Hello! I went through incorrect questions on last 2 tests, and found 2 math topics to focus on: Number Properties (specially Divisibility & Primes), and Inequalities. Done going through all the Number Properties questions from OG11, 12 and OG Quant supplement. Might go through "Integer Properties" category of questions from my BTG account.
Going to do the same for Inequalities today.
Finished Modifiers from MGMAT SC. Need to go through my notes again, and practice some more questions from OGV for those topics.
Another practice test tomorrow morning.
Happy studying you guys!!!
Showing posts with label Number Properties. Show all posts
Showing posts with label Number Properties. Show all posts
Tuesday, January 18, 2011
Tuesday, January 11, 2011
CP #25
MGMAT CAT#2
The greatest common factor of 16 and the positive integer n is 4, and the greatest common factor of n and 45 is 3. Which of the following could be the greatest common factor of n and 210?
Solution
The greatest common factor (GCF) of two integers is the largest integer that divides both of them evenly (i.e. leaving no remainder).
One way to approach the problem is to consider what the GCFs stated in the question stem tell us about n:
The greatest common factor of n and 16 is 4. In other words, n and 16 both are evenly divisible by 4 (i.e., they have the prime factors 2 × 2), but have absolutely no other factors in common. Since 16 = 2 × 2 × 2 × 2, n must have exactly two prime factors of 2--no more, no less.
The greatest common factor of n and 45 is 3. In other words, n and 45 both are evenly divisible by 3, but have absolutely no other factors in common. Since 45 = 3 × 3 × 5, n must have exactly one prime factor of 3--no more, no less. Also, n cannot have 5 as a prime factor.
So, n must include the prime factors 2, 2, and 3. Furthermore, n is not divisible by 8, 9, or 5.
The number 210 has the following prime factorization: 210 = 2 x 3 x 5 x 7. Thus n and 210 are both divisible by 2 and 3, and both might be divisible by 7, but it is not the case that both numbers are divisible by 4 or by 5:
(A) 3: missing a 2
(B) 14: missing a 3
(C) 30: n cannot be divisible by 5
(D) 42: Correct. n and 210 are both divisible by 2 and 3, and n could be divisible by 7.
(E) 70: n cannot be divisible by 5 and must be divisible by 3.
The correct answer is D.
The greatest common factor of 16 and the positive integer n is 4, and the greatest common factor of n and 45 is 3. Which of the following could be the greatest common factor of n and 210?
 | 3 | |
| | 14 | |
| | 30 | |
| | 42 | |
| | 70 |
Solution
The greatest common factor (GCF) of two integers is the largest integer that divides both of them evenly (i.e. leaving no remainder).
One way to approach the problem is to consider what the GCFs stated in the question stem tell us about n:
The greatest common factor of n and 16 is 4. In other words, n and 16 both are evenly divisible by 4 (i.e., they have the prime factors 2 × 2), but have absolutely no other factors in common. Since 16 = 2 × 2 × 2 × 2, n must have exactly two prime factors of 2--no more, no less.
The greatest common factor of n and 45 is 3. In other words, n and 45 both are evenly divisible by 3, but have absolutely no other factors in common. Since 45 = 3 × 3 × 5, n must have exactly one prime factor of 3--no more, no less. Also, n cannot have 5 as a prime factor.
So, n must include the prime factors 2, 2, and 3. Furthermore, n is not divisible by 8, 9, or 5.
The number 210 has the following prime factorization: 210 = 2 x 3 x 5 x 7. Thus n and 210 are both divisible by 2 and 3, and both might be divisible by 7, but it is not the case that both numbers are divisible by 4 or by 5:
(A) 3: missing a 2
(B) 14: missing a 3
(C) 30: n cannot be divisible by 5
(D) 42: Correct. n and 210 are both divisible by 2 and 3, and n could be divisible by 7.
(E) 70: n cannot be divisible by 5 and must be divisible by 3.
The correct answer is D.
CP #24
MGMAT CAT#2
Is the positive integer N a perfect square?
(1) The number of distinct factors of N is even.
(2) The sum of all distinct factors of N is even.
Answer Explanation:
(1) SUFFICIENT: The factors of any number N can be sorted into pairs that multiply to give N. (For instance, the factors of 24 can be paired as follows: 1 and 24; 2 and 12; 3 and 8; 4 and 6.) However, if N is a perfect square, one of these ‘pairs’ will consist of just one number: the square root of N. (For example, if N were 49, it would ahve the factor pair 7 × 7) Since all of the other factors can be paired off, it follows that if N is a perfect square, then N has an odd number of factors. (If N is not a perfect square, then all of its factors can be paired off, so it will have an even number of factors.) This statement then implies that N is not a perfect square.
(2) SUFFICIENT: Let N be a perfect square. If N is odd, then all factors of N are odd. Therefore, by the above reasoning, N has an odd number of odd factors, so their sum must be odd. If N is even, let M be the product of all the odd prime factors (as many times as they appear in N – not distinct) of N, which is also a perfect square. (For instance, if N = 100, then M =5 × 5 = 25.) Then the sum of factors of M is odd, by the above reasoning. Furthermore, all other factors of N (i.e., that don’t also divide M) are even. The sum total is thus odd + even = odd.
Therefore, the sum of the factors of any perfect square is odd, so this statement implies that N is not a perfect square.
This statement can also be investigated by trying several cases of even perfect squares (4, 16, 36, 64, 100), noting that in each case the sums of the factors are odd, and generalizing.
The correct answer is D.
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I picked numbers to get to my answer, but made a silly mistake by omitting "1" as a distinct factor in Statement II.
And to the MGMAT explanation - hunh?!
Is the positive integer N a perfect square?
(1) The number of distinct factors of N is even.
(2) The sum of all distinct factors of N is even.
Answer Explanation:
(1) SUFFICIENT: The factors of any number N can be sorted into pairs that multiply to give N. (For instance, the factors of 24 can be paired as follows: 1 and 24; 2 and 12; 3 and 8; 4 and 6.) However, if N is a perfect square, one of these ‘pairs’ will consist of just one number: the square root of N. (For example, if N were 49, it would ahve the factor pair 7 × 7) Since all of the other factors can be paired off, it follows that if N is a perfect square, then N has an odd number of factors. (If N is not a perfect square, then all of its factors can be paired off, so it will have an even number of factors.) This statement then implies that N is not a perfect square.
(2) SUFFICIENT: Let N be a perfect square. If N is odd, then all factors of N are odd. Therefore, by the above reasoning, N has an odd number of odd factors, so their sum must be odd. If N is even, let M be the product of all the odd prime factors (as many times as they appear in N – not distinct) of N, which is also a perfect square. (For instance, if N = 100, then M =5 × 5 = 25.) Then the sum of factors of M is odd, by the above reasoning. Furthermore, all other factors of N (i.e., that don’t also divide M) are even. The sum total is thus odd + even = odd.
Therefore, the sum of the factors of any perfect square is odd, so this statement implies that N is not a perfect square.
This statement can also be investigated by trying several cases of even perfect squares (4, 16, 36, 64, 100), noting that in each case the sums of the factors are odd, and generalizing.
The correct answer is D.
-----------
I picked numbers to get to my answer, but made a silly mistake by omitting "1" as a distinct factor in Statement II.
And to the MGMAT explanation - hunh?!
Thursday, November 4, 2010
CP#17
Question (from MGMAT Practice Test #1)
For positive integer k, is the expression (k + 2)(k2 + 4k + 3) divisible by 4?
(1) k is divisible by 8.
Solution
The quadratic expression k2 + 4k + 3 can be factored to yield (k + 1)(k + 3). Thus, the expression in the question stem can be restated as (k + 1)(k + 2)(k + 3), or the product of three consecutive integers. This product will be divisible by 4 if one of two conditions are met:
If k is odd, both k + 1 and k + 3 must be even, and the product (k + 1)(k + 2)(k + 3) would be divisible by 2 twice. Therefore, if k is odd, our product must be divisible by 4.
If k is even, both k + 1 and k + 3 must be odd, and the product (k + 1)(k + 2)(k + 3) would be divisible by 4 only if k + 2, the only even integer among the three, were itself divisible by 4.
The question might therefore be rephrased “Is k odd, OR is k + 2 divisible by 4?” Note that a ‘yes’ to either of the conditions would suffice, but to answer 'no' to the question would require a ‘no’ to both conditions.
(1) SUFFICIENT: If k is divisible by 8, it must be both even and divisible by 4. If k is divisible by 4, k + 2 cannot be divisible by 4. Therefore, statement (1) yields a definitive ‘no’ to both conditions in our rephrased question; k is not odd, and k + 2 is not divisible by 4.
(2) INSUFFICIENT: If k + 1 is divisible by 3, k + 1 must be an odd integer, and k an even integer. However, we do not have sufficient information to determine whether k or k + 2 is divisible by 4.
The correct answer is A.
Query
I think the answer is D. In Statement (2), for (k+1)/3 to be an odd INTEGER, k+1 has to be divisible by 3. Also, k must be even. Therefore, k could be 2,8,... And if k is an even integer, (k + 1)(k + 2)(k + 3) will not be divisible by 4 for any of these values of k. Therefore, Statement (2) also gives us the answer. Please tell me where I'm going wrong.
For positive integer k, is the expression (k + 2)(k2 + 4k + 3) divisible by 4?
(1) k is divisible by 8.
| (2) | k + 1  3 | is an odd integer. |
Solution
The quadratic expression k2 + 4k + 3 can be factored to yield (k + 1)(k + 3). Thus, the expression in the question stem can be restated as (k + 1)(k + 2)(k + 3), or the product of three consecutive integers. This product will be divisible by 4 if one of two conditions are met:
If k is odd, both k + 1 and k + 3 must be even, and the product (k + 1)(k + 2)(k + 3) would be divisible by 2 twice. Therefore, if k is odd, our product must be divisible by 4.
If k is even, both k + 1 and k + 3 must be odd, and the product (k + 1)(k + 2)(k + 3) would be divisible by 4 only if k + 2, the only even integer among the three, were itself divisible by 4.
The question might therefore be rephrased “Is k odd, OR is k + 2 divisible by 4?” Note that a ‘yes’ to either of the conditions would suffice, but to answer 'no' to the question would require a ‘no’ to both conditions.
(1) SUFFICIENT: If k is divisible by 8, it must be both even and divisible by 4. If k is divisible by 4, k + 2 cannot be divisible by 4. Therefore, statement (1) yields a definitive ‘no’ to both conditions in our rephrased question; k is not odd, and k + 2 is not divisible by 4.
(2) INSUFFICIENT: If k + 1 is divisible by 3, k + 1 must be an odd integer, and k an even integer. However, we do not have sufficient information to determine whether k or k + 2 is divisible by 4.
The correct answer is A.
Query
I think the answer is D. In Statement (2), for (k+1)/3 to be an odd INTEGER, k+1 has to be divisible by 3. Also, k must be even. Therefore, k could be 2,8,... And if k is an even integer, (k + 1)(k + 2)(k + 3) will not be divisible by 4 for any of these values of k. Therefore, Statement (2) also gives us the answer. Please tell me where I'm going wrong.
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