What is the lowest positive integer

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What is the lowest positive integer

by vikkimba17 » Thu Mar 23, 2017 11:37 pm
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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by Jay@ManhattanReview » Thu Mar 23, 2017 11:43 pm
vikkimba17 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
Hi vikkimba17,

The question basically asks for the LCM of 1, 2, 3, 4, 5, 6 and 7. LCM is the smallest possible number divisible by its numbers.

LCM of 1, 2, 3, 4, 5, 6 and 7 = 420.

The correct answer: A

Hope this helps!

Relevant book: Manhattan Review GMAT Math Essentials Guide

-Jay
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by Brent@GMATPrepNow » Fri Mar 24, 2017 5:11 am
vikkimba17 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
-------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) = [spoiler]420 = A[/spoiler]

Cheers,
Brent
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by DavidG@VeritasPrep » Fri Mar 24, 2017 6:29 am
For those who love charts, you can approach this problem as follows:

First: take the prime factorization of each number
Next: Assemble the prime factorizations into neatly ordered columns. 6, for example, is 2*3, so in our '6' row, we'll have one 2 and one 3.

See below:


Image

Now: go down each column and place the largest number of the column in the associate cell for the LCM. The largest value in the '4' column, for example, is 2^2, so that value would be a part of our LCM. See below:


Image

Now simple multiply the values in the LCM row: 2^2*3*5*7 = 420. The answer is A
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by DavidG@VeritasPrep » Fri Mar 24, 2017 6:34 am
Better yet, back-solve. Because we're looking for the smallest possible value that will work, we'll start with the smallest answer choice. The first one that's divisible by all the integers from 1 to 7 inclusive must be our answer.

A) 420
1) All integers are divisible by 1
2) Even, so divisible by 2
3) If the sum of the digits of a number is divisible by 3 then the number itself is also divisible by 3. 4 + 2 + 0= 6, so 420 is divisible by 3.
4) If the last two digits form a number that is divisible by 4, the number is divisible by 4. 20 is divisible by 4, so 420 is divisible by 4
5) Any number ending in 0 or 5 is divisible by 5, so 420 is divisible by 5
6) Any even number divisible by 3 is divisible by 6. 420 is divisible by 2 and 3, so it's divisible by 6
7) You can just eyeball it. 420/7 = 60, so it's divisible by 7.

And we're done. The answer is A
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by Jeff@TargetTestPrep » Thu Mar 30, 2017 3:35 pm
vikkimba17 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 (LCM) of 1, 2, 3, 4, 5, 6, and 7; however, it may be easier 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) are a number divisible by 4, we know 4 divides into 420.

Since 420 ends in a zero, we know 5 divides into 420.

Since 420 is divisible 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.

Answer: A

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by [email protected] » Thu Mar 30, 2017 5:29 pm
Hi vikkimba17,

This is a great question to TEST THE ANSWERS on, but you should note what is specifically asked for: we're asked for the LOWEST integer that is divisible by all of those integers. Under typical circumstances, you would start with Answer B or D - but here, 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: A

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by hoppycat » Wed Jun 14, 2017 8:18 am
How many number properties questions like this roughly will I get on the test? They seem easier than probability

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by DavidG@VeritasPrep » Wed Jun 14, 2017 8:57 am
hoppycat wrote:How many number properties questions like this roughly will I get on the test? They seem easier than probability
There isn't a set number. I checked my most recent official practice test and saw that there were four number property questions and two probability questions. So know that there's a range of possibilities, but this sample seemed fairly representative of what you might expect.
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by Brent@GMATPrepNow » Wed Jun 14, 2017 7:44 pm
hoppycat wrote:How many number properties questions like this roughly will I get on the test? They seem easier than probability
I agree with David.
I'd say that number properties (I call them integer properties) questions are perhaps the 4th most common question type after Statistics, Algebra and Geometry questions.

Cheers,
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by Matt@VeritasPrep » Thu Jun 22, 2017 5:35 pm
Brent@GMATPrepNow wrote:
hoppycat wrote:How many number properties questions like this roughly will I get on the test? They seem easier than probability
I agree with David.
I'd say that number properties (I call them integer properties) questions are perhaps the 4th most common question type after Statistics, Algebra and Geometry questions.

Cheers,
Brent
Are stats and geometry that common these days? It seems to me that arithmetic and algebra, in some guise, account for the lion's share of the problems, with stats and geometry being ≤ 5 problems each (stats maybe even lower).