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Comment on Consecutive Integers
Hello
In below question
N = abc where a, b and c are the hundreds, tens and units digit respectively. If a, b and c are non-zero consecutive numbers such that a < b < c, then which of the following must be true?
I.N is always divisible by 2
II.N is always divisible by 3
III.N is divisible by 6 only if b is odd.
If we apply the logic (n-1)n(n+1) then it does not hold true for above case.
Can you help me understand why not applicable to above case.
Regards,
Abhimanyu
Be careful. abc does not
Be careful. abc does not represent the product of three numbers.
Each variable (a, b, and c) represents a DIGIT in the 3-digit number abc. As such the rules related to the product (n-1)n(n+1) do not hold.
Then how it can be solved
The question: N = abc where a
The question: N = abc where a, b and c are the hundreds, tens and units digit respectively. If a, b and c are non-zero consecutive numbers such that a < b < c, then which of the following must be true?
I. N is always divisible by 2
II. N is always divisible by 3
III. N is divisible by 6 only if b is odd.
------------------------------------------
IMPORTANT:
Property #1) If an integer is divisible by 3, then the sum of its integers will be divisible by 3.
Property #2) If an integer is divisible by 6, then it must be divisible by 3 AND 2.
To get a feel for the properties of these 3 consecutive integers, let's...
Let x = a the 1st number (aka a)
So, x+1 = the 2nd number (aka b)
And x+2 = the 3rd number (aka c)
So, the sum a + b + c = x + (x+1) + (x+2) = 3x + 3 = 3(x + 1)
Aha, so the SUM of any 3 consecutive integers must be divisible by 3 (since 3(x+1) is definitely divisible by 3)
Now let's examine the statements...
I. N is always divisible by 2
This need not be true
For example, one possible value of N is 345, in which case N is NOT divisible by 2
II. N is always divisible by 3
This IS true
We already showed that, since the SUM of any 3 consecutive integers must be divisible by 3, then we know that N is definitely divisible by 3
III. N is divisible by 6 only if b is odd.
If b is odd, then c is EVEN, which means N is divisible by 2
Since N is also divisible by 3, we can conclude that N IS divisible by 6
So statements II and III must be true.
Will the GMAT always
Yes, on test day, the
Yes, on test day, the question will explicitly state the kinds of consecutive numbers.
Consecutive integers ...-3, -2, -1, 0, 1, 2, 3, ...
Consecutive EVEN integers: ...-6, -4, -2, 0, 2, 4, 6, ...
Consecutive ODD integers: ...-5, -3, -1, 1, 3, 5,...
Hey Brent,
Question Link: https://gmatclub.com/forum/if-n-is-a-positive-integer-is-n-3-n-divisible-by-109941.html
In the question above, I just wanted to confirm that the answer is A because the integers 0*1*2 (E-O-E) is still divisible by 4 as 0 is divisible by 4 (or any number)?
Thanks in advance!
Neel
Question Link: https:/
Question Link: https://gmatclub.com/forum/if-n-is-a-positive-integer-is-n-3-n-divisible...
You are correct; 0 IS divisible by 4 (in fact, 0 divisible by all integers)
If x and y are integers, we can say that x is divisible by y, if we can write x = yk, where k is some integer.
For example, 12 is divisible by 3, because we can write 12 = (3)(4), and 4 is an integer.
And 70 is divisible by 5, because we can write 70 = (5)(14), and 14 is an integer.
Likewise, 0 is divisible by 0, because we can write 0 = (4)(0), and 0 is an integer.
Does that help?
Cheers,
Brent
Hi
Is Zero divisible into any number with no remainder?
Yes, that's correct. However,
Yes, that's correct. However, I've never seen an official GMAT question that requires this property. In fact, the GMAT test-makers typically restrict the values in Integer Properties question to POSITIVE integers only.
It is an interesting fact.
Agreed!
Agreed!
Hi Brent, would you happen to
Question link: https:/
Question link: https://gmatclub.com/forum/the-sum-of-4-different-odd-integers-is-64-wha...
Statement 1: The integers are consecutive odd numbers
Let's examine some consecutive odd numbers: 1, 3, 5, 7, 9, 11, 13....
Notice that each integer is 2 greater than the number before it.
So, if we let x = the first number...
then x + 2 must be the next integer
and x + 2 + 2 must be the integer after that.
And so on.
Does that help?
Cheers,
Brent
Here's my full solution: https://gmatclub.com/forum/the-sum-of-4-different-odd-integers-is-64-wha...
Makes a lot more sense.
Hi Brent, could you please
(Question: https://gmatclub.com/forum/if-k-is-a-positive-integer-what-is-the-remainder-when-k-2-k-3-k-242852.html)
Bunuel states that: "There is a rule saying that The product of n consecutive integers is always divisible by n!."
However, in this video you tell us that the Rule states that "Every nth # is divisible by n."
From my understanding, being divisible by n! and being divisible by n are two very different operations.
Thank you!
Bunuel's rule doesn't
Bunuel's rule doesn't contradict the rule in the video lesson. In fact, his law is very similar to mine.
Keep in mind that, within the product of n consecutive integers, we also have the product of n-1 consecutive integers.
For example, consider the product of 5 consecutive integers: (3)(4)(5)(6)(7)
This product must be divisible by 5
Also, within these 5 consecutive integers, we have the product of 4 consecutive integers, (3)(4)(5)(6), which means (3)(4)(5)(6) must be divisible by 4, which means (3)(4)(5)(6)(7) is divisible by 4 as well.
Likewise, within these 5 consecutive integers, we have the product of 3 consecutive integers, (3)(4)(5), which means (3)(4)(5) must be divisible by 3, which means (3)(4)(5)(6)(7) is divisible by 3 as well.
And so on.
So.....
Let's say K = the product of n consecutive integers.
As such, K must be divisible by n
Also, within the n consecutive integers, there are n-1 consecutive integers.
As such, K must be divisible by n-1
Also, within the n consecutive integers, there are n-2 consecutive integers.
As such, K must be divisible by n-2
etc
So, K must be divisible by n, n-1, n-2, n-3, . . . . 2, 1
In other words, K must be divisible by n!
Cheers,
Brent
Awesome! Thank you for going
Hi Brent, quick question here
Question link https://gmatclub.com/forum/if-x-y-and-z-are-consecutive-integers-such-that-x-y-z-156400.html
I am reading Bunuel's and Pacifist85 explanations, and both of them use 0 as the first multiple of 10. That is a surprise for me because I used to think multiples of 10 are: 10, 20, 30 and so on. Can then we say that 0 is the first multiple of any number since we can divide 0 by any number?
Thanks!
0 is a multiple of all
0 is a multiple of all integers.
That said, pretty much all official GMAT questions involving integer properties will limit the values to positive integers.
https://gmatclub.com/forum
Hi Brent,
Can you please show me what process to follow to solve this problem please!
Thanks
Fatima-Zahra
Here's one approach: https:/
Here's one approach: https://gmatclub.com/forum/what-is-the-sum-of-the-integers-from-100-to-2...
Cheers,
Brent
Hey Brent,
is there a rule that allows us to conclude that (k+1)(k)(2k+1) is DIVISIBLE by 4?
Regarding this Q:
https://gmatclub.com/forum/if-n-is-a-positive-integer-is-n-3-n-divisible-by-109941.html
And generally, is a product of 3 consecut. numbers also always divisible by 6?
Thanks,
Philipp
Link: https://gmatclub.com
Link: https://gmatclub.com/forum/if-n-is-a-positive-integer-is-n-3-n-divisible...
There's no rule that says (k+1)(k)(2k+1) is DIVISIBLE by 4
Notice that, if k = 2, then (k+1)(k)(2k+1) = (3)(2)(5) = 30, and 30 isn't divisible by 4.
There is a rule that says the product of n consecutive integers MUST be divisible by n, n-1, n-2,..., 2 and 1
For example, the product of 6 consecutive integers MUST be divisible by 6, 5, 4, 3, 2 and 1
Likewise, the product of 3 consecutive integers MUST be divisible by 3, 2 and 1
If the product is divisible by 3 and 2, the product is also divisible by 6.
Cheers,
Brent
So in the question the answer
Yes, if there are two even
Yes, if there are two even numbers as part of a product, then that product must be divisible by 4.
Is it considered a
I mean in descending order
Or when in any GMAT questions the word "consecutive" appears we then must consider it as an incremental trend?
Thank you in advance,
If we CAN take a set of
If we CAN take a set of integers and arrange those integers in ascending order so that each integer is one greater than the integer before it, then we have a set of consecutive integers.
So, for example {4, 7, 6, 5} is a set of consecutive integers.
I mention this because I've seen questions that state x, y and z are consecutive integers. In these cases, we can't immediately assume that x < y < z.
It COULD be the case that x = 4, y = 2 and z = 3.
Cheers,
Brebt
Thank you
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