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 4 of 14.
Rrai said:
3 years ago
1&2 option will be eliminated as they are less than the denominator, and in 3rd option even * odd = even.
And the denominator is even + odd = odd, and even no. Can't divide the odd so it will also be eliminated.
And the denominator is even + odd = odd, and even no. Can't divide the odd so it will also be eliminated.
(74)
B.nandu said:
1 decade ago
2^32+1 is divisible by all whole numbers.
So here we multiply the power of 32 with whole numbers. i.e,
32 x 1=32.
32 x 2=64.
32 x 3=96.
We have this option 96. So 2696+1 is correct one.
So here we multiply the power of 32 with whole numbers. i.e,
32 x 1=32.
32 x 2=64.
32 x 3=96.
We have this option 96. So 2696+1 is correct one.
Tarun said:
10 years ago
We know that (x^n + 1) is completely divisible by (x+1), when n is natural number.
So, (2^96 + 1) = ((2^32) ^2 +1).
By using given statement it is divisible by (x+1).
= (2^32 + 1).
So, (2^96 + 1) = ((2^32) ^2 +1).
By using given statement it is divisible by (x+1).
= (2^32 + 1).
Neha garg said:
1 decade ago
2^32 means (2 ^4)^8 this will come 6.
Because 2*2=4
2*2*2=8
2*2*2*2= 16
If we multiply 2^5 then again unit digit will b 2, then 4 soon.
6+1= 7
and 2^96 =
(2^4)24= 6
6+1 = 7.
Because 2*2=4
2*2*2=8
2*2*2*2= 16
If we multiply 2^5 then again unit digit will b 2, then 4 soon.
6+1= 7
and 2^96 =
(2^4)24= 6
6+1 = 7.
Nishu kumari said:
9 years ago
a^n + b^n is divisible by a + b only when n is odd. Suppose 2^32 is a then (2^32) ^3 = 2^96 i.e. a^3 which is odd means a^3 + b^3 here b=1. i.e, a^3 + b^3 is divisible by a + b.
Bappa said:
1 decade ago
I am confusing. (2^32 + 1) is completely divisible by (2^96 + 1) OR (2^96 + 1) is completely divisible by (2^32 + 1). Sorry, perhaps I can't understood the question.
Kamal said:
6 years ago
Suppose 2^32 = X.
so the first equation is X+1.
we can break 2^96 in 2 ^ 32^3 (mean (2 power 32 )power 3 )
So we can write X^3 + 1 because 2^32 =x already define.
so the first equation is X+1.
we can break 2^96 in 2 ^ 32^3 (mean (2 power 32 )power 3 )
So we can write X^3 + 1 because 2^32 =x already define.
Nagaraj said:
1 decade ago
It's a simple thing;
((2^96)+1)/((2^32)+1)
=((x^3)+1)/(x+1)
=(x+1)(x^2-x+1)/(x+1); here easy to delete the (x+1);
So we get only a (x^2-x+1)as anatural number;
((2^96)+1)/((2^32)+1)
=((x^3)+1)/(x+1)
=(x+1)(x^2-x+1)/(x+1); here easy to delete the (x+1);
So we get only a (x^2-x+1)as anatural number;
Sandeep kumar jangir said:
1 decade ago
2^32 = x.
So (2^32+1) = x+1.
And option A, B n C have less power compared to given value.
So 2^96 = 2^(32*3). Which is completely divisible by 2^32.
So (2^32+1) = x+1.
And option A, B n C have less power compared to given value.
So 2^96 = 2^(32*3). Which is completely divisible by 2^32.
Selvakumar O S said:
1 year ago
I used the logic by considering 32 as 2 and it's 3 times Multiple 96 as 6.
So, I used this value in this equation.
2^6+1/2^2+1.
=> 65/5 = 13.
So, I used this value in this equation.
2^6+1/2^2+1.
=> 65/5 = 13.
(5)
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