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by Brent@GMATPrepNow » Fri Sep 15, 2017 9:58 am
vaibhav101 wrote:The difference between the squares of two consecutive odd integers is always divisible by :
A 3
B 6
C 7
D 8
E 9
The word "ALWAYS" tells us that we can just test any two consecutive odd integers.

How about 3 and 1
3² - 1² = 9 - 1 = 8
Check the answer choices...only answer choice D works.
Answer: D

Not convinced? Try another pair of consecutive odd integers
How about 5 and 3
5² - 3² = 25 - 9 = 16
Check the answer choices...only answer choice D works.

Or, how about 9 and 7
9² - 7² = 81 - 49 = 32
Check the answer choices...only answer choice D works.

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by Brent@GMATPrepNow » Fri Sep 15, 2017 10:03 am
vaibhav101 wrote:The difference between the squares of two consecutive odd integers is always divisible by :
A 3
B 6
C 7
D 8
E 9
We can also PROVE that the answer is D

Notice that all ODD integers can be written in the form 2k + 1, where k is an integer.
So, let 2k+1 = the smaller ODD integer.
This means 2k+3 = the bigger ODD integer

Now let's examine the difference between their squares
We get: (2k + 3)² - (2k + 1)²
Expand: (4k² + 12k + 9) - (4k² + 4k + 1)
Simplify: 8k + 8
Factor: 8(k+1)

As we can see, 8(k+1) is a multiple of 8, which means it will ALWAYS be divisible by 8

Answer: D

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Brent
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Re: NUMBERS

by Scott@TargetTestPrep » Sun Feb 09, 2020 5:33 am
vaibhav101 wrote: ↑
Fri Sep 15, 2017 9:47 am
 The difference between the squares of two consecutive odd integers is always divisible by :
A 3
B 6
C 7
D 8
E 9
Solution:

Some examples are:

9 - 1 = 8

25 - 9 = 16

49 - 25 = 24

Thus, we see that each of these numbers is divisible by 8.

Alternate Solution:

Every odd integer can be expressed as 2n + 1 for some integer n. The odd integer coming after 2n + 1 is 2n + 3; thus we can write the difference between the squares of two consecutive odd integers as:

(2n + 3)^2 - (2n + 1)^2 = (2n + 3 + 2n + 1)(2n + 3 - 2n -1) = (4n +4)(2) = 8n + 8 = 8(n + 1)

Since 8(n + 1) has a factor of 8, it must be divisible by 8.

Answer: D

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