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Re: What is the lowest positive integer that is divisible by [#permalink]
11 Sep 2012, 08:24

1

This post received KUDOS

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 1 420 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. _________________

Re: What is the lowest positive integer that is divisible by [#permalink]
11 Sep 2012, 09:07

1

This post received KUDOS

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 _________________

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

My choice is E.But its wrong. My Explanation:lowest positive integer that is divisible by each of the integers 1 through 7 means 7!. 7!=5040.Hence (E) However this answer is wrong. I am not able to understand the OG12 explanation

Re: What is the lowest positive integer that is divisible by [#permalink]
13 Jan 2013, 22:43

4

This post received KUDOS

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.

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

My choice is E.But its wrong. My Explanation:lowest positive integer that is divisible by each of the integers 1 through 7 means 7!. 7!=5040.Hence (E) However this answer is wrong. I am not able to understand the OG12 explanation

Merging similar topics. Please ask if anything remains unclear.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to the rule #3: the name of a topic (subject field) MUST be the first 40 characters (~the first two sentences) of the question. _________________

Re: What is the lowest positive integer that is divisible by [#permalink]
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.

Re: What is the lowest positive integer that is divisible by [#permalink]
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

Re: What is the lowest positive integer that is divisible by [#permalink]
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?

Re: What is the lowest positive integer that is divisible by [#permalink]
29 Dec 2013, 10:20

Expert's post

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?

The 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 prime-factorization. Then multiply all the factors (pick the highest power of the common factors). _________________

What is the lowest positive integer that is divisible by [#permalink]
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?

gmatclubot

What is the lowest positive integer that is divisible by
[#permalink]
24 Nov 2014, 12:46

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