Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 3)
3.
It is being given that (232 + 1) is completely divisible by a whole number. Which of the following numbers is completely divisible by this number?
Answer: Option
Explanation:
Let 232 = x. Then, (232 + 1) = (x + 1).
Let (x + 1) be completely divisible by the natural number N. Then,
(296 + 1) = [(232)3 + 1] = (x3 + 1) = (x + 1)(x2 - x + 1), which is completely divisible by N, since (x + 1) is divisible by N.
Discussion:
137 comments Page 2 of 14.
Sasikumar dpi said:
6 years ago
Easy to understand this Answer.
So,
2^32 = 32x32 = 1024 + 1(this question value)
2^16 = 16x16 = 256 + 1 (not covered value)
2^16 = 16x16 = 256 - 1 (not covered value)
2^23 = 23x23 = 529 x 7 (not covered value)
2^96 = 96x96 = 9216 + 1(so this option is fully covered for question value so we can choose D option)
So,
2^32 = 32x32 = 1024 + 1(this question value)
2^16 = 16x16 = 256 + 1 (not covered value)
2^16 = 16x16 = 256 - 1 (not covered value)
2^23 = 23x23 = 529 x 7 (not covered value)
2^96 = 96x96 = 9216 + 1(so this option is fully covered for question value so we can choose D option)
(7)
Shreya said:
7 years ago
Hi;
Actually, if any no is of the form a^n +b^n if n is odd.
So, here if in 2^96 + 1 we take n=3 it will become (2^32)^3 + (1^32)^3..to tell you we took n=3 so that we get 32 after dividing...now from the above-mentioned formula (2^32)^3 + (1^32)^3 will always be divisible by 2^32 + 1^32 as n is odd.
Therefore 2^32 + 1 completely divides 2^96 + 1.
Actually, if any no is of the form a^n +b^n if n is odd.
So, here if in 2^96 + 1 we take n=3 it will become (2^32)^3 + (1^32)^3..to tell you we took n=3 so that we get 32 after dividing...now from the above-mentioned formula (2^32)^3 + (1^32)^3 will always be divisible by 2^32 + 1^32 as n is odd.
Therefore 2^32 + 1 completely divides 2^96 + 1.
Kranthi said:
8 years ago
Remember the condition that (x^n+1^n) is divisible by (x+1). Given number (2^32+1) is divisor. He said which of the following number is divided by this number i.e. (2^32+1). The dividend should be in the form of (x^n+1^n). Let x=2^32(from divisor).
Substituae it in dividend. [(2^32)^n+1^n].
Substitube n= 1,3,5,... 4 th option suits when n=3.
Substituae it in dividend. [(2^32)^n+1^n].
Substitube n= 1,3,5,... 4 th option suits when n=3.
Aviii said:
11 months ago
Let, (2^32) = (2^n)
If ( 2^n + 1 ) is divisible by a whole number, then higher powers of that same base will also be divisible by that same number.
(2^nm + 1 ) will also be divisible by the same number, where "m" is an integer.
Since, (2^96 + 1 ) = ( 2^32*3 + 1)
Therefore, (2^96 + 1) is divisible by the same number.
"Option D " is correct!
If ( 2^n + 1 ) is divisible by a whole number, then higher powers of that same base will also be divisible by that same number.
(2^nm + 1 ) will also be divisible by the same number, where "m" is an integer.
Since, (2^96 + 1 ) = ( 2^32*3 + 1)
Therefore, (2^96 + 1) is divisible by the same number.
"Option D " is correct!
(6)
Pavan.k said:
1 decade ago
Elimination method...
I use ^ symbol as power
2^32+1 (we can also written as) 2^32+1^32..here the bases are equal so add the bases. we get 3.therefore options b c are eliminated.
Now coming to A power 16 is not divisible by 3 where as power 96 in D. we get 3.
D) 2^96+1 we can write as 2^96+1^96.powers =.so add the bases
I use ^ symbol as power
2^32+1 (we can also written as) 2^32+1^32..here the bases are equal so add the bases. we get 3.therefore options b c are eliminated.
Now coming to A power 16 is not divisible by 3 where as power 96 in D. we get 3.
D) 2^96+1 we can write as 2^96+1^96.powers =.so add the bases
Mandar said:
1 decade ago
In simple words.
Guy's just look at the power of each no. It is lesser than the (2^32+1) they ask us to completely divisible by (2^32+1).
So Only (2^96+1) Has The Greater power Than the other given No. So Answer is (2^96+1).
Because to divide NR. completely power of the NR. Or no in NR. it has to be greater than Dr.
Guy's just look at the power of each no. It is lesser than the (2^32+1) they ask us to completely divisible by (2^32+1).
So Only (2^96+1) Has The Greater power Than the other given No. So Answer is (2^96+1).
Because to divide NR. completely power of the NR. Or no in NR. it has to be greater than Dr.
Anshul said:
1 decade ago
Put X =3 in place of X in Expression (X+1)(X^2-x-1),
Then 4 * (9-3-1).
= 4 * 5.
= 20.
Which is divided by (x+1) means 4.
Hence BY if "a" is divided by "b"
b is divided by C.
Then a is divided by C.
Hence x^3+1 is divided by X+1 , & X^3+1 is 2^96+1.
Then 4 * (9-3-1).
= 4 * 5.
= 20.
Which is divided by (x+1) means 4.
Hence BY if "a" is divided by "b"
b is divided by C.
Then a is divided by C.
Hence x^3+1 is divided by X+1 , & X^3+1 is 2^96+1.
Soumya said:
1 decade ago
(2^32+1).
Take x = 2^32, Then (2^32+1) = (x+1) ------> Eq1.
2^96 = (2^32) ^3, ie x^3.
Then (2^96+1) = (x^3+1).
We know that.
(x^3+1) = (x+1) (x^2-x+1) ------> Eq2.
Then dividing eq2 by eq1.
(x+1) (x^2-x+1) / (x+1) = (x^2-x+1).
If we give any value to x we get only natural numbers.
Take x = 2^32, Then (2^32+1) = (x+1) ------> Eq1.
2^96 = (2^32) ^3, ie x^3.
Then (2^96+1) = (x^3+1).
We know that.
(x^3+1) = (x+1) (x^2-x+1) ------> Eq2.
Then dividing eq2 by eq1.
(x+1) (x^2-x+1) / (x+1) = (x^2-x+1).
If we give any value to x we get only natural numbers.
Tej said:
6 years ago
When we get to handle a big no then just assume for small no;
eg. 2^1(n)+1 always divides 2^3(n)+1;
This pattern followed for all the powers of 2,
For simplest one took n=1
2^1+1=3
2^3+1=9;
For n=2.
2^2+1=5
2^6+1=65;
So we can see the pattern.
Likewise, it follows the last option.
eg. 2^1(n)+1 always divides 2^3(n)+1;
This pattern followed for all the powers of 2,
For simplest one took n=1
2^1+1=3
2^3+1=9;
For n=2.
2^2+1=5
2^6+1=65;
So we can see the pattern.
Likewise, it follows the last option.
Shyam said:
8 years ago
Let 2^32 be x.
2^32+1 = x+1----------->(1)
2^96+1 = (2^32)^3+1
= x^3+1
= (x+1)(x^2-x+1)--------->(2)
By (2)÷(1)
2^96+1/2^32+1 = (x+1)(x^2-x+1)/(x+1).
= x^2-x+1.
So, 2^96+1 Is completely divisible by 2^32+1.
2^32+1 = x+1----------->(1)
2^96+1 = (2^32)^3+1
= x^3+1
= (x+1)(x^2-x+1)--------->(2)
By (2)÷(1)
2^96+1/2^32+1 = (x+1)(x^2-x+1)/(x+1).
= x^2-x+1.
So, 2^96+1 Is completely divisible by 2^32+1.
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