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What is the lowest positive integer that is divisible by
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11 Sep 2012, 03:43
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Re: What is the lowest positive integer that is divisible by
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11 Sep 2012, 03:43
SOLUTIONWhat is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive?(A) 420 (B) 840 (C) 1,260 (D) 2,520 (E) 5,040 The integer should be divisible by: 2, 3, 4(=2^2), 5, 6(=2*3), and 7. The least common multiple of these integers is LCM=2^2*3*5*7=420. Answer: A.
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Re: What is the lowest positive integer that is divisible by
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13 Jan 2013, 22:43
You need LCM of first 7 numbers, NOT factorial. If a number is divisible by 6 then its also divisible by 2 & 3. You dont have to count 2 & 3 again when you consider factor as 6. \(1=1^1\) \(2=2^1\) \(3=3^1\) \(4=2^2\) \(5=5^1\) \(6=2^1 *3^1\) \(7=7^1\) \(LCM =1^1 * 2^2 * 3^1 * 5^1 * 7^1= 420\) Hence choice(A) is the answer.
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Re: What is the lowest positive integer that is divisible by
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11 Sep 2012, 08:24
Factors of 420 are: 1, 2, 2, 3, 5, 7
Now 420 should be divided by each of 1, 2, 3, 4, 5, 6, 7 From the factors which we get above by factorization and combining them to make all numbers from 1 to 7, 420 divided by 1420 divided by 2 (picking up from factors) 420 divided by 3 (picking up from factors) 420 divided by 4 (picking up two 2s and multiplying to make it 4) 420 divided by 5 (picking up from factors) 420 divided by 6 (picking up 2 and 3 from factors and multiplying to make it 6) 420 divided by 7 (picking up from factors) All leaves remainder as ZERO. No need to go further. Answer is A.
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Re: What is the lowest positive integer that is divisible by
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11 Sep 2012, 09:04
its A.. 1*2*3*2*5*7=420...
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Re: What is the lowest positive integer that is divisible by
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11 Sep 2012, 09:07
Here are the integers from 1 to 7 including: 1, 2, 3, 4, 5, 6, 7 So the lowest positive integer divisible by every single numbers set forth above would have to be divisible by 7,5,4,3 simultaneously or 7*5*4*3=420 please, correct me if I went awry
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Re: What is the lowest positive integer that is divisible by
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12 Sep 2012, 11:26
What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive? (A) 420 (B) 840 (C) 1,260 (D) 2,520 (E) 5,040 The question is basically asking the LCM (smallest multiple) of number from 1 to 7 (both inclusive) LCM of 1,2,3,4,5,6,7 = 420 Answer A Hope it helps
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Re: What is the lowest positive integer that is divisible by
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17 Jan 2013, 02:58
PraPon wrote: You need LCM of first 7 numbers, NOT factorial. If a number is divisible by 6 then its also divisible by 2 & 3. You dont have to count 2 & 3 again when you consider factor as 6.
\(1=1^1\) \(2=2^1\) \(3=3^1\) \(4=2^2\) \(5=5^1\) \(6=2^1 *3^1\) \(7=7^1\)
\(LCM =1^1 * 2^2 * 3^1 * 5^1 * 7^1= 420\)
Hence choice(A) is the answer. how about the 4 and the 2? if 4 is divisible by 2 and 2, then should we not consider it as well? I find this confusing.



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Re: What is the lowest positive integer that is divisible by
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17 Jan 2013, 08:31
the question stem says that the number is divisible by each of the integers from 1 through 7..then,simply,it means that each number is a factor of our lowest common multiple..hence 420 holds water
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Re: What is the lowest positive integer that is divisible by
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29 Dec 2013, 10:01
Bunuel wrote: SOLUTION
What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive?
(A) 420 (B) 840 (C) 1,260 (D) 2,520 (E) 5,040
The integer should be divisible by: 2, 3, 4(=2^2), 5, 6(=2*3), and 7. The least common multiple of these integers is LCM=2^2*3*5*7=420.
Answer: A. Hi Bunuel, I do understand the LCM but can you explain WHY when we have four sets of 2's we eliminate TWO of them? For our two 3's, we eliminate one of them and I understand why  we might have the "same" 3. But when we eliminate two 2's, it's not as intuitive to me. Is it simply that, when we have an even number of the same integer, we remove half of them when we calculate the LCM? If so, what would've happened if we had three 2's or five 3's?



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Re: What is the lowest positive integer that is divisible by
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29 Dec 2013, 10:20
aeglorre wrote: Bunuel wrote: SOLUTION
What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive?
(A) 420 (B) 840 (C) 1,260 (D) 2,520 (E) 5,040
The integer should be divisible by: 2, 3, 4(=2^2), 5, 6(=2*3), and 7. The least common multiple of these integers is LCM=2^2*3*5*7=420.
Answer: A. Hi Bunuel, I do understand the LCM but can you explain WHY when we have four sets of 2's we eliminate TWO of them? For our two 3's, we eliminate one of them and I understand why  we might have the "same" 3. But when we eliminate two 2's, it's not as intuitive to me. Is it simply that, when we have an even number of the same integer, we remove half of them when we calculate the LCM? If so, what would've happened if we had three 2's or five 3's? From here: mathnumbertheory88376.htmlThe lowest common multiple or lowest common multiple (lcm) or smallest common multiple of two integers a and b is the smallest positive integer that is a multiple both of a and of b. To find the LCM, you will need to do primefactorization. Then multiply all the factors ( pick the highest power of the common factors).
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What is the lowest positive integer that is divisible by
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24 Nov 2014, 12:46
Bunuel wrote: What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive? (A) 420 (B) 840 (C) 1,260 (D) 2,520 (E) 5,040 Practice Questions Question: 40 Page: 157 Difficulty: 600 With this approach, if LCM of these numbers(from 1 to 7) was not available in options, than precious time could have been wasted while calculating LCM. So in my opinion the quickest and most accurate approach would be to check the divisibility of each given option (by starting with lowest one) by using following rules: 3>>>>sum of digits should be divisible by 3 6>>>>integer should be divisible by both 2 and 3 9>>>>sum of digits should be divisible by 9 4>>>>should be divisible by 2 twice 5>>>>should be 0 or 5 in the end 7>>>>check divisibility in normal way Please guide me if i am wrong?



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Lowest positive integer that is divisible by each...
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15 Sep 2015, 12:03
OG 13/2015  PS  40  157: What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive? A) 420 B) 840 C) 1260 D) 2520 E) 5040 Seeking a better explanation for the answer and a faster way to solve. In my approach I attempted to figure out if each term was divisible by each 1, 2, 3, 4, 5, 6, and 7, starting with D and B. I see now this was not the right approach. In reviewing the explanation provided from MGMAT, I am still confused and would appreciate clarification on how to solve.



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What is the lowest positive integer that is divisible by
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15 Sep 2015, 12:11
KOS75 wrote: OG 13/2015  PS  40  157: What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive? A) 420 B) 840 C) 1260 D) 2520 E) 5040 Seeking a better explanation for the answer and a faster way to solve. In my approach I attempted to figure out if each term was divisible by each 1, 2, 3, 4, 5, 6, and 7, starting with D and B. I see now this was not the right approach. In reviewing the explanation provided from MGMAT, I am still confused and would appreciate clarification on how to solve. Search for a question before posting. Topics merged. Lowest positive integer = LCM of (1,2,3,4,5,6,7) = LCM (1,2,3,2^2,5,2*3,7) = 1*2^2*3*5*7 = 1*4*3*5*7=420. For LCM of 2 numbers, you need to break down the numbers to their prime factors and then the LCM will be composed of the product of prime factors raised to the heighest power in either of the 2 numbers. In the given question, you had 2 as the highest power of 2 and 1 as the highest power of 3,5,7



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Re: What is the lowest positive integer that is divisible by
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16 Sep 2015, 04:36
KOS75 wrote: OG 13/2015  PS  40  157: What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive? A) 420 B) 840 C) 1260 D) 2520 E) 5040 Seeking a better explanation for the answer and a faster way to solve. In my approach I attempted to figure out if each term was divisible by each 1, 2, 3, 4, 5, 6, and 7, starting with D and B. I see now this was not the right approach. In reviewing the explanation provided from MGMAT, I am still confused and would appreciate clarification on how to solve. We simply need to find the LCM of 1,2,3,4,5,6,7 here and that is 420. LCM of a set of numbers is the product of highest powers of primes numbers in that set. In this case LCM = 1*2^2*3*5*7 = 420 Moreover, if you are testing options in a question that asks you to find the lowest value, you should always start with the lowest one. This ways if the answer is the lowest option (as is the case here), yo do not need to look further.



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Re: What is the lowest positive integer that is divisible by
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27 May 2016, 06:22
Bunuel wrote: What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive?
(A) 420 (B) 840 (C) 1,260 (D) 2,520 (E) 5,040 We need to determine the smallest number that is divisible by the following: 1, 2, 3, 4, 5, 6, and 7 That is, we need to find the least common multiple of 1, 2, 3, 4, 5, 6, and 7; however, it may be easiest to use the answer choices and the divisibility rules. Let’s start with answer choice A, 420. Since 420 is an even number we know 2 divides into 420. Since the digits of 420 add to 6 (a multiple of 3), we know 3 divides into 420. Since the last two digits of 420 (20) is divisible by 4, we know 4 divides into 420. Since 420 ends in a zero, we know 5 divides into 420. Since 420 is divisibly by both 2 and 3, we know 6 divides into 420. Finally, we need to determine whether 420 is divisible by 7. While there is no easy divisibility rule for 7, we do know that 7 divides evenly into 42, so it must also divide evenly into 420. Thus, we have determined that 420 is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive. The answer is A.
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Re: What is the lowest positive integer that is divisible by
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27 Feb 2018, 08:05
Bunuel wrote: SOLUTION
What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive?
(A) 420 (B) 840 (C) 1,260 (D) 2,520 (E) 5,040
The integer should be divisible by: 2, 3, 4(=2^2), 5, 6(=2*3), and 7. The least common multiple of these integers is LCM=2^2*3*5*7=420.
Answer: A. Bunuel if you counted total FOUR 2s here > 2, 3, 4(=2^2), 5, 6(=2*3), and 7 then why did you count here only TWO 2s ? LCM=2^2*3*5*7=420. PLEASE EXPLAIN



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Re: What is the lowest positive integer that is divisible by
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27 Feb 2018, 09:17
dave13 wrote: Bunuel wrote: SOLUTION
What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive?
(A) 420 (B) 840 (C) 1,260 (D) 2,520 (E) 5,040
The integer should be divisible by: 2, 3, 4(=2^2), 5, 6(=2*3), and 7. The least common multiple of these integers is LCM=2^2*3*5*7=420.
Answer: A. Bunuel if you counted total FOUR 2s here > 2, 3, 4(=2^2), 5, 6(=2*3), and 7 then why did you count here only TWO 2s ? LCM=2^2*3*5*7=420. PLEASE EXPLAIN Please read the whole thread and follow the links: https://gmatclub.com/forum/whatisthe ... l#p1311043
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Re: What is the lowest positive integer that is divisible by
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28 Feb 2018, 00:01
Bunuel wrote: What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive? (A) 420 (B) 840 (C) 1,260 (D) 2,520 (E) 5,040 Practice Questions Question: 40 Page: 157 Difficulty: 600 IMO A basically this question indirectly asking for LCM(lowest common factor) and if you find LCM then it is 420.




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