Data Sufficiency

Thanks Night Reader, i get all this but a basic doubt:

the question no where says n is an integer or for that matter its positve.
am i missing something here?

Night reader wrote: an easy entry; it tests a^n - b^n reformation --> every time n={n is integer, 2,3,4, ... +infinity} (a-b) is one factor present ahead of the reformation
a^2 - b^2=(a-b)(a+b), a^3 - b^3=(a-b)(a^2 +ab + b^2)

st(1) given the explanation above, this is Not Sufficient, as n can be odd OR even;
st(2) a^n + b^n is not divisible by a + b, if n=1 --> (a^1 + b^1)/(a + b) is quite divisible BUT when n=2 or 4 or 6 ... (a^2 + b^2)/(a + b). In fact a^3 + b^3=(a+b)((a^2 -ab + b^2), hence n must be even for st(2) to be true and our answer is No, n is not odd, it's even. st(2) is Sufficient - B

Is n odd ?

1. a^n - b^n is divisible by a - b
2. a^n + b^n is not divisible by a + b

rohu27 wrote:re-opening an old thread (this was todays BTG ath question)

can someone explain the solution wth exact steps? unable to figure out

i think we assume the real but terminal numbers for a precision here; if both a and b are -ve you can always put the -ve sign out of the parentheses, with one -ve and +ve you just switch the values in their places.

harsh.champ wrote:Is n odd ?

1. a^n - b^n is divisible by a - b
2. a^n + b^n is not divisible by a + b

1. Take n=2

=( a^2-b^2) = (a+b) (a-b). Yes definitely divisible by (a-b)

Take n=1

= a-b . Yes definitely divisible by (a-b)

So n can be odd or even. Both Satisfies 1. NS.

2. Take n =1

a^n+ b^n = a+b ; which according to 2 is definitely not divisible by a+b, Not possible. Therefore n cannot be equal to 1.

2 Happens only when n is even. When n is odd, a^n + b^n is divisible by a + b. So n is even . Sufficient.

@rohu, i'm sorry for didn't address your concern correctly

we should not assume n=0 at least, because 0 is even and a^0 + b^0=2 not True for all values of (a+b). So I would like to recall my previous answer in this thread as n couldn't be even safely. BUT n could not be odd for sure; the answer only --> No n is not odd, please delete from your memory as I did from mine n is even, it's not always true here

Is n odd ?

1. a^n - b^n is divisible by a - b
2. a^n + b^n is not divisible by a + b

ankur.agrawal wrote:

harsh.champ wrote:Is n odd ?

1. a^n - b^n is divisible by a - b
2. a^n + b^n is not divisible by a + b

1. Take n=2

=( a^2-b^2) = (a+b) (a-b). Yes definitely divisible by (a-b)

Take n=1

= a-b . Yes definitely divisible by (a-b)

So n can be odd or even. Both Satisfies 1. NS.

2. Take n =1

a^n+ b^n = a+b ; which according to 2 is definitely not divisible by a+b, Not possible. Therefore n cannot be equal to 1.

2 Happens only when n is even. When n is odd, a^n + b^n is divisible by a + b. So n is even . Sufficient.

Agree. This is actually the smartest and fastest way to solve it.

stmt1:
If n=1
a^1-b^1=(a-b), divisible by (a-b).

If n=2
a^2-b^2 =(a+b)(a-b), divisible by (a-b). Stmt 1 NS.

stmt2:
If n=1
a^1+b^1 = a+b, divisible by (a+b).

If n=2, so we have (a^2+b^2), and if
a=b=1. yes, divisible by (a+b).
a=b=2. yes, divisible by (a+b).
a=b=10. yes, divisible by (a+b).
a!=b, no, not divisible by (a+b).

(a^2+b^2) is always divisible by (a+b) if a=b. Are we supposed to assume a doesn't equal b? Further, are a and b assumed to be non-zero positive integers?

The answer is B.

Paradoxial situation in Maths question

What papgust says

papgust wrote:
Rules for a^n + b^n:

--> NEVER divisible by (a-b)
--> Divisible by (a+b), when n is ODD
--> NOT Divisible by (a+b), when n is EVEN

which is valid

What IAN says
If, say, a=20 and b=5, then 20^2 + 5^2 most certainly *is* divisible by 20+5.

which is again valid

This means formula is not always valid... what to do
Hate Maths ?? NO

Consider this

If say, a=0 and b=0, then 0^2 + 0^2 is 0 and a+b = 0
then if we consider the second statement that a^n + b^n is divisible by a+b,
the statement itself is invalid because 0/0 is an invalid expression.
This will imply the question is giving us an invalid statement, let alone it is true or not.
The answer then becomes E(which is wrong)

The conclusion is assuming values for a,b will not be the right approach in all the questions because it will lead us to wrong answers. Since it is a Data Sufficiency question, so have take into consideration the entire domain of values. Using the formula takes care of the entire set and we get the right answer in questions where there no restriction on values to be considered.

how can the answer be B.i am still not able to work out with values.like for statement 2 a^n+b^n if i take values such as 2^2+3^2 then n is not odd but for value 3^3+4^3 it also satisfies and n is odd.please anyone explain it to me picking values

@prashant
From statement 2 .... Lets take a=2 and b=3 and n=2. a+b=5
a^n+b^n=4+9=13 which is not divisible by a+b. Similarly, a=2 and b=3, a+b=5 and n=3 then a^n+b^n=8+27=35 which is divisible by a+b. Hope ur clear now!!

If u've read Stuart's post earlier on this thread he urges everyone to not memorise rules. In my case these rules were forced into my head in 10th standard algebra. lol. If it is helpful, then you may wanna by-heart the following rules

a^n-b^n is always divisible by a-b. Irrespective of whether n is odd or even
a^n-b^n is divisible by a+b if n is EVEN ONLY. Never if n is odd.
a^n+b^n is divisibly by a+b when n is ODD ONLY. Never if n is even.

selected option B as the answer.
It is good in a way to memorize some rule related to this as that way you can quickly arrive at the answer.

I think the answer should be B.

(B) is answer.
a^3+b^3 is divisible by a+b but a^2+b^2 is not.

Thank u sooooooooooooo much!

papgust wrote:

can't understand the logic behind why the statement B IS CORRECT. Ajith! can you explain? I can't make a logic to fit in without trying real numbers. Is this some kind of rule in maths.

Rules for a^n - b^n:

--> Always divisible by (a-b)
--> Divisible by (a+b), when n is EVEN
--> NOT Divisible by (a+b), when n is ODD

Rules for a^n + b^n:

--> NEVER divisible by (a-b)
--> Divisible by (a+b), when n is ODD
--> NOT Divisible by (a+b), when n is EVEN

---

Coming to the question,

n odd?

A. a^n - b^n is divisible by a - b

when n is ODD/EVEN, (a-b) divides (a^n - b^n)
Insufficient.

B. a^n + b^n is not divisible by a + b

ONLY when n is EVEN. When n is ODD, a^n + b^n is divisible by a + b.
So, n is EVEN. Sufficient.

Hence B

Is n odd ?

1. a^n - b^n is divisible by a - b
2. a^n + b^n is not divisible by a + b


Explanation with formula:

Statement 1:


[a^n - b^n] is divisible by (a-b), for all values of n.

It means that whether n is odd or even, it does not matter...


Statement 2:

[a^n - b^n] is divisible by (a+b), only when n is even... do not ask too many questions as these are formule...

If it is not divisible by (a+b), then definitely n is odd because the rule says that (a^n - b^n) is always divisible by (a-b)...

so when it is not divisible by (a+b) and is divisible by (a-b), it means that 'n' is odd...


Guyzzz!!! honestly do not ask how is this possible.. This is definitely a 700 level question...

These are formulae just like you say (a + b) ^2 = a^2 + 2 X a X b + b^2

do not ask questions as this is a formula based question...