Manhattan GMAT Challenge Problem of the Week – 9 Nov 2010:

by on November 9th, 2010

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Question

The function p(n) on non-negative integer n is defined in the following way: the units digit of n is the exponent of 2 in the prime factorization of p(n), the tens digit is the exponent of 3, and in general, for positive integer k, the digit in the place of n is the exponent on the kth smallest prime (compared to the set of all primes) in the prime factorization of p(n). For instance, . What is the smallest positive integer that is not equal to p(n) for any permissible n?

(A) 1

(B) 29

(C) 31

(D) 1,024

(E) 2,310

The hardest part about this problem is understanding how p(n) is defined. After reading the definition, focus on the example given. We are told that . Study this example: the 102 shows up in the exponents of the primes: .

Try constructing another example or two. Use small numbers with just units and tens digits, since you know explicitly what to do with units and tens digits. For instance, p(11) is equal to . Likewise, p(12) is equal to , in fact.

Now we are asked for the smallest integer that is NOT equal to p(n) for any permissible n. In other words, you cannot construct the prime factorization of this number using the process above.

We should now think of limitations on the results. Since what we’re doing is putting the digits of n into separate exponents, the highest that any particular exponent can go is 9 (the largest digit in the base-10 system). This means that is not constructible, since only the units digit of n goes in as the exponent of 2.

Since 2 is the smallest prime, is the smallest number that you can’t reach with this process. .

Notice that you can in fact output 1 as p(n) if you put in n = 0, since .