# Manhattan GMAT Challenge Problem of the Week – 23 July 2012

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## Question

A decade is defined as a complete set of consecutive nonnegative integers that have identical digits in identical places, except for their units digits, with the first decade consisting of the smallest integers that meet the criteria, the second decade consisting of the next smallest integers, etc. A decade in which the prime numbers contain the same set of units digits as do the prime numbers in the second decade is the

A. fifth

B. seventh

C. eighth

D. ninth

E. eleventh

## Answer

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 , 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.**

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## 1 comment

Kalpana Sharma on August 23rd, 2012 at 1:14 am

very good question and the twisted language makes it even more interesting.