in like 1:30 mins but I chickened out with the difficulty on how long will it take and guessed the wrong answer haha

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