Data Sufficiency - 45

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.

Data Sufficiency - 42

Is a-3b an even number?

1). b=3a+3
2). b-a is an odd number

Answer: C

From statement (1): Given that b=3a+3
Thus a-3b=a-3(3a+3) = -8a-9 which may be even, odd, integer, non-integer, rational etc ... Hence insufficient

From statement (2): Given that b-a is an odd number implies b is of the form b=(2k+1)+a where k is an integer
Thus a-3b= a-3[(2k+1)+a] = -2a -6k-3 which may be even, odd, integer, non-integer, rational etc ..Hence insufficient

Taking statement (1) and (2) together: -8a-9=-2a-6k-3 for some integer k
or -6a=-6k+6=-6(k+1) implies a=k+1
Thus a is an integer, either odd or even

Now statement (2) tells us that b is also an integer and that exactly one of {a,b} is even
If a is even and b is odd, a-3b is odd
If b is even and a is odd a-3b is odd

Thus (1) and (2) combined tell us that a-3b is an odd number...hence sufficient

Data Sufficiency - 38

If q is a integer, is q^4 a multiple of 64?

(1) q^4 is not a multiple of 128.
(2) q^2 has 27 factors, 7 of which are less than or equal to 10

Answer: A

OE:
From statement (1): Given q is an integer, thus q can be written as product of distinct prime factors, where the power of 2 must be a whole number(a non-negative integer). This implies that the power of 2 in the prime factorization of q^4 must be a multiple of 4. As 2^7 is not a factor of q^4, the highest power of 2 that could be a factor of q^4 is 2^4.
Hence q^4 is not a multiple of 64=2^6......sufficient

From statement (2): Given that q^2 has 27 factors. Thus q^2 could be of the form
(i) a^26, or
(ii)b^2*c^8 or
(iii) a^2*b^2*c^2 where a b and c are distinct prime numbers.

Because q^2 has 7 factors less than 11, (i) is impossible.

As for (ii) 2^8*3^2 has exactly 7 factors under 11 (1,2,3,4,6,8,9), in which case q^4 would be a multiple of 64. 3^8*2*2 is not a possibility for q^2 (it only has 6 factors under 1: 1,2,3,4,6,9)

Regarding (iii) if q^2=2^2*3^2*5^2, q^2 would have 8 factors under 11- 1,2,3,4,5,6,9,10 and q^2=2^2*3^2*7^2 would have 7 factors under 11: 1,2,3,4,6,7,9) In this case, q^4 would not be a multiple of 64. Thus (2) is insufficient

hence the answer A

Manhattan Challenge Problem of the week! - 04/30/07

How many terminating zeroes does 200! have?

(A) 40
(B) 48
(C) 49
(D) 55
(E) 64

Answer : C

OE - To determine how many terminating zeroes a number has, we need to determine how many times the number can be divided evenly by 10. (For example, the number 404000 can be divided evenly by 10 three times, as follows:

404000/10 = 40400
40400/10 = 4040
4040/10 = 404

We can see that the number has three terminating zeroes because it is divisible by 10 three times.) Thus, to arrive at an answer, we need to count the factors of 10 in 200! Recall that

200! = 200 * 199 * 198 ........... * 4 * 3 * 2 * 1

Each factor of 10 consists of one prime factor of 2 and one prime factor of 5. Let’s start by counting the factors of 5 in 200!. Starting from 1, we get factors of 5 at 5, 10, 15, . . . , 190, 195, and 200, or every 5th number from 1 to 200. Thus, there are 200/5 or 40 numbers divisible by five from 1 to 200. Therefore, there are at least 40 factors of 5 in 200!.

We cannot stop counting here, because some of those multiples of 5 contribute more than just one factor of 5. Specifically, any multiple of 52 (or 25) and any multiple of 53 (or 125) contribute additional factors of 5. There are 8 multiples of 25 (namely, 25, 50, 75, 100, 125, 150, 175, 200). Each of these 8 numbers contributes one additional factor of 5 (in addition to the one we already counted) so we now have counted 40 + 8, or 48 factors of 5. Finally, 125 contributes a third additional factor of 5, so we now have 48 + 1 or 49 total factors of 5 in 200!.

Let us now examine the factors of 2 in 200!. Since every even number contributes at least one factor of 2, there are at least 100 factors of 2 in 200! (2, 4, 6, 8 . . .etc). Since we are only interested in the factors of 10 — a factor of 2 paired with a factor of 5 — and there are more factors of 2 than there are of 5, the number of factors of 10 is constrained by the number of factors of 5. Since there are only 49 factors of 5, each with an available factor of 2 to pair with, there are exactly 49 factors of 10 in 200!.

It follows that 200! has 49 terminating zeroes, and the correct answer is C.

Problem Solving - 19

How many 5 digit numbers can be created if the following terms apply: the leftmost digit is even, the second is odd, the third is a non even prime and the fourth and fifth are two random digits not used before in the number?

a) 2520
b) 3150
c) 3360
d) 6000
e) 7500


Answer: Given OA - A is incorrect ... ans is 2688 which is not an option in the choices...

When 2nd and 3rd digit gets repeated:

The first digit will be a non zero even (2, 4, 6, 8) = 4 ways
3rd digit is a non even prime = (3, 5, 7) = 3 ways
2nd digit is a REPEAT of that prime: 1 way
the fourth digit has not been used: 8 ways
the fifth digit has not been used: 7 ways

Hence 4*3*8*7 = 672 ways

Now, the non repeating case:

1st digit will be a non zero even (2, 4, 6, 8) = 4 ways
3rd digit (3, 5, 7) = 3 ways
2nd (no repeat and odd) = 4 ways
4th digit = 7
5th digit = 6

Hence 4*3*4*7*6 = 2016

Total number of ways = 2016 + 672 = 2688 ways


Note:
1). 1 is neither a prime nor a composite number.
2). 2 is the only even prime number.

DS Question - 22

n is a positive integer. What is the remainder when n is divided by 6?

(1) n is a multiple of 3.
(2) When n is divided by 2, the remainder is 1.

Answer: C

From statement (1) - n can be 3, 6, 9, 12...

Therefore:

(i) If n is odd, the reminder is 3 on dividing by 6.
(ii) If n is even, the reminder is 0 on dividing by 6.

Hence Insufficient.

From statement (2) - When n is divided by 2, the remainer is 1. Hence this is an odd integer as when an odd number is divided by 2, the remainder is 1.

n = 2*k + 1

(i) If n = 7 then n/6 = 6*1 + 1
(ii) If n = 9 then n/6 = 6*1 + 3

Hence insufficient.

taking statement (1) and (2) together - From Statement (1) we know that we need to know that whether n is odd or even in order to be able to conclude.Statement (2) gives us this information.

Hence Sufficient

Hence, C is the answer.

Manhattan Challenge Problem of the week ! - Jan 8

If x = 3 ^ 21 and y = 6 ^ 55, what is the remainder when xy is divided by 10?

(A) 2

(B) 3

(C) 4

(D) 6

(E) 8


Answer : The correct answer is E.

Since every multiple of 10 must end in zero, the remainder from dividing xy by 10 will be equal to the units’ digit of xy. In other words, the units’ digit will reflect by how much this number is greater than the nearest multiple of 10 and, thus, will be equal to the remainder from dividing by 10. Therefore, we can rephrase the question: “What is the units’ digit of xy?”

Next, let’s look for a pattern in the units’ digit of 3 ^ 21. Remember that the GMAT will not expect you to do sophisticated computations; therefore, if the exponent seems too large to compute, look for a shortcut by recognizing a pattern in the units' digits of the exponent:

3 ^ 1 = 3
3 ^ 2 = 9
3 ^ 3 = 27
3 ^ 4 = 81
3 ^ 5 = 243

As you can see, the pattern repeats every 4 terms, yielding the units digits of 3, 9, 7, and 1. Therefore, the exponents 3 ^ 1, 3 ^ 5, 3 ^ 9, 3 ^ 13, 3 ^ 17, and 3 ^ 21 will end in 3, and the units’ digit of 3 ^ 21 is 3.

Next, let’s determine the units’ digit of 6 ^ 55 by recognizing the pattern:

6 ^ 1 = 6
6 ^ 2 = 36
6 ^ 3 = 256
6 ^ 4 = 1,296

As shown above, all positive integer exponents of 6 have a units’ digit of 6. Therefore, the units' digit of 6 ^ 55 will also be 6.

Finally, since the units’ digit of 3 ^ 21 is 3 and the units’ digit of 6 ^ 55 is 6, the units' digit of 3 ^ 21 × 6 ^ 55 will be equal to 8, since 3 × 6 = 18.

Therefore, when this product is divided by 10, the remainder will be 8.

Problem Solving - 9

9). AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1

(B) 3

(C) 7

(D) 9

(E) Cannot be determined


Answer - D

Explanation -- AB + CD = AAA
AB and CD are two digit numbers, hence AAA must be 111
Therefore 1B + CD = 111
B can assume any value between 3 and 9
If B = 3, then CD = 111-13 = 98 and C = 9
If B = 9, then CD = 111-19 = 92 and C = 9
Hence for all values of B between 3 & 9 ---- C = 9
Thus the answer is D i.e (C = 9)

Manhattan Challenge Problem of the week ! july 24

Is the positive integer x odd?

(1) x = y2 + 4y + 6, where y is a positive integer.

(2) x = 9z2 + 7z - 10, where z is a positive integer.

(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.

Answer - The correct answer is B.

For complete solution click on the link below.
Manhattan challenge problem

Problem Solving - 8

How many factors of 2940 are NOT factors of 112?

(A) 22

(B) 30

(C) 32

(D) 34

(E) None of these

Answer - B

Prime factors of 2940 = (2 ^ 2) * 3 * 5 * (7 ^ 7)

Hence total number of factors of 2940

= (2 + 1) * ( 1 + 1) * (1 + 1) * ( 2 + 1) = 36
(by the method of counting number of factors)

Prime factors of 112 = (2 ^ 4) * 7

Therefore common factors for the above two numbers are

2^2 * 7

Total number of factors for (2 ^ 2) * 7 = (2 + 1) * ( 1 + 1) = 6

Hence total factors of 2940 which are not factors of 112 are

36 - 6 = 30, Hence B is the answer.

DS Question - 12

Does x + y = xy?

(1) x is neither a positive integer nor a negative integer

(2) y is neither a positive integer nor a negative integer

Answer - E - x and y can be either 0 or fractions.. which makes both conditions insufficient.

DS Question - 10

When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product nt is divided by 15 ?

(1) n- 2 is divisible by 5.

(2) t is divisible by 3.

Answer - C

Statement (1) alone - no sufficient information about t
Statement (2)alone - no sufficient information about n

From the ques --- n = a*3 + 2

t = b*5 + 3

Using Statement (1) --- n = p*5 + 2

Thus n = 15*k + 2

Using Statement (2) --- t = q*3 + 3

Thus t = 15*j + 3

nt = 15*k(15*j + 3) + 30*j + 6

=> always leaves a remainder of 6 when divided by 15.

Problem Solving - 4

What is the sum of all three digit positive integers, the sum of whose digits is a multiple of 3?

a) 900

b) 165150

c) 165000

d) 329199

e) 328900

Answer - B

Sum of n numbers = n/2(2a+(n-1)d)

We want to add the numbers from 102 to 999 that are divisible by 3.

We divide by 3 to get 34 to 333, or 300 numbers in total.

Here n = 300, a = 34, d= 1

Sum = 1/2*300*(333+34) = 150 * 367 = 55,050

Multiplying by 3 again gives 165,150 .

TIP

To add the first n integers from 1 to n
add together 1 + n = n+1
then 2 + (n-1) = n + 1 until you get to n/2 (depending on whether n is odd or even)
Ultimately we have n/2 pairs that all add up to (n+1)
So the sum is 1/2 * n * (n+1)
If you don't start at 1, or have a gap of more than 1 --- slightly adjust the formula.

DS Question - 8

If X and Y are nonzero integers, what is the remainder when X is divided by Y?

(1) When X is divided by 2y, the remainder is 4

(2) When X+Y is divided by y, the remainder is 4

Answer - B

Statement (1) ---- x = (2y)M +4

if y > 4 the remainder of x/y will be 4

if y= 1 or 2 or 4 there will be no remainder,

if y=3, remainder will be 1,

so it depends on value of y...

so insufficient

Statement (2) -----(x+y)/y=M+4

It is clear that y/y=1 therefore (x/y) has to be 4 - Thus

sufficient

Problem Solving - 3

There are two buses A and B. On Monday, bus A departs at 3pm and bus B at 4pm. After this, bus A departs every 10 hours and bus B every 15 hours. What is the earliest day they will depart at the same time?

A). Tuesday

B). Wednesday

C). Thursday

D). Sunday

E). so long as they continue operating in this manner, they will never leave at the same time.

Answer - E is the right choice.

The LCM of 10 hours and 15 hours = 30 hours,
30 hours = 6 hours mod 24 hour (day)

After that period they will just repeat the cycle.

Problem solving - 1

Problem Solving - Numbers

If (3^4)(5^6)(7^3)=(35^n)(x), where x and n are both positive integers, how many different positive values of n are there?

A) 1

B) 2

C) 3

D) 4

E) 6

Answer - C

(3^4)(5^6)(7^3)=(35^n)(x)

Now 35 = 5 * 7

=> 35^n = 5^n * 7^n

3^4 * 5^3 * (5^3 * 7^3) = (35^n) * x

So x is a multiple of 3^4 * 5^3 n => 3 n is a positive integer So n=1, 2, 3

DS Question - 2

If d represents the hundredths digit and e represents the thousandths digit in the decimal .4de, what is the value of this decimal rounded to the nearest tenth?

(1) d – e is equal to a positive perfect square.

(2) sqrt (d) > e*e

(A) Statement (1) alone is sufficient, but statement(2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement(1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOTsufficient.


Answer - E

From statement (1), we know that d – e must equal a positive perfect square. This means that d is greater than e. In addition, since any single digit minus any other single digit can yield a maximum of 9, d – e could only result in the perfect squares 9, 4, or 1.

However, this leaves numerous possibilities for the values of d and e respectively. For example, two possibilities are as follows:

d = 7, e = 3 (d – e = the perfect square 4)
d = 3, e = 2 (d – e = the perfect square 1)

In the first case, the decimal .4de would be .473, which, when rounded to the nearest tenth, is equal to .5. In the second case, the decimal would be .432, which, when rounded to the nearest tenth, is .4.

Thus, statement (1) is not sufficient on its own to answer the question.

Statement (2) tells us that sqrt d = e². Since d is a single digit, the maximum value for d is 9, which means the maximum square root of d is 3. This means that e² must be less than 3. Thus the digit e can only be 0 or 1.

However, this leaves numerous possibilities for the values of d and e respectively. For example, two possibilities are as follows:

d = 9, e = 1
d = 2, e = 0

In the first case, the decimal .4de would be .491, which, when rounded to the nearest tenth, is equal to .5. In the second case, the decimal would be .420, which, when rounded to the nearest tenth, is .4.

Thus, statement (2) is not sufficient on its own to answer the question.


Taking both statements together, we know that e must be 0 or 1 and that d – e is equal to 9, 4 or 1.

This leaves the following 4 possibilities:

d = 9, e = 0
d = 5, e = 1
d = 4, e = 0
d = 1, e = 0

These possibilities yield the following four decimals: .490, .451, .440, and .410 respectively. The first two of these decimals yield .5 when rounded to the nearest tenth, while the second two decimals yield .4 when rounded to the nearest tenth.

Thus, both statements taken together are not sufficient to answer the question.

The correct answer is E: Statements (1) and (2) TOGETHER are NOT sufficient.