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 12 of 14.
Aryan Verma said:
1 decade ago
@Nagaraj Your Way of making understand is good dear.
Harshita said:
1 decade ago
I can't understand. Please explain again.
Manoj said:
1 decade ago
I can't understand. Explain.
Lekha said:
1 decade ago
I can't understand. Please explain again. !
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.
Divyarekha said:
1 decade ago
Very good explanation.
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 )
Mala said:
1 decade ago
Its confusing. Is there any simple methods?
Manohar said:
1 decade ago
I cannot understand please explain again.
Arun said:
1 decade ago
I understand but I want another example.
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