Integers: Rule of Divisibility by 11,7

Now, for the general concept important for GMAT integer questions: Multiple of n ± multiple of n = MUST be a multiple of n. multiple of n ± NOT multiple of n = CANNOT be a multiple of n. How about the case of: NOT multiple of n ± NOT a multiple of n? Can we decide on a rule for the result?

2 and 1 are NOT multiples of 4, and the result of 2+1=3 isn't a multiple of 4 either. However, 3 and 1 are NOT multiples of 4, but the result of 3+1=4 IS a multiple of 4. Therefore, NOT multiple of n ± NOT a multiple of n - May or May NOT be a multiple of n. To sum up: Multiple of n ± multiple of n = MUST be a multiple of n. multiple of n ± NOT multiple of n = CANNOT be a multiple of n. NOT multiple of n ± NOT a multiple of n - May or May NOT be a multiple of n

Rule of divisibility by 11 - This one's a little complicated, but still better than long division. A number is divisible by 11 if the difference between the sum of its digits in the odd places and the sum of its digits in the even places is divisible by 11. In mathematical formula form: (sum of digits in odd places) - (sum of digits in even places) = a number that is divisible by 11.

Let's look at 121,792 and see whether its divisible by 11. The sum of digits in 1st, 3rd and 5th places (odd places: 121,792) = 1+1+9 = 11 The sum of digits in 2nd, 4th and 6th places (even places: 121,792) = 2+7+2 = 11 The difference between these two sums is 11-11=0, which is divisible by 11 (0 divided by 11 is 0, which is an integer). Therefore the original number 121,792 is divisible by 11.

The rule of divisibility for 7 is a little complicated, but fun: 1) Multiply the number of hundreds the number has by 2. 2) Add the remaining two digit number. 3) If the result is divisible by 7, the original number is divisible by 7. 5040 is divisible by 7: 1) There are 50 "hundreds" in 5040. Multiply 50 by 2 to get 100. 2) Add the remaining two-digit number 40 to get 140. 3) 140 is divisible by 7, so 5040 is divisible by 7

5040 is divisible by 7: 1) There are 50 "hundreds" in 5040. Multiply 50 by 2 to get 100. 2) Add the remaining two-digit number 40 to get 140. 3) 140 is divisible by 7, so 5040 is divisible by 7

If N is the product of all positive integers less than 31, then what is the greatest integer k for which is an integer?

The limitation is that N/18k needs to be an integer. So that all the powers of 18 in the denominator must be reduced by "18s" in the numerator N. The number of 18s N can be divided by is simply the number of 18 building blocks it contains. For example, if N contained one 18 building block, it could be divided only by 181. If it contained two 18 building blocks, it could be divided by 182, and so on. The question is effectively asking "N can be divided by how many 18s?". 2) N is the product of all positive integers less than 31: i.e., N=1∙2∙3∙4...∙30. (the multiplication of all the numbers between 1 and 30). It seems that N has only one 18 in it, so k should be 1, but don't fall into that trap: 9 and 2 can make another 18. So can 6 and 3. You could theoretically count all of the different ways of making an 18 in N, but that will take too long. Try a more systematic approach: 3) Further simplify the question. Which do you have more of, the 3s or the 2s? Every even number in N is divisible by 2: 8 alone accounts for three 2s (8=23), which could be paired up with 3s to make another 18. In short, you have many more 2s than 3s. You can now count how many multiples of 3 are there in N, and know that for every pair of 3s you will have a 2 to pair up and make an 18. The question now becomes: "how many 3s are there in N?" 4) List the multiples of 3 in N: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30. Ignore the twos, and find how many powers of 3 there are. Correct. There are 10 multiples of 3 less than 31: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 Ignore the twos, and find how many powers of 3 there are. 3: one power of 3. 6 = 3⋅2: one power of 3. 9 = 3⋅3: two powers of 3. 12 = 3⋅4: one power of 3. 15 = 5⋅3: one power of 3. 18 = 3⋅3⋅2: two powers of 3. 21 = 7⋅3: one power of 3. 24 = 8⋅3: one power of 3. 27 = 33 : three powers of 3. 30 = 3⋅10: one power of 3. The total is 14 powers of 3, which can be used to make 7 "18s", as each 18 demands a pair of 3s. Thus, the answer is C.

Is 22 divisible by 4?

Let's try to formulate the rule behind this intuition: multiple of 4 ± multiple of 4 - MUST be a multiple of 4. multiple of 4 ± NOT multiple of 4 - CANNOT be a multiple of 4. So 20±8 must be a multiple of 4, because all we're doing is adding or subtracting multiples of 4. However, 20± 6 CANNOT be a multiple of 4, because the result "lands" somewhere between other multiples of 4.

If N is the product of all multiples of 3 between 1 and 100, what is the greatest integer m for which is an integer?

Incorrect. 1) First step: understand what the question is asking: Break down N: N=3⋅6⋅9⋅...30⋅33..60⋅63..90⋅93....⋅96⋅99 - all the multiples of 3 within the range. The limitation is that N/10m needs to be an integer. So that all the powers of 10 in the denominator must be reduced by "10s" in the numerator N. The question is effectively asking "N can be divided by how many 10s?" 2) Don't make the mistake of focusing only on the multiples of 10 in N (30, 60, 90). Remember that a 10 can also be 'constructed' from the building blocks of 2·5 - for example, 15 and 6 (both included in N) multiplied together equal 90, which is another power of 10 than can eliminate a 10 on the denominator. You could theoretically count all of the different ways of making a 10 in N, but that will take too long, and you might miss something. Try a more systematic approach: 3) Further simplify the question. Which do you have more, the 5s or the 2s? Note that there are many more 2-factors than 5-factors (every even number in N will have a 2), so there is no need to count all the 2s - the 5s (that are also multiples of 3) are scarce, so count only them. You can now count how many multiples of 5 are there in N, and know that for every 5 you will have a 2 to pair up and make a 10. The question now becomes: "how many 5s are there in N?" Since N is composed of multiples of 3, you are effectively looking for multiples of both 5 and 3: The multiples of 5 that are also multiples of 3 are basically multiples of 5·3=15. There are 6 multiples of 15 between 1 and a 100: 15, 30, 45, 60, 75, 90. Focus on those, and count how many powers of 5 are in them. Correct. Count the number of 5s: 15, 30, 45, 60, and 90 each have one '5'. However, 75=3⋅25 = 3⋅52 has two building blocks of 5. This gives you a total of seven '5' building blocks, to which you will be able to find a '2' from some other factor on N, so N can be divided by 10 seven times at most.

If m is a two-digit number, what is the remainder when m is divided by 3? (1) m+1 is divisible by 3. (2) m is positive, and the sum of its two digits is 8. Correct. Stat.(1): If m+1 is divisible by 3, then m-2 is also divisible by 3. In this case, dividing m by 3 can only leave a remainder of 2. You could also plug in m+1 = 12, so m = 11 = 3·3+2 or m+1 = -12, so m= -13 = -5·3+2... all such m's leave a remainder of 2 when divided by 3. (Do keep in mind that all remainders are positive, even when the divisor or dividend is negative.) Therefore, Stat.(1)->S->AD. Stat.(2): Mark m's digits as XY, so that X+Y = 8. According to the Rule of Divisibility by 3, m isn't divisible by 3, but m+1 is divisible by 3, because the sum of its digits is X + (Y+1) = 9, which is divisible by 3. Now, as you've established, if m+1 is divisible by 3, then m-2 is also divisible by 3 and m/3 leaves a remainder of 2. Alternative method for Stat.(2): Manually test all cases - there aren't that many. List all two-digit numbers for which the sum of the figures is 8: 17, 26, 35, 44, 53, 62, 71, 80. The nearest multiples of 3 are 15, 24, 33, 42, 51, 60, 69, 78, respectively. All of the numbers you listed as m are two more than the nearest multiple of 3, which means they all leave a remainder of 2 when divided by 3. Stat.(2)->S->D.

Incorrect. The issue is remainder. When any number, including m, is divided by 3, the remainder may be 0, 1, or 2. Yes, Stat.(2) is Sufficient. But what about Stat.(1)? Note that remainders are always positive, even when the divisor or the dividend is negative. For example, try plugging in -22 for m (where m+1 = -22+1 = -21, which is divisible by 3). If m and n are integers, where n≠0, then the remainder is a non-negative integer r such that m = qn + r (where q is the quotient), and with 0 ≤ r < |n|. What this means in plain English is that -22/3 = -8 r2. "What the ......???" you ask. Here's the breakdown: m = -22 n = 3 r = 0, 1, or 2 If you multiply q = -7 by 3, you get -21. However, -22 = -21 + -1, and this violates 0 ≤ r < |n|. Hence, -1 cannot be the remainder. If you multiply q = -8 by 3, you get -24. -22 = -24 + 2, which satisfies both restrictions. Hence, -22/3 gives a quotient of -8 with a remainder of 2. Plug-in some different positive and negative numbers and see what you come up with.

In the above example we counted odd and even place from left to right, i.e 100,000s digit -1st place, units digit - last place (6th place). However, for the purpose of this particular rule of divisibilty the direction of counting makes no difference. We could just as easily count right to left, i.e units digit - 1st place, 100,000 digit last. As long as the difference between the sums of odd and even placed integers is divisble by 11, the original number will be divisible by 11.

Is 902 divisible by 11?

To sum up:

Common divisors - definition: Two non-zero integers have a common divisor if they are both divisible by the same positive integer. Remember this about the Greatest Common Divisor of two integers: Two primes will always have a G.C.D of 1. A G.C.D of 1 does not mean the two integers are prime - just that they have no common divisor greater than 1. Example: 8 and 9 have a G.C.D of 1. An integer can serve as the G.C.D of itself and another integer. Example: 6 and 12 have a G.C.D of 6.

Integers: GCD - Greatest Common Divisor

Common divisors - definition: Two non-zero integers have a common divisor if they are both divisible by the same positive integer. For example, 2 is a common divisor of 6 and 8, because both 6 and 8 are divisible by 2. All integers are divisible by 1, so 1 is always a common divisor of any two integers. However, some GMAT integers questions ask about the Greatest Common Divisor of two or more integers, and whether the G.C.D is greater than 1 or not. In the above case, 2 is also the Greatest Common Divisor of 6 and 8.

If m and k are non-zero integers, is m a multiple of k? (1) (m2+m)/k is an integer. (2) m=2k2−3k

Correct. According to Stat.(1), m(m+1)/k = integer. Plugging in m=3 and k=6 gives 3·(3+1)/6 = 2. However, 3 is not a multiple of 6, which makes the answer "No." Plugging in m=4 and k=2 gives 4·(4+1)/2 = 10. In this case, 4 is a multiple of 2, which makes the answer "Yes." Therefore, Stat.(1)->Maybe->IS->BCE. According to Stat.(2), m=2k2−3k. Both terms on the right hand side are multiples of k. Multiple of k - multiple of k = multiple of k. Hence, m must also be a multiple of k. Therefore, Stat.(2)->Yes->S->B.

One final fun fact regarding the Greatest Common Divisor is demonstrated in the following question: Which of the following is the greatest common divisor of 6 and 12? 1 2 3 6 12

Correct. Since both 6 and 12 are divisible by 6, 6 is a common factor of 12. And Since 6 cannot be divisible by any number greater than itself, it is also the Greatest common factor of 12. Remember that any integer is also a factor of itself, and therefore any integer could serve as the Greatest Common Divisor of itself and another integer.

k=2n+7, where n is an integer greater than 1. If k is divisible by 9, which of the following MUST be divisible by 9? 2^n-8 2^n-2 2^n 2^n+4 2^n+5

Plugging in is the method of choice for this question in general, as it employs variables in the answer choices. However, the good numbers here are difficult to find (here's a clue: n=7). Take a step back and look at what the question is providing: 2n+7 is divisible by 9, which means that 2n+7 is also a multiple of 9. Multiple of 9 ± multiple of 9 = MUST be a multiple of 9. Therefore, the right answer must be a "multiple of 9 away" from 2n+7 in order to give a result that is a multiple of 9. The only answer that fits the bill is 2n-2, which is basically (2n+7)-9.

Good Info

Since prime numbers are divisible only by themselves and 1, any two primes will only have a G.C.D of 1. However, note that the reverse isn't necessarily true: The fact that two integers have a G.C.D of 1 does not necessarily indicate that they are Primes. This is a common misconception, easily disproved by example: 8 and 9 have a G.C.D of 1, even though neither of them is Prime.