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## Kinh Nghiệm Hướng dẫn What is the lowest positive integer that is divisible by each of the integers 1 through 7? 2022 Mới Nhất

Pro đang tìm kiếm từ khóa What is the lowest positive integer that is divisible by each of the integers 1 through 7? 2022 được Update vào lúc : 2022-10-24 00:10:00 . Với phương châm chia sẻ Thủ Thuật về trong nội dung bài viết một cách Chi Tiết Mới Nhất. Nếu sau khi tìm hiểu thêm nội dung bài viết vẫn ko hiểu thì hoàn toàn có thể lại phản hồi ở cuối bài để Mình lý giải và hướng dẫn lại nha.

Mẹo Hướng dẫn What is the lowest positive integer that is divisible by each of the integers 1 through 7? 2022

Quý khách đang tìm kiếm từ khóa What is the lowest positive integer that is divisible by each of the integers 1 through 7? được Cập Nhật vào lúc : 2022-10-24 00:05:19 . Với phương châm chia sẻ Kinh Nghiệm về trong nội dung nội dung bài viết một cách Chi Tiết 2022. Nếu sau khi tìm hiểu thêm nội dung nội dung bài viết vẫn ko hiểu thì hoàn toàn hoàn toàn có thể lại phản hồi ở cuối bài để Ad lý giải và hướng dẫn lại nha.

First find the least common multiple (LCM) of all the integers 1 to 16. One way to find the LCM’s is to use prime factorization.

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1 = **1**

2 = 2

3 = 3

4 = 2 * 2

5 = **5**

6 = 2 * 3

7 = **7**

8 = 2 * 2 * 2

9 = **3 * 3**

10 = 2 * 5

11 = **11**

12 = 2 * 2 * 3

13 = **13**

14 = 2 * 7

15 =

3 * 5

16 = **2 * 2 * 2 * 2**

The LCM of all these will be the product of all the different factors. When a factor appears more than once in list of factors, use the the longest string of this factor. So, LCM = 1*2*2*2*2*3*3*5*7*11*13 = **720720**.

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Math Expert

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

11 Sep 2012, 03:43

00:00

Question Stats:

75%

(01:04) correct 25% (01:17) wrong based on 3426 sessions

### Hide Show timer Statistics

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

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Math Expert

Joined: 02 Sep 2009

Posts: 86916

Re: What is the lowest positive integer that is divisible by each of the

[#permalink]

11 Sep 2012, 03:43

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.

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Re: What is the lowest positive integer that is divisible by each of the

[#permalink]

11 May 2022, 09:43

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

——-ASIDE————————–

A lot of integer property questions can be solved using prime factorization.

For questions involving divisibility, divisors, factors and multiples, we can say:

If N is divisible by k, then k is “hiding” within the prime factorization of

N

Consider these examples:

24 is divisible by 3 because 24 = (2)(2)(2)(3)

Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)

And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)

And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7)

—–NOW ONTO THE QUESTION—————–

What is the lowest

positive integer that is divisible by each of the integers 1 through 7, inclusive?

Let K = that lowest positive integer

This means that there’s a 2 “hiding” within the prime factorization of K, a 3 “hiding” within the prime factorization of K, a 4 “hiding” within the prime factorization of K, etc.

So, let’s begin with a 2 “hiding” within the prime factorization of K.

This means that K =

(2)(other numbers)

Also, if there’s a 3 “hiding” within the prime factorization of K, then we need to add a 3 like so: K = (2)(3)

There’s a 4 “hiding” within the prime factorization of K.

Since 4 = (2)(2), then we need to add a SECOND 2 to get: K = (2)(2)(3)

There’s a

5 “hiding” within the prime factorization of K, so we’ll add a 5 to get: K = (2)(2)(3)(5)

There’s a 6 “hiding” within the prime factorization of K.

Since 6 = (2)(3), we can see that we ALREADY have a 6 “hiding” in the prime factorization: K = (2)(2)(3)(5)

There’s

a 7 “hiding” within the prime factorization of K, so we’ll add a 7 to get: K = (2)(2)(3)(5)(7)

We have now ensured that K is divisible by every integer from 1 to 7 inclusive. This means that we have found the LEAST possible value of K that satisfies the given conditions.

So, K = (2)(2)(3)(5)(7) = 420

Answer: A

Cheers,

Brent

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

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 each of the

[#permalink]

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 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.

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Re: What is the lowest positive integer that is divisible by each of the

[#permalink]

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 each of the [#permalink]

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 each of the

[#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.

Manager

Joined: 04 Jan 2013

Posts: 67

Re: What is the lowest positive integer that is divisible by each of the

[#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

Posted from my mobile device

Manager

Joined: 12

Jan 2013

Posts: 110

Re: What is the lowest positive integer that is divisible by each of the

[#permalink]

29 Dec 2013, 10:01

Bunuel wrote:

SOLUTION

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?

Math Expert

Joined: 02 Sep 2009

Posts: 86916

Re: What is the lowest positive integer that is divisible by each of the

[#permalink]

29 Dec 2013, 10:20

aeglorre wrote:

Bunuel wrote:

SOLUTION

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: math-number-theory-88376.html

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).

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Re: What is the lowest positive integer that is divisible by each of the

[#permalink]

24 Nov 2014, 12:46

Bunuel wrote:

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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Joined: 09 Jun 2015

Posts: 4

Re: What is the lowest positive integer that is divisible by each of the

[#permalink]

15 Sep 2015, 12:03

OG 13/2015 – PS – 40 – 157:

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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Re: What is the lowest positive

integer that is divisible by each of the [#permalink]

15 Sep 2015, 12:11

KOS75 wrote:

OG 13/2015 – PS – 40 – 157:

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 each of the

[#permalink]

16 Sep 2015, 04:36

KOS75 wrote:

OG 13/2015 – PS – 40 – 157:

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 each of the [#permalink]

28 Nov 2015, 14:56

Hi All,

This is a great question to TEST THE ANSWERS on, but you have to pay careful attention to what is asked for:

we’re asked for the LOWEST integer that is divisible by all of those integers. As such, you should start with Answer A. If “A” is not correct, then you TEST B, and so on. There is a Number Property rule that you can use to help you to avoid some of the ‘excess math.’

Any number that is divisible by 6 is ALSO divisible by 2 and 3

So, how long would it take you to determine if 420 is divisible by 7, 5, 4 and 6 (or 3 and 2)?

Final Answer:

GMAT assassins aren’t born,

they’re made,

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Re: What is the lowest positive integer that is divisible by each of the

[#permalink]

27 May 2022, 06:22

Bunuel wrote:

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 each of the [#permalink]

26 Sep 2022, 05:20

We have to find least positive integer that is divisible by 1,2,3,4,5,6,7

Basically question is asking

what is LCM of 1…..7

Answer:== 420. A

Note :- Had it been asking the largest positive integer we would have chosen the answer where the choice was divisible by LCM i.e 420

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

26 Sep 2022, 06:12

Lowest positive integer that is divisible by each of the integers 1 through 7, inclusive: LCM of [1, 2, 3, 4, 5, 6, 7]

=>

420

Answer A

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Re: What is the lowest positive integer that is divisible by each of the

[#permalink]

09 Oct 2022, 03:01

Lowest positive integer divisible by each of integers from 1 to 7 will be the product of prime factors (2 * 3 * 5 * 7) or its

multiple.

=> 2 * 3 * 5 * 7 = 210 [lowest multiple of 210 is A = 420 ]

Answer A

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Re: What is the lowest positive integer that is divisible by each of the

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09 Oct 2022, 03:01

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