p^a * q^b * r^c * s^d = x, where x is a perfect square. If p, q, r, and s are prime integers, are they distinct?

(1) 18 is a factor of ab and cd

(2) 4 is not a factor of ab and cd

Answer: B

OE: When a perfect square is broken down into its prime factors, those prime factors always come in "pairs." For example, the perfect square 225 (which is 15 squared) can be broken down into the prime factors 5 * 5 * 3 * 3. Notice that 225 is composed of a pair of 5's and a pair of 3's.

The problem states that x is a perfect square. The prime factors that build x are p, q, r, and s. In order for x to be a perfect square, these prime factors must come in pairs. This is possible if either of the following two cases hold:

Case One: The exponents a, b, c, and d are even. In the example 3^2 5^4 7^2 11^6, all the exponents are even so all the prime factors come in pairs.

Case Two: Any odd exponents are complemented by other odd exponents of the same prime. In the example 3^1 5^4 3^3 11^6, notice that 3^1 and 3^3 have odd exponents but they complement each other to create an even exponent (3^4), or "pairs" of 3's. Notice that this second case can only occur when p, q, r, and s are NOT distinct. (In this example, both p and r equal 3.)

Statement (1) tells us that 18 is a factor of both ab and cd. This does not give us any information about whether the exponents a, b, c, and d are even or not.

Statement (2) tells us that 4 is not a factor of ab and cd. This means that neither ab nor cd has two 2's as prime factors. From this, we can conclude that at least two of the exponents (a, b, c, and d) must be odd. As we know from Case 2 above, if paqbrcsd is a perfect square but the exponents are not all even, then the primes p, q, r and s must NOT be distinct.

The correct answer is B: Statement (2) alone is sufficient, but statement (1) alone is not sufficient.

## Wednesday, March 19, 2008

### Problem Solving - 45

If p is the smallest positive integer such that (p^3)/3920 is also an integer, what is the sum of the digits of p?

(A) 5

(B) 7

(C) 9

(D) 11

(E) 13

Answer: A

m = p*p*p/ 3920

m = p*p*p/ 2*2*2*2*5*7*7

m = p*p*p/ [(4^2)*5*(7^2)]

Hence p must be equal to 4*5*7

thus p = 140

Hence sum of digits = 1+4+0 = 5

(A) 5

(B) 7

(C) 9

(D) 11

(E) 13

Answer: A

m = p*p*p/ 3920

m = p*p*p/ 2*2*2*2*5*7*7

m = p*p*p/ [(4^2)*5*(7^2)]

Hence p must be equal to 4*5*7

thus p = 140

Hence sum of digits = 1+4+0 = 5

Labels:
Integers,
Problem Solving

## Tuesday, March 18, 2008

### Problem Solving - 44

Samar tried to type his new 7-digit phone number on a form, but what appeared on the form was 39269, since the '4' key on his computer no longer works. His secretary has decided to make a list of all of the numbers that could be Samar's new number. How many numbers will there be on the list?

(A) 21

(B) 24

(C) 25

(D) 30

(E) 36

Answer: A

It is clear from the question that there are two 4's missing as it is a seven digit number.

Total number of ways of choosing places for these two missing 4's in these 7 digits is 7C2 = 21 Once you fix the place for these two 4's rest all numbers will occupy remaining places.

(A) 21

(B) 24

(C) 25

(D) 30

(E) 36

Answer: A

It is clear from the question that there are two 4's missing as it is a seven digit number.

Total number of ways of choosing places for these two missing 4's in these 7 digits is 7C2 = 21 Once you fix the place for these two 4's rest all numbers will occupy remaining places.

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