Lesson: Consecutive Integers

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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
gmat-admin's picture

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
gmat-admin's picture

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 explicitly specify if they're asking for consecutive integers, consecutive EVEN integers, and consecutive ODD integers? I got your first reinforcement prob wrong because I thought a, b, c could also be 1, 3, 5, or 2, 4, 6, hence c-a = 2 cannot always equal 2
gmat-admin's picture

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
gmat-admin's picture

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

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