Around the World in 80 Questions (Day 9): If n is a positive integer

answer is D.

STATEMENT 1 : n divided by 6 gives the remainder of 1.
so n is odd. (3-x)^n = ((-1)*(x-3))^n = (-1)^n * (x-3)^n = -((x-3)^n) ------ since n is odd
therefore the required value is 0. this statment is sufficient

STATEMENT 2 : 21 divided by n gives the same remainder as 20 divided by n.
21 and 20 are co prime. and n is positive. thus n is 1. placing 1 in the equation, we can determine the value to be 0. this statment is sufficient

EACH ALONE IS SUFFICIENT. SO ANSWER IS D

1) n divided by 6 gives the remainder of 1.

n = 6p+1, i.e. n can be 1,7,13,19,... (all odd)

(x - 3)^n + (3 - x)^n = (x - 3)^n + (-1)^n * (x - 3)^n

since n is odd, They cancel out and the expression becomes equal to 0

Sufficient

2) 21 divided by n gives the same remainder as 20 divided by n

Only number that will leave same remainder will be 1 with the remainder as 0. So n = 1

Putting n = 1, the expression becomes equal to 0.

Sufficient

Ans D

statement 1
condition gives n=5
(x-3)^5+(3-x)^5
by binomial th .
figure will come 0
statement 1 is sufficient
statement 2
by condition it gives n=1
x-3+3-x=0
statement 2 is suff
answer is d

(x - 3)^n + (3 - x)^n

(x - 3)^n + (-1)^n(x - 3)^n

(1) n divided by 6 gives the remainder of 1.

n is odd, hence (1)^n = -1

(x - 3)^n - (x - 3)^n = 0

The value is zero. Sufficient.

(2) 21 divided by n gives the same remainder as 20 divided by n.

20 and 21 are consecutive numbers, hence co-primes. Only common factor = 1

n is odd, hence (1)^n = -1

(x - 3)^n - (x - 3)^n = 0

Sufficient.

IMO D

Asked: If n is a positive integer, what is the value of (x - 3)^n + (3 - x)^n ?
(x - 3)^n + (3 - x)^n = (x-3)^n + (-1)^n(x-3)^n = (x-3)^n { 1 + (-1)^n }

(1) n divided by 6 gives the remainder of 1.
n = 6k + 1;
(x - 3)^n + (3 - x)^n = (x-3)^n + (-1)^n(x-3)^n = (x-3)^n { 1 + (-1)^n } = (x-3)^{6k+1} {1 + (-1)^{6k+1} } = 0
Since 1+ (-1)^6k*(-1) = 1 - 1 = 0
SUFFICIENT

(2) 21 divided by n gives the same remainder as 20 divided by n.
n = 1
(x - 3)^n + (3 - x)^n = (x-3)^n + (-1)^n(x-3)^n = (x-3)^n { 1 + (-1)^n } = (x-3) {1 + -1 } = 0
SUFFICIENT

IMO D

If n is a positive integer, what is the value of (x - 3)^n + (3 - x)^n ?
= (x-3)^n(1+ (-1)^n)
In case of n being odd number the value of (x - 3)^n + (3 - x)^n will be zero

(1) n divided by 6 gives the remainder of 1.
It means n is odd number hence the value of expression will be zero
Statement 1 is sufficient

(2) 21 divided by n gives the same remainder as 20 divided by n.
being n is positive integer hence if we divide 21 by n or 20 by n the reminder is zero
it is only possible only if n=1
Hence n is odd
the value of expression (x - 3)^n + (3 - x)^n will be zero
Statement 2 is sufficent

Option D is correct

We need to find: \((x - 3)^n + (3 - x)^n\).
This can be written as: \((x - 3)^n + ((-1)^n)*(x-3)^n \)
Hence, when n is even, the value of the above function will be \(2*(x-3)^n\)
But when n is odd, the value of the above function will be 0.

(1) n=6*(some integer)+1. Hence, n will always be odd. Sufficient.
(2) 21=k1*n+R and 20=k2*n+R
hence, the above 2 equation give us: 1=n(k1-k2). Since n, k1 and k2 all are integers, all have to be equal to 1 to satify the equation.
Hence, n is odd. Sufficient.
Hence D is right option.

If n is a positive integer, what is the value of (x - 3)^n + (3 - x)^n ?

(1) n divided by 6 gives the remainder of 1.
(2) 21 divided by n gives the same remainder as 20 divided by n.

n>0 and is an integer
(x-3)^n + (3-x)^n
which is nothing but (x-3)^n + (-1)^n (3-x)^n
or (3-x)^n *( 1 + (-1)^n) ??

Case I: n divided by 6 gives remainer 1
so n = 6k +1, so n will end up with 7, 3, 9,5,1 which are all odd
now we know that n is always odd
so -1^n will be -1
hence the entire expression of question stem becomes 0
hence this is sufficient

Case II: 21 divided by n gives the same remainder as 20 divided by n
21 = nx + a
20 = ny + a
subtracting , we get 1 = n(x-y)
so n has to be 1
even otherwise we do not get any value of n other than 1
so n is odd and hence the entire expression becomes o
hence this is sufficient

D is the answer

S1:- n = 6a + 1 i.e. n is always odd

Now,\( (x-3)^n + (3-x)^n = (x-3)^n + (-1)^n*(x-3)^n = (x-3)^n*( 1 + (-1)^n) = (x-3)^n*( 1 + (-1)) \)(since n is odd) = 0

S1 sufficient.

S2:- If n=1, 21/n and 20/n both give 0 as a remainder.
So, putting n=1 in the expression we again get the value equals to 0.

S2 is sufficient.

Hence, D is correct choice.

If n is a positive integer, what is the value of (x - 3)^n + (3 - x)^n ?

if n= odd, the value of (x-3)^n- (x-3)^n = 0

(1) n divided by 6 gives the remainder of 1.
this means n=1, 7, 13, etc.
Hence the answer = 0, hence sufficient

(2) 21 divided by n gives the same remainder as 20 divided by n.
Two consecutive numbers cannot be commonly divisible by any number except 1, this implies that two consecutive number cannot have the same reminder unless it is divided by and remainder =0,
if n=1, then answer =0. Hence sufficient.

Given : n is a positive integer
To find : Value of \((x - 3)^n + (3 - x)^n\)

i.e \((x - 3)^n + (-1(x-3))^n\)

\((x - 3)^n + (-1)^n (x-3)^n\)

\((x-3)^n (1 + (-1)^n )\)

From the above eq. we can see that if n is odd the second term will give result 0. In that case the value of whole expression will become 0.

(1) n divided by 6 gives the remainder of 1.

n = 6k + 1
This means that n is an odd integer.
As stated above, the value of expression will be 0.

Sufficient.

(2) 21 divided by n gives the same remainder as 20 divided by n.

The only way two consecutive numbers when divided by a single number 'n' gives same remainder is when the number 'n' is 1 .
We now know that n is odd.
So, the value of expression will be 0.

Sufficient.

Answer : D

(X-3)^n+(3-x)^n can be rewritten as 2.(X-3)^N for N Even and Zero for N as odd.

Statement 1: N divided by 6 giving a remainder as 1 means N is Odd. So answer is Zero. Sufficient
Statement 2: 21 divided by n gives same remainder as 20 divided by n. 2 consecutive numbers giving same remainder can happen only for divisior as 1. so N =1. and hence N is Odd. for odd the answer is Zero. Sufficient.

D is the answer.

If n is a positive integer, what is the value of (x - 3)^n + (3 - x)^n ?

(1) n = 6k+1 where k = positive integer
so n = odd = 1, 7, 13...
if n = odd then, (x - 3)^n + (3 - x)^n = (x - 3)^n - (x - 3)^n = 0. Sufficient.

(2) only possible value of n = 1 = GCF (20,21)
if n = 1, (x - 3)^1 + (3 - x)^1 = (x - 3) - (x - 3) = 0. Sufficient.

Either of (1) or (2) is sufficient.

IMO D.

S1) \(n=6k+1=1,7,13,19...\)(odd)

we can say the exp is \(0 \)(suff)

S2) \(gcd(20,21)=1\). The only value common remainder is \(0\) when divided by \(n=1\).
Hence exp is \(0\) (Suff)

D)

If n is a positive integer, what is the value of (x - 3)^n + (3 - x)^n ?

(1) n divided by 6 gives the remainder of 1.
n = 6k + 1; where k is non negative integer ; Also, n is odd integer
for n =1 , the expression is x -3 + 3 -x = 0
for n = 7, the expression is again 0 as all the terms in the expression (x-3)^7 will cancel of the terms in the expression (3-x)^7
hence, statement 1 is sufficient
(2) 21 divided by n gives the same remainder as 20 divided by n.
the only possible value of n is 1
21/1 gives remainder 0
20/1 gives remainder 0
and for n=1, the expression =0
hence, statement 2 is sufficient
Answer D

st.1)
possible value of n are 1, 7, 13, 19, 25....

one thing that we notice here is n is always odd. additionally, (x-3)^n+(3-x)^n is an expression in which each term will cancel out each other when n is odd. thus value of the expression will be zero

Sufficient

st.2)

now only possible value of n for which 20, and 21 will have the same remainder is 1, and the remainder will be 0. value of expression when (3-x)^1+(x-3)^1= 3-x+x-3 =0

Sufficient

IMO D

Given "n" is a positive integer. Also we can simplify the given expression as:-
\((x - 3)^n + (3 - x)^n = (x - 3)^n + (-1)^n*(x - 3)^n\)

\( = (1 + (-1)^n)*(x - 3)^n\)

So if "n" is odd, we get:- \((x - 3)^n + (3 - x)^n = 0\)

(1) n divided by 6 gives the remainder of 1:-

For positive "n", the given condition is possible for :- n = 1, 7, 13, 19, 25, .....etc.
In all these conditions, "n" is odd. And so as shown earlier, we can obtain a definite value of the given expression, which is "0"

Hence, statement (1) alone is sufficient.


(2) 21 divided by n gives the same remainder as 20 divided by n:-

The only way this condition is possible is if, n = 1.

We can also prove this, we can write, 21 = a*n +r and 20 = b*n + r. Here "a" and "b" are some integers, and "r" is the common remainder.
So, 21 - a*n = 20 - b*n.
Therefore, (a - b) * n = 1, which is only possible when a - b =1 and n = 1.

So we get, n = 1, where "n" is odd and hence given expression = 0

Hence, statement (2) alone is sufficient.

Hence correct answer is option D.