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

given that 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 can be 1,7,13,19,25,31,37,43
n is always odd
no matter what value of x is the sum of (x - 3)^n + (3 - x)^n with n as odd will always give sum as 0
sufficient
#2
21 divided by n gives the same remainder as 20 divided by n.

value of n can range from 1
we have odd say powers for (x - 3)^n + (3 - x)^n such that for all odd numbers the sum is 0 ,
sufficient

OPTION D is correct

n is a positive integer. Required value: (x -3)^n + (3 - x)^n

1. n/6 gives remainder of 1. So, n is odd.
the expression is:
(x-3)^7 + (-1 * (x-3))^7 = (x-3)^7 + {(-1)^7 * (x-3)^7} = (x-3)^7 - (x-3)^7 =0

So, 1 is sufficient.

2.

If n is a positive integer, what is the value of (x - 3)^n + (3 - x)^n ?
=> (x-3)^n + [-1 (x-3)]^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 = 6Q + 1 => n is odd
=> (x-3)^n * [1 + (-1)^n] => (x-3)^n * [1 - 1] => (x-3)^n * (0) = 0 ----> SUFFICIENT

(2) 21 divided by n gives the same remainder as 20 divided by n.
=> 21 = Zn + r & => 20 = nT + r
=> 21 - Zn = 20 - nT
=> 1 = n (Z-T) ---> NOT SUFFICIENT

Answer = A

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.
Question is basically asking if n is odd or even. If odd then the value is 0

(1) n/6 give remainder of 1 meaning n is an odd number. Sufficient.
(2) 21/n gives same remainder of 20/n meaning n is 1 or -1. sufficient to answer

Answer is (D)

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

The Answer is E
Statement 1 says that n=6q+1. If we substitute the values in the equation given in the stem, We cannot simplify further since there is no information about the value of x or n. Hence insufficient
Statement 2 says that the remainders when dividing 21 and 20 by n are the same. However we still don't have enough information to find the value of the equation in the stem.
Combining both, we still don't have enough information to find the value of the equation in the stem. Hence E

(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

Both statements together are needed to answer the question. The first one alone isn't sufficient but it narrows down our answer for value of n to be such that n+1 is a multiple of 6. Then we can test the possible values that fit in statement 2 accordingly. Answer C

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(1) We know that n is odd and n>=7 but we don't know about x -> Insufficient

(2) The same remainder must be 0 -> n=1 -> the function results in 6 -> Sufficient

Answer B

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.