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

Vincen wrote: ↑

Tue Jun 30, 2020 7:16 am

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

[spoiler]OA=A[/spoiler]

Source: Official Guide

-------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

Vincen wrote: ↑

Tue Jun 30, 2020 7:16 am

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

[spoiler]OA=A[/spoiler]

Source: Official Guide

Solution:

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