LCM's and Harder Integers Questions

The answer, as we'll soon see, is: both. To sum up: Solving questions asking for the greatest number that n must be divisible by: 1) Use the information in the question to find the minimum list of building blocks that you know for sure n MUST be divisible by. Eliminate all answer choices greater than that list. Eliminate any answer choices NOT on that list, if there are any. 2) From the remaining answer choices, eliminate the smaller numbers and choose the greatest.

The initial steps remain the same: 1) Break down 60 to its building blocks. a2=2^2×3^1×5^1×n 2) a is squared, so every factor of a needs to repeat twice. Therefore, n MUST include at least 3^1×5^1=15 in order to complete everything on the right side to the power of 2. Now use POE: D 30 and E 60 are eliminated because you must come down to the minimum number of building blocks that you know for sure n MUST be divisible by. We're left with A, B, and C, all of which include numbers that n must be divisible by. Now the whole "greatest number that n must be divisible by" comes into effect: eliminate A and B, since C is the greatest of the list of numbers that n MUST be divisible by. The answer is C.

Looking at the broader picture is necessary for solving questions of this sort: the question deals with what n must be divisible by. By selecting 15, we're not ruling out the fact that n IS divisible by 30 or 60 - we're saying that this fact does not constitute a MUST. Any n that IS divisible by 60 (e.g. 60 itself, or 120, or 6,000) will also satisfy the rest of the question's requirements for n (i.e. will also be divisible by 3×5). But when asked which of the following n MUST be divisible by, we have to come down to the minimum number of building blocks that we know for sure n MUST be divisible by: 3 and 5 - no more, no less.

To sum up what we've seen so far: 1) When asked which of the following n MUST be divisible by, come down to the minimum number of building blocks that you know for sure n MUST be divisible by. Everything not in the list of minimum building blocks falls under "CAN" - not "MUST". 2) The prime factors of any integer that is a power of another integer come in pairs, triplets, quadruplets etc. according to the power. If a is an integer, and a2 is an integer, then a2's prime factors must come in pairs.

Integers: Integer 'Must be True' Questions Back to our equation: a2=2^2×3^1×5^1×n - the right side has 2^2, but is still missing a 3 and 5 to complete everything to the second power. These must be "supplied" by n, so n MUST include 3^1×5^1. Can you now answer the question? If a2=60n, where a and n are integers, then n MUST be divisible by which of the following: Correct. 15=3×5. We've seen that the 3 and 5 are necessary in order to complete all of the right side's components to the second power, so n must "supply" them and must therefore be divisible by 15.

We move now to some of the tougher questions the integer subject has to offer. Take a look at the following problem: If a2=60n, where a and n are integers, then n MUST be divisible by which of the following: A 12 B 15 C 20 D 30 E 60 First, note that the problem supplies an equation. As with all equations, whatever is on one side must equal the other. Let's analyze right side. Find the prime factors of 60 using the factor tree - 60=2×2×3×5. The equation now looks like this: a2=2^2×3^1×5^1×n It's clear that a must be an integer with at least 2, 3, and 5 as its prime factors. But note an even more interesting fact: since the left side presents a2, every factor of a must appear twice on the right side. If, for example, a=2×3×5, then a2=a×a=(2×3×5)⋅(2×3×5) = 2^2×3^2×5^2 - twice for each factor. If the left side had a3, the right would need to include every factor three times; a4 on the left = four times on the right, etc.

Remember this question? If a2=60n, where a and n are integers, then n MUST be divisible by which of the following: A 12 B 15 C 20 D 30 E 60 Oh Yeah. That was an interesting one. What about it?

We now revisit the question, but with a twist: If a2=60n, where a and n are integers, which of the following is the greatest integer that n MUST be divisible by? A 3 B 5 C 15 D 30 E 60 The phrasing is confusing here. On the one hand, the question asks for the greatest integer that MUST divide n. On the other hand, remember that MUST be questions involving divisibility require us to come down to the minimum number of building blocks that you know for sure n MUST be divisible by. So which is it - the greatest, or the minimum?

Can i see another explanation for why the remainder 0.35y, other than the algebraic proof above?

When dividing 7 by 2, the result is 3.5. What exactly does that mean? Dividing 7 by 2 is like asking "how many times does 2 fit into 7". The answer is 3 and a half times, or in other words, 3 times with room for another 'half of 2' left: 3 times 2 is 6, which leaves a remainder of 1. Since 1 is half of 2, the remainder is 'half of 2'. Now, suppose x/y=3.5. This means that y fits in x exactly 3.5 times; in other words, you can place 3 y's in x, and still have room for another half of y. As before, this 'half of y' is the remainder. Therefore, in this case, the remainder would be 0.5y. Now, apply the same logic on the question at hand: x/y = 5.35 This means that y fits in x 5 times, with room left for 0.35y, So the remainder is 0.35y.

If a and b are integers, is ab/8 an integer? (1) 2a=b (2) ab/4 is an integer.

You grossly underestimated the time this question took you. You actually solved it in 4 minutes and 15 seconds. Correct. This is a DS Yes\No question. Answering a definite "Yes" or a definite "No" means Sufficient. If the answer is sometimes "Yes" and sometimes "No", it means Maybe, which means Insufficient. The issue is divisibility. For a·b/8 to be an integer, a·b must be divisible by 8. Stat. (1): Plug in good numbers for a and b, such as a=2 and b=4. These numbers satisfy stat. (1), but a⋅b=8 is divisible by 8, so the answer is "Yes". However, is this always true, for any number? Try to find values of a and b that satisfy the equation and yield an answer of "No". If a=1, then b=2, and a⋅b=2, which is not divisible by 8, yielding an answer of "No". No definite answer, so Stat.(1)->Maybe->IS->BCE. Stat. (2): If a·b/4 is an integer, then a·b is divisible by 4. Plug in for a·b as a whole : if a·b=4, then a·b is not divisible by 8 and the answer is "No"; however, if a·b=8, then a·b is divisible by 8 and the answer is "Yes". No definite answer, so Stat.(2)->Maybe->IS->CE. Stat. (1+2): from stat. (2), a·b is divisible by 4. This eliminates the possibility that a=1 and b=2, as 1·2 is not divisible by 4. Under test conditions, after plugging in several more values of a and b that satisfy both statements, you can reach the conclusion that a·b is divisible by 4 and the answer is "Yes". Stat. (1+2)->Yes->S->C. If you're not sure, take a step back and work the problem algebraically: Plug in b=2a into stat. (2): ab/4 = a·2a/4 = 2a^2/4 = a^2/2 is an integer. This means that a^2 is divisible by 2. Since a is an integer, a cannot be √2, so a^2 must be divisible by 4 and a must be divisible by 2. Plug this information into stat. (1): Since b=2a, if a is divisible by 2, then b must be divisible by 2·2, and a·b must be divisible by 2·4=8.

What is the lowest positive integer that is divisible by each of the odd integers between 15 and 21, inclusive? 3×17×19×21 5×17×19×23 7×15×17×19 7×15×19×21 15×17×19×21

You overestimated the time this question took you. You actually solved it in 1 minutes and 19 seconds. Correct. To find the LCM (Least Common Multiple) of 2 or more numbers - 1. Break down each integer into its "building blocks" using the factor tree. 2. Build a list comprised of the least number of prime "building blocks" required to "build" each of the integers. 3. Multiply the building blocks in the list to find the LCM. Remember that each prime number is included the minimum number of times required to build the LCM. The odd integers between 15 and 21 inclusive are 15, 17, 19, and 21. 17 and 19 are primes, thus you just have to factor 15 and 21: The LCM is composed of the minimum factors required to make all 4 odd integers. 3 is only counted once, as both 15 and 21 only need one "3" as a building block. The LCM is thus --> 3×5×17×19×7 = --> 7×15×17×19

x/y = 5.35 If x and y are positive integers, then what is the value of x? (1) When y is divided by x, the remainder is 20. (2) When x is divided by y, the remainder is 7.

You slightly underestimated the time this question took you. You actually solved it in 4 minutes and 47 seconds. Incorrect. This is a "what is the value of..." DS question. In this type of question, a statement will be sufficient only if it leads to a single value of the variable (or expression) you're asked about. The question stem provides a single equation with x and y. In order to find the value of x, you need the value of y. From glancing at the statements, the issue is remainders. Do some preliminary work: Manipulate the equation into Recall the definition of the remainder: Remainder is the distance (in units) from the dividend to the nearest multiple of the divisor that is smaller than the dividend. The nearest multiple of y that is smaller than x is 5y. Whatever's left, i.e. 0.35y, is the remainder when dividing x by y. Stat. (2): According to Stat. 2, when x is divided by y the reminder is 7. hence 0.35y = 7. By solving the equation you may find the value of y, and from that, the value of x. Therefore, Stat.(2)->S. Yes, Stat.(2) is Sufficient. But what about Stat.(1)? Correct. Stat. (1): although the information so far only discusses the remainder when x is divided by y, there is still something that can be determined about the remainder when y is divided by x. From the facts that x and y are both positive, and the fact that x is greater than 5 times y, you can determine a crucial insight - x is greater than y. What is the remainder when dividing a number in a number greater then itself? well, The remainder is equal to the entire smaller number. Let's elaborate on this concept using 2 simple examples: Imagine dividing 20 eggs into cartons of 8 eggs each (This is the verbal equivalent of 20/8). No problem, right? 16 out of the 20 eggs fill up 2 whole cartons, leaving a remainder of 4 eggs. Now, let's see what happens when dividing a smaller number by a greater one: Imagine dividing 8 eggs into cartons of 20 eggs each (the verbal equivalent of 8/20). Seeing that we cannot fill even a single carton, all 8 eggs remain unsorted. In other words, all 8 eggs are literally the remainder. According to Stat. 1, When y is divided by x the remainder is 20. As we just saw, when dividing a smaller number by a greater one, the remainder is equal to the smaller number. Therefore, if x is greater than y, it follows that y equals 20. Having discovered the value of y, you may now plug this value in the original equation to find the value of x. Stat.(1)->S->AD. Stat. (2): According to the original equation, x=5y+0.35y, meaning 0.35y is the remainder of dividing x by y. Using Stat. 2, you may create the equation 0.35y = 7. By solving the equation you can find the value of y; plug in the value of y into the equation x=5y+0.35y and you can find the value of x. Therefore, Stat.(2)->S.

Discrepancy means inconsistency, contradiction or difference.

As in many other words, in this case the prefix dis- has a negative context. A few well-known examples are appear-disappear, ability-disability, and connect-disconnect. Examples: 1. Leg length discrepancy (LLD) is a condition of unequal lengths of the legs. 2. A learning disability is often responsible for the discrepancy between a child's high intelligence and poor academic achievements. In the GMAT verbal section, you are likely to see the word discrepancy in Critical Reasoning questions, specifically in a Critical Reasoning question type called Paradox Questions. Example: Which of the following, if true, best reconciles the seeming discrepancy described above?

Note that each prime number is included the minimum number of times required to build the LCM. We need two 3s to build a 36, but we already included those in the stronger requirement of three 3s to build a 27. LCM=108(2,2,3,3,3) To sum up our general method for LCM questions:

Break down each integer into its "building blocks" using the factor tree. Build a list comprised of the least number of prime "building blocks" required to "build" each of the integers. Multiply the building blocks in the list to find the LCM. Remember that each prime number is included the minimum number of times required to build the LCM. the LCM of 12, 27 and 36 is 108:

What is the lowest possible common multiple of 2 distinct integers, each greater than 251?

Correct. If you notice the "distinct" in the question stem, E seems the quickest and easiest answer - the two next integers greater than 251. However, that would've been too easy, right? The GMAT will not present you with dumb questions - if you've just solved a question in 40 seconds, then you must have missed on something. The question now becomes, can you find a common multiple of two distinct integers greater than 251, that is smaller than the staggering product of 252⋅253? these integers are the smallest - how can you go below the product of the two? Now is the time to think about what you're being asked, but get creative. You don't need the product of the two integers - just a small number that is divisible by both integers, i.e. the LCM. The problem with 252 and 253 is that they're completely distinct from each other - no common factors, so the LCM in this case HAS to be the product of the the two. But what if we take two integers with more common factors? Even if you take 252 and 254, their LCM should be lower than 252⋅253 as 252 and 254 will share a common factor of 2, which only need be counted once. Since the LCM is comprised of a list of the minimum number of factors required to make each number, the smallest LCM of two integers will occur when the two integers have the most in common - so that as many factors appear in both integers, and are thus counted only once. The general principle behind this question is this: The smallest possible LCM of two integers is reached when the two integers have the greatest number of factors in common, factors which are only counted once in the LCM. This should lead you to using 252 (the smallest possible integer), and a multiple of 252, so that the two integers will have the greatest number of common factors that are only counted once in the LCM. Thus, you can use 252 and (252⋅2), so that the common multiple will be the greater of the two: (2⋅252) is divisible by both itself and by 252, and is thus the LCM of 252 and itself. 252⋅2 = 504, which happens to be the correct answer.

Which of the following is odd? 117−220−121+222−123+224 117−221−123+225−123+227 1113-799 1047-519 1-99

Correct. Rules for adding / subtracting Even and Odd numbers: Even ± Even = Even (e.g. 2+2=4; 4-2=2) Even ± Odd = Odd (e.g. 2+1=3; 2-1=1) Odd ± Odd = Even (e.g. 1+1=2; 3-1=2) Don't bother calculating the exact result of the answer choices - that's a waste of time and a careless mistake waiting to happen.Plug in easy numbers (2 for even, 1 for odd) to simulate the rules of adding/subtracting Even and Odd integers. Eliminate answer choices that give an even result. O-E-O+E-O+E=1-2-1+2-1+2=1. Since 1 is odd, the result of the above calculation is also Odd.

What is the lowest positive integer that is divisible by both 5,000,000 and 256? 5,000,000 10,000,000 20,000,000 50,000,000 100,000,000

Correct. The lowest positive integer divisible by 5,000,000 and 256 is the LCM (Lowest Common Multiple) of the two numbers. Break down each integer into its "building blocks" using the factor tree. Build a list comprised of the least number of prime "building blocks" required to "build" each of the integers. Multiply the building blocks in the list to find the LCM. Remember that each prime number is included the minimum number of times required to build the LCM. The prime factors of 5,000,000 and 256 are 5,000,000 = 5 x 106 = 5 x (2 x 5)6 = 26 x 56 x 5 = 26 x 57 256 = 28 Build a list of the minimum factors needed to make up both 5,000,000 and 256: You need: 28 to make up 256 (which already include the 26 needed for 5,000,000) And 57, which are the remaining factors needed to make up 5,000,000. The LCM is the product of these factors: 28 x 57 = 2 x (27 x 57) = 2 x (2 x 5)7 = 2 x (10)7 = 20,000,000 Hence, this is the correct answer.

If r·s≠0, then what is the value of r/s + s/r? (1) r·s=8 (2) r/s=2

Correct. This is a "what is the value of..." DS question. In this type of question, a statement will be sufficient only if it leads to a single value of the variable (or expression) you're asked about. The issue is Fractions. Use the bowtie to add the two fraction: r/s + s/r = r^2+s^2/sr After using the bowtie, the fraction in the question stem looks like this: Stat. (1) provides the denominator of the combined fraction. However, you're still missing the values of s^2 and r^2, so Stat.(1)->IS->BCE. Stat. (2): if r/s = 2, then the reciprocal s/r must equal the reciprocal of 2: 1/2. Therefore, r/s+s/r = 2+1/2 . This is a single value for the sum of the fractions, so Stat.(2)->S->B.

What is the value of y ? (1) y is an integer and 10≤y<13 (2) 11<y<13

Correct. This is a "what is the value of..." DS question. In this type of question, a statement will be sufficient only if it leads to a single value of the variable (or expression) you're asked about. Plug in and see if you can get more than one value. First, consider stat. (1) alone: there are 3 integers that satisfy the inequality: 10, 11, 12. No single value can be determined for y, so Stat. (1)->IS->BCE. Next, consider stat. (2) alone: There is only one integer between 11 and 13 and that is 12. However, the fact that y is an integer in the first place is only known from statement (1). Alone, stat. (2) allows an infinite values of y as a fraction between 11 and 13, such as 11.5, 12.01, or 12.95. No single value can be determined for y, so Stat. (2)->IS->CE. Since neither statement is sufficient on its own, consider both statements together: from stat. (1) you know that y is an integer. From stat. (2) you know that y is between 11 and 13. There is indeed only one integer between 11 and 13, and that is 12. This determines a single value for y, so Stat. (1+2)->S->C.

If x and y are integers, is y divisible by 2? (1) xy=1 (2) x is negative.

Correct. This is a DS Yes\No question. Answering a definite "Yes" or a definite "No" means Sufficient. If the answer is sometimes "Yes" and sometimes "No", it means Maybe, which means Insufficient. The issue is odd and even powers. To prove the statements Insufficient, plug in x and y for which y is and is not divisible by 2. Remember that when a negative number is raised to an even power the result is always positive. Stat (1): Plug in x=1, y=1 into statement 1: 11=1. y=1 is not an integer divisible by 2. giving a "no". Try with a different set of numbers - for instance x=1, y=2. In this case 12=1, but y is an integer divisible by 2, giving an answer of "yes". Since you received two different answers with two different sets of numbers, the answer is "maybe". Therefore, Stat.(1)->Maybe->Insufficient->BCE Stat (2): You can choose whichever y you like since statement 2 tells you nothing of y. In this case, you can pick y=1, or y=2 and the statement will not be affected. Since y=1 is not divisible by 2, and y=2 is, you receive two different answers for two different sets of numbers you plug in. Therefore Stat.(2)->Maybe->IS->CE Stat. (1)+(2): Examine both statements together. To do so, plug the same numbers into both statements and then the question stem. According to stat. (2), x is negative, and according to statement 1, xy = 1. Pick a negative x, for instance, x=-1, and a y that complies with stat (1). Remember that when a negative number is raised to an even power the result is always positive. Therefore in this case y has to be a multiple of 2, and the answer is "yes". The two statements are also satisfied by plugging in y=0; this doesn't change the answer, seeing as zero is indeed divisible by 2. Therefore Stat.(1+2)->Yes->Sufficient->C

If x, y and z are positive integers, is it true that x2 is divisible by 16? (1) 5x = 4y (2) 3x2 = 8z

Correct. This is a Yes/No DS question. Answering a definite "Yes" or a definite "No" means Sufficient. If the answer is sometimes "Yes" and sometimes "No", it means Maybe which means Insufficient. The issue is divisibility. For x2 to be divisible by 16, x must be divisible by 4. According to Stat. (1), --> 5x = 4y Rewrite the equation as 5x/4 = y. Since y is an integer, it follows that 5x must be divisible by 4. Since 5 is not divisible by 4, it follows that x is divisible by 4, thus x2 is divisible by 4·4 = 16, and the answer is a definite 'yes'. Stat.(1)->S->AD. According to Stat. (2), --> 3x2 = 8z First isolate the z to see what the divisibility of x2 should be: z = 3x2/8. Since 3 is not divisible by 8, x2 must be divisible by 8. You would then think that x2 could be equal to 8, 16, 24 and so on. However, you are told that x must be an integer, which means that x2 must be a perfect square (as x2 = 8, for example, yields x = √8, which is NOT an integer) . x2 could, however, be 16, 64, 256, etc... All of these are divisible by 16, which gives us a consistent 'yes'. Therefore, Stat.(2) is sufficient. Stat.(2)->S->D.

If m, n and k are positive integers, is m divisible by 3? (1) 2m2 = 75k (2) 51m = 60n

Correct. This is a Yes/No DS question. Answering a definite "Yes" or a definite "No" means Sufficient. If the answer is sometimes "Yes" and sometimes "No", it means Maybe which means Insufficient. The issue is whether m is divisible by 3. Remember that an integer n is divisible by a smaller integer g if n is divisible by all of g's distinct prime factors. For example, if n=3g, then n is divisible by 3 AND by all of g's distinct prime factors. According to Stat. (1), --> 2m2 = 75k m2 has to be divisible by 75, which is 3·5·5. Since the square root of m2 is m - an integer, the prime factors of m2 must come in pairs (√m2 = m cannot be 5√3, for instance). Therefore, m2 must be divisible by at least another 3, and m is at least √(3·3·5·5) = 3·5. Therefore, m has to be divisible by 3, and the answer is "Yes". Stat.(1)->Yes->S->AD. According to Stat. (2), --> 51m = 60n --> n = 51m/60 = 17m/20 Since n is an integer, m must cancel the 20 in the denominator. Thus m must be divisible 2·2·5, but it does not have to be (though it may be) divisible by 3. No definite answer, so Stat.(2)->Maybe->IS->A.

What is the smallest integer that is greater than 500 and is divisible by 8, 30 and 45?

Correct. To find the LCM (Least Common Multiple) of 2 or more numbers - 1. Break down each integer into its "building blocks" using the factor tree. 2. Build a list comprised of the least number of prime "building blocks" required to "build" each of the integers. 3. Multiply the building blocks in the list to find the LCM. Remember that each prime number is included the minimum number of times required to build the LCM. Found the LCM of the three numbers? Good. Now find the first multiple over 500 of that LCM, and you got it. Factor the three integers: Note that 2 is counted 3 times , because 8 requires 3 "2"s to make. These three "2's already include the 2 needed to make 30. The same goes for the two "3"s needed to make a 45 - they already include the single 3 needed to make a 30. The LCM is 2×2×2×3×3×5=360. This isn't the answer, though, since you're looking for a multiple that is over 500, thus the next multiple of 360 --> 720 is the multiple you're looking for.

If a is an even integer and a/18 is an odd integer, which of the following is NOT an even integer? a2/54 a2/12 a2/9 a2/6 a2/2

Correct. Variables in the answer choices? Plug in an integer for a. a can be 54, for example, since 54 is even and 54=3×18, so 54/18=3, which is odd. a = 18 would work as well. Try to eliminate answers until you're left with only one non-eliminated option. Now, for a number to be divisible by 12, it has to be divisible by the prime building blocks of 12=2*2*3. Let's factor 54: 54=2*3*3*3, so 54²=2*2*3*3*3*3*3*3. Dividing this by 12=2*2*3 will eliminate the 2*2 and leave us with a product of 3s, which is odd. The same will happen with any other plug-in.

To reconcile means to resolve, to settle, to bring back to harmony.

Example: Conflict management is used to help people and organizations reconcile their differences with others. You are likely to encounter this word in Critical Reasoning Paradox questions.

Consider this GMAT question: Which of the following is the smallest number that is divisible by 12, 27 and 36? A 72 B 108 C 135 D 972 E 11,664

How would you approach this problem? Options are: •Multiply 12 by 27 by 36. •Multiply 27 by 36. •Reverse PI - divide each answer choice by all three numbers. •Is there an elegant way The question is basically asking you to find the LCM (Least Common Multiple) of three integers. The following is a general method of solving GMAT LCM problems: Begin by breaking down each number into its building blocks using the factor tree: To be divisible by 12, the LCM must include a 2, 2 and 3 as its prime factors. To be divisible by 27, the LCM must include three 3s as its prime factors. To be divisible by 36, the LCM must include two 2s and two 3s as its prime factors. Now, construct the LCM using the least number of building blocks required to "build" all three integers. The strongest requirements are two 2s (to make 36) and three 3s (to make 27). Therefore, the LCM must have 2, 2, 3, 3, 3 as a minimum of its building clocks. Finally, multiply the minimal building blocks together to find the LCM - 2×2×3×3×3=4×9×3=36×3=108. Therefore, the answer is B.

If x and y are integers, is x even? (1) x+y=x−y (2) x+y=x·y

Incorrect. This is a DS Yes/No question. The issue is whether x is even. Answering a definite "Yes" or a definite "No" means Sufficient. If the answer is sometimes "Yes" and sometimes "No", it means Maybe, which means Insufficient. Isolate x, then try to plug in even and odd numbers for y and x to show that the statements are insufficient. When plugging-in, ask yourself "is this always true, for any number?" According to Stat. (1), --> x+y=x-y --> 2y=0 --> y=0, but what is x? It can be anything. Stat.(1)->IS->BCE. According to Stat. (2), --> x+y=x·y Plug in y=2, x=2: --> 2 + 2 = 2·2, so x can be even. Can x be odd? Plug in odd values for x, but you will not be able to find a corresponding value for y that satisfies the equation in (2). After two or three attempts, take a step back and see if you can find a general rule: If x is odd and y is odd, then their sum is even, but their product is odd. This can't be according to statement 2. If x is odd and y is even, then their sum is odd, but their product is even. This too can't be according to statement 2. Therefore, x cannot be odd, and must be even. Stat.(2)->S->B.

What is the lowest possible common multiple of 2 distinct integers, each greater than 67? 68 69 136 68^2 68.69 Correct. The general principle behind this question is this: The smallest possible LCM of two integers is reached when the two integers have the greatest number of factors in common, factors which are only counted once in the LCM. This should lead you to using 68 (the smallest possible integer), and a multiple of 68, so that the two integers will have the greatest number of common factors that are only counted once in the LCM. Thus, you can use 68 and (68⋅2), so that the common multiple will be the greater of the two: (2⋅68) is divisible by both itself and by 68, and is thus the LCM of 68 and itself. 68⋅2 = 136, which happens to be the correct answer.

Incorrect. This is indeed a possible common multiple of two distinct integers which are greater than 67, but this is not the least possible common multiple of two such numbers. If you notice the "distinct" in the question stem, E seems the quickest and easiest answer - the two next integers greater than 67. However, that would've been too easy, right? The GMAT will not present you with dumb questions - if you've just solved a question in 40 seconds, then you must have missed on something. The question now becomes, can you find a common multiple of two distinct integers greater than 67, that is smaller than the staggering product of 68⋅69? these integers are the smallest - how can you go below the product of the two? Now is the time to think about what you're being asked, but get creative. You don't need the product of the two integers - just a small number that is divisible by both integers, i.e. the LCM. The problem with 68 and 69 is that they're completely distinct from each other - no common factors, so the LCM in this case HAS to be the product of the the two. But what if we take two integers with more common factors? Even if you take 68 and 70, their LCM should be lower than 68⋅69 as 68 and 70 will share a common factor of 2, which only need be counted once. Since the LCM is comprised of a list of the minimum number of factors required to make each number, the smallest LCM of two integers will occur when the two integers have the most in common - so that as many factors appear in both integers, and are thus counted only once.

What is the lowest positive integer that is divisible by each of the even integers between 1 and 10, inclusive? 60 100 120 160 240

Incorrect. To find the LCM (Least Common Multiple) of 2 or more numbers - 1. Break down each integer into its "building blocks" using the factor tree. 2. Build a list comprised of the least number of prime "building blocks" required to "build" each of the integers. 3. Multiply the building blocks in the list to find the LCM. Remember that each prime number is included the minimum number of times required to build the LCM. 240 is indeed a multiple of 2,4,6,8 and 10, but it is not the smallest multiple.