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 3 of 14.
Nitesh Jindal said:
1 decade ago
It is not comfusing. ok. try to understand the logic here.
lets assume 2^32 is x.
Now 2^96 = {2^32 }^3 .As 2^32 is x so 2^96 is x^3.
Then x^3 +1 is furter solved by formula
as a^3 + b^3 = (a+b)(a^2 -ab +b^3) .
therefore x^3 + 1 = (x+1)(x^2 - x + 1 )
lets assume 2^32 is x.
Now 2^96 = {2^32 }^3 .As 2^32 is x so 2^96 is x^3.
Then x^3 +1 is furter solved by formula
as a^3 + b^3 = (a+b)(a^2 -ab +b^3) .
therefore x^3 + 1 = (x+1)(x^2 - x + 1 )
Divyarekha said:
1 decade ago
Very good explanation.
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.
Lekha said:
1 decade ago
I can't understand. Please explain again. !
Manoj said:
1 decade ago
I can't understand. Explain.
Harshita said:
1 decade ago
I can't understand. Please explain again.
Aryan Verma said:
1 decade ago
@Nagaraj Your Way of making understand is good dear.
Pranjal said:
1 decade ago
It's a simple thing;
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.
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.
Arjun said:
1 decade ago
[(x^n) + 1] will be divisible by (x + 1) only when n is odd.
Hope this will help you guys.
Hope this will help you guys.
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
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