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What is the lowest positive integer that is divisible by [#permalink]
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29 Jul 2008, 22:57
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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
What is the fastest way to solve...



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Re: Integers [#permalink]
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30 Jul 2008, 00:05
The lowest possible number divisible by the integers 1 through 7 will have the least possible number of primes. In this case:
2 (divisible by 2) 3 (divisible by 3) 2 (divisible by 4  you already have one 2) 5 (divisible by 5)  (no need for a six, you already have a 2 and a 3) 7 (divisible by 7)
(2^2)(3)(5)(7) = 420 = A



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Re: Integers [#permalink]
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30 Jul 2008, 00:12
asdert wrote: The lowest possible number divisible by the integers 1 through 7 will have the least possible number of primes. In this case:
2 (divisible by 2) 3 (divisible by 3) 2 (divisible by 4  you already have one 2) 5 (divisible by 5)  (no need for a six, you already have a 2 and a 3) 7 (divisible by 7)
(2^2)(3)(5)(7) = 420 = A How come 6 we don't use 2? We did for "4". 2 would be the lowest possible integer divisible by 6... Maybe the simplest way would have been for me to start w/ A and divide by integers 17 and worked my way up.



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Re: Integers [#permalink]
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30 Jul 2008, 03:21
420.
Fastest way:
Multiply all the prime numbers between 1 to 7 in their highest powers.
2^2*3*5*7 =420



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Re: Integers [#permalink]
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30 Jul 2008, 06:11
Still don't understand why we don't use prime #2 for 6 whereas we did for 4...



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Re: Integers [#permalink]
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30 Jul 2008, 06:42
droopy57 wrote: Still don't understand why we don't use prime #2 for 6 whereas we did for 4... LCM of all 7 numbers LCM OF 1, 2, 3, 4, 5, 6, 7 = LCM OF 1 , 2, 3, 2^2, 5, 2*3, 7 = 1 * 2^2 * 3 * 5 * 7 = 420 Hope this helps
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Re: Integers [#permalink]
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30 Jul 2008, 08:30
x2suresh wrote: droopy57 wrote: Still don't understand why we don't use prime #2 for 6 whereas we did for 4... LCM of all 7 numbers LCM OF 1, 2, 3, 4, 5, 6, 7 = LCM OF 1 , 2, 3, 2^2, 5, 2*3, 7 = 1 * 2^2 * 3 * 5 * 7 = 420 Hope this helps Sorry to beat the dead horse but.... ok so LCM "LCM of 6" would be #6 or otherwise #2*#3. Since we have #2 & #3 already represented in "LCM of 2, 3", "LCM of 6" would be duped. "LCM of 4" is #4 or #2*#2. Are we including the other #2 from "LCM of 2" since #4 or any other combination does not make up #4?



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Re: Integers [#permalink]
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30 Jul 2008, 09:12
The multiple can be found by mutiplying the factors: 1x2x3x4x5x6x7
as 4 = 2x2, 2 should be omitted as 6 = 3x2 , 6 should be omitted
Ans = 1x3x4x5x7 = 420



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Re: Integers [#permalink]
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30 Jul 2008, 09:22
rahulgoyal1986 wrote: 420.
Fastest way:
Multiply all the prime numbers between 1 to 7 in their highest powers.
2^2*3*5*7 =420 i agreed with this approach



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Re: Integers [#permalink]
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30 Jul 2008, 09:42
droopy57 wrote: Sorry to beat the dead horse but....
ok so LCM
"LCM of 6" would be #6 or otherwise #2*#3. Since we have #2 & #3 already represented in "LCM of 2, 3", "LCM of 6" would be duped.
"LCM of 4" is #4 or #2*#2. Are we including the other #2 from "LCM of 2" since #4 or any other combination does not make up #4? Droopy, try to see it this way: You need to include as many primes needed to build the divisors. You invlude two 2s because that's what you need to make it divisible by 4. You don't need to include those of 6 because you already have what you need, a 2 and a 3. This would be different if you needed the number to be divisible by 18. In this case, you would need an extra 3 (18 is made our of 2, 3, 3 and since you already have a 2 and a 3...) That makes sense? Also, although I wouldn't take this approach, I would use divisibility rules if I wanted to divide all the answer choices. However, I would advice for you to understand the concept, since you will definitely see questions regarding prime factorization, prime box... Now let me ask you this. What's the least number that can be divisible by the integers 18 and 12? How many more primes and which ones do you need?



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Re: Integers [#permalink]
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30 Jul 2008, 10:25
asdert wrote: droopy57 wrote: Sorry to beat the dead horse but....
ok so LCM
"LCM of 6" would be #6 or otherwise #2*#3. Since we have #2 & #3 already represented in "LCM of 2, 3", "LCM of 6" would be duped.
"LCM of 4" is #4 or #2*#2. Are we including the other #2 from "LCM of 2" since #4 or any other combination does not make up #4? Droopy, try to see it this way: You need to include as many primes needed to build the divisors. You invlude two 2s because that's what you need to make it divisible by 4. You don't need to include those of 6 because you already have what you need, a 2 and a 3. This would be different if you needed the number to be divisible by 18. In this case, you would need an extra 3 (18 is made our of 2, 3, 3 and since you already have a 2 and a 3...) That makes sense? Also, although I wouldn't take this approach, I would use divisibility rules if I wanted to divide all the answer choices. However, I would advice for you to understand the concept, since you will definitely see questions regarding prime factorization, prime box... Now let me ask you this. What's the least number that can be divisible by the integers 18 and 12? How many more primes and which ones do you need? Least # would be 420. We do not need any more primes b/c primes 3 and 4 are sufficient to be divisible by 12. {3, 4, 5, 7}= 420 On the other hand, if the number was 24, we would need an extra 2 to make it divisible by 24. {2, 3, 4, 5, 7}= 840



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Re: Integers [#permalink]
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30 Jul 2008, 11:14
droopy57 wrote: 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
What is the fastest way to solve... you have to find the least common multiple for all the integers from 1 to 7. So let's look at each of the number and try to break them down to their prime factors: \(2=\)\(2^2\) \(3=\)\(3^1\) \(4=\)\(2^2\) \(5=\)\(5^1\) \(6=\)\(2^1*3^1\) \(7=\)\(7^1\) Then, because we're looking for the least common multiple, you should consider only the prime factors that have the highest number of exponents and then multiply them by each other. Therefore; \(2^2*3^1*5^1*7^1= 420\) Therefore, answer is A



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Re: Integers [#permalink]
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30 Jul 2008, 12:34
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droopy57 wrote: asdert wrote: Now let me ask you this. What's the least number that can be divisible by the integers 18 and 12? How many more primes and which ones do you need? Least # would be 420. We do not need any more primes b/c primes 3 and 4 are sufficient to be divisible by 12. {3, 4, 5, 7}= 420 You missed an extra 2 needed for the number to be divisible by 8, but it seems like you got the concept good luck.



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Re: Integers [#permalink]
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30 Jul 2008, 12:41
asdert wrote: droopy57 wrote: asdert wrote: Now let me ask you this. What's the least number that can be divisible by the integers 18 and 12? How many more primes and which ones do you need? Least # would be 420. We do not need any more primes b/c primes 3 and 4 are sufficient to be divisible by 12. {3, 4, 5, 7}= 420 You missed an extra 2 needed for the number to be divisible by 8, but it seems like you got the concept good luck. Thanks everyone. Asdert, your explanation cleared things cleared things up very nicely. i misread the question to be 1~7. anyways kudos to you



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Re: Integers [#permalink]
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31 Jul 2008, 10:39
droopy57 wrote: 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
What is the fastest way to solve... 1 to 7 means:1x2x3x4x5x6x7.. 4=2^2 6=2x3. So if a number could be divisible by 1 to 7. that is :1x2x2x3x5x7=420



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Re: Integers [#permalink]
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03 Aug 2008, 07:52
droopy57 wrote: 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
What is the fastest way to solve... Its just a tricky question of LCM , We got to take LCM of 1 through 7 =420 is Answer
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