### Discussion :: Numbers - General Questions (Q.No.3)

Pavi said: (Jul 26, 2010) | |

I cannot understand, please explain. |

Devi said: (Sep 17, 2010) | |

I cannot understand how replace (2^96+1) = [ (2^32) ^3]. |

Rahul said: (Sep 26, 2010) | |

Hi Dear, 2^32^3 here multiplication of power will take place. So 32x3 = 96 Ok. |

A.Athiban Sakthivel said: (Nov 2, 2010) | |

But its too complicated procedure. Is not there more simple procedure to solve this problem? |

Divya said: (Nov 19, 2010) | |

I can't to understand your explanation. Can you explain clearly and basically................ |

Priya said: (Nov 24, 2010) | |

When we make 2^32 is X, We try to simplify to get the answer split 2^96 = 2^(32X3) so it is = X^3 When using the formula for X^3+1, we can identify that number is divisible by 2^32 |

Mohsin said: (Dec 15, 2010) | |

Ohh Its confusing !!!!! I cant understand y take (2^96+1)?????????????????? |

Mickeey said: (Dec 18, 2010) | |

this is some what called as hit and trial method, you have to try the options which are similar to the value (2^32+1) in the question. remaining, you can easily eliminate by seeing. in the above, options A and D are similar to the expression given in the question. suppose the given four options are similar, you have to try all... this is also called as trial and error method... |

Anusha said: (Dec 29, 2010) | |

Is there any simple procedure? |

Rahul Wadhwa said: (Jan 16, 2011) | |

Really its very confusing!!!!!!!!!!!!!!!!!!! |

Jaiveer said: (Jan 19, 2011) | |

I could not understand so pls help me for full procedure |

Richa said: (Jan 30, 2011) | |

Unable to undestand. Kindly help. |

Nagaraj said: (Jan 31, 2011) | |

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; |

Karthik said: (Feb 9, 2011) | |

How do u directly write (2^96+1) ? |

Karthi said: (Feb 21, 2011) | |

Give some explanations to understand clearly. |

Rajan said: (Mar 15, 2011) | |

Please explain briefly this sum. |

Jitendra said: (Mar 20, 2011) | |

Please help in another example similar to this. |

Arun said: (Apr 5, 2011) | |

I understand but I want another example. |

Manohar said: (May 11, 2011) | |

I cannot understand please explain again. |

Mala said: (Jul 4, 2011) | |

Its confusing. Is there any simple methods? |

Nitesh Jindal said: (Jul 9, 2011) | |

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 ) |

Divyarekha said: (Jul 16, 2011) | |

Very good explanation. |

Neha Garg said: (Sep 12, 2011) | |

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. |

Lekha said: (Sep 24, 2011) | |

I can't understand. Please explain again. ! |

Manoj said: (Nov 2, 2011) | |

I can't understand. Explain. |

Harshita said: (Feb 7, 2012) | |

I can't understand. Please explain again. |

Aryan Verma said: (Feb 8, 2012) | |

@Nagaraj Your Way of making understand is good dear. |

Pranjal said: (Apr 6, 2012) | |

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. |

Arjun said: (May 14, 2012) | |

[(x^n) + 1] will be divisible by (x + 1) only when n is odd. Hope this will help you guys. |

Pavan.K said: (May 31, 2012) | |

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 |

Madhu said: (Aug 15, 2012) | |

I have one solution. Let 232 = x. Then, (232 + 1) = (x + 1). Let (x + 1) be completely divisible by the natural number N. Then [(232)3 + 1] = (296 + 1) =(x3 + 1) = (x + 1)(x2 - x + 1), which is completely divisible by N, since (x + 1) is divisible by N. |

S.Kishore Sulthana said: (Aug 31, 2012) | |

Given :(2)^32+1 is a whole number we find that: which number from the following option is divisible by this whole number. I have one best solution let 2^32=X, now,given: (2^32)+1=X+1, THEN CHECK any option, first OPTION A, (2^16)+1, IT IS TOO small number than given number,so it is not divisible by the given number. therefore,it is wrong. SECOND OPTION B, (2^16)-1, IT IS also a TOO small number than given number,so it is not divisible by the given number. therefore,it is wrong. THIRD OPTION C,7*(2^23) Here (2^23) is too small number than the given number. so it is not divisible by given number then Final FOURTH OPTION D, (2^96)+1, here 2^96 is biggest number than (2^32) so (2^96)+1 is divisible by (2^32)+1, EXPLANATION: let (2^32)=X, here let write(2^32)^3=2^96(for our convenience) option4:(2^96)+1, therefore, [(X)^3]+1 we know the formula:(a^3+b^3)=(a+b)(a^2-ab+b^2) from the above :write [(x^3)+(1)^3)=(x+1)(x^2-x+1) here (x+1)(x^2-x+1)is divisible by (X+1) that is given number.. we know that (X+1)=(2^32)+1=given number.. Therefore the option4 is divisible by given number. so,it is a correct answer friends... |

Bappa said: (Mar 21, 2013) | |

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. |

Hasan said: (May 20, 2013) | |

Suppose X = 2^32 then 2^32+1 = X+1. 2^96 + 1 = (2^32)^3 + 1 = X^3 + 1. .'. (2^32 = X) Let X^3+1/X+1 (X+1)(X^2 - X + 1) / (X+1) =(X^2 - X + 1). Which is completely divisible by N, since (x + 1) is divisible by N. |

Mandar said: (Jul 15, 2013) | |

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. |

Anshul said: (Aug 10, 2013) | |

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. |

Srujana said: (Nov 1, 2013) | |

I can't understand how to replace (2^96+1) please explain? |

D.Raja said: (Jan 22, 2014) | |

I don't understand this answer please explain briefly. How to replace this answer (2^96+1)? |

Navin Kumar Kamti said: (Mar 4, 2014) | |

Its very simple try to understand friend. Here, Let suppose x+1 divisible by n where n is natural number as like 0. 1.2.3.... If put x=1 then, x+1=1+1=2. Which is divisible by n. As like, 2/2=0. Thus, 2^32=x suppose. Then we can write, x+1= 2^23+1. Now (2^23)^3+1) = 2^96+1. Now we can write, Formula of (x^3+1) = (x+1)(x^2-x+1) which is divisible by n nis natural no As like 0.1.2.....n since x+1 divisible by n. |

Senthil Kumar said: (Jul 4, 2014) | |

I cannot understand how this (2^23)^3+1 is created. Please explain sir. |

Sandeep Kumar Jangir said: (Aug 17, 2014) | |

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. |

Anonymoustype said: (Oct 29, 2014) | |

2^32 + 1 = 4294967297 = 641 * 6700417. Now [A] (2^16+1) = 65537 is a prime. [B] (2^16-1) = 3*5*17*257. [C] (7*2^23) only is divisible by 7 and a bunch of even numbers. [D] (2^96+1) = 79228162514264337593543950337 = 641 * 6700417 * 18446744069414584321. So there. |

Deepa said: (Nov 11, 2014) | |

How can we know (x^3+1) is divisible by n? |

Deependra said: (Nov 21, 2014) | |

It is very easy question guys. Try again to understand the logic of mathematics which applied in this problem. |

Satyajit Chakraborty said: (Jan 12, 2015) | |

Simply remember this: (a^n +b^n) is always divisible by same value of 'a' and 'b' if 'n' is odd in number. Here 'n' denotes power of a number. |

Soumya said: (Apr 11, 2015) | |

(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. |

B.Nandu said: (May 3, 2015) | |

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. |

Nani said: (Jun 9, 2015) | |

Why we don't consider (x^2 - x + 1)? |

Chethanya said: (Jul 9, 2015) | |

Please explain in simple method, because I'm from arts background. |

Ram said: (Jul 31, 2015) | |

I can't understand. |

Priya said: (Aug 1, 2015) | |

It is difficult. But if you we observe carefully we can get it. Think of it more and more. |

Varsha Pathak said: (Aug 3, 2015) | |

(296 + 1) = [(232)3 + 1] = (x3 + 1) = (x + 1)(x2 - x + 1) please explain this part. |

Ashwini said: (Aug 12, 2015) | |

I can't understand please explain briefly? |

Dibya said: (Aug 21, 2015) | |

Why it will not be divisible by 2^16+1? As if 2^32 then why not 2^16? |

Bryan said: (Sep 4, 2015) | |

Above mentioned are totally absurd. Few of them having tried good but not upto the mark. Now see this carefully. We are trying to get our given no. 2^32+1 from the choices given and I tried all d choices. And atlast 2^96+1 is easily broken to get 2^32+1. Now I take an example to explain it in detail. Since 2^32+1 is completely div by N therefore N<=2^32+1 (less than equal to). Ex: If 4 is completely div by x then x is always less than or equal to 4. Now take first option 2^16+1 here it will be completely div N when N<=2^16+1 but wt if N>2^16+1 since value of N is ranging from 0 to 2^32+1 so wt if N>2^16+1 then 2^16+1 is not completely div. Since we exactly don't know value of N we are checking its range to determine. Therefore option A is not answer. Now similarly B is not answer not. Option D is answer because all the range of N completely divides it which is explained how by indiabix. If you don't understand this answer just try to focus on range of N. And again see my example of 4 completely divided. |

Pari said: (Sep 21, 2015) | |

Please give a simple method with which one can get answer in like just 40 secs. Please do help. |

Radhika said: (Sep 24, 2015) | |

I can't understand please again reply. |

Manoj said: (Oct 18, 2015) | |

I can understand the problem. But is there any simplest method to solve this? |

Manoj said: (Oct 18, 2015) | |

I can understand the problem. But is there any simplest method to solve this? |

Kapis said: (Oct 25, 2015) | |

Please give me a basic and simple methods. |

Varun said: (Nov 14, 2015) | |

I am not understand. Please explain again. |

Divya said: (Nov 17, 2015) | |

Multiplication of 36*2 gives correct solution. As friends said use hit and trial method. |

Vaithi said: (Nov 27, 2015) | |

I am not clear about the answer. Please help me. |

Harshada said: (Dec 18, 2015) | |

Problem is that how we can directly get this option i.e. (2^96+1). |

Rajendra said: (Jan 6, 2016) | |

Take 2^32 (2 power 32 as x). Given 2^32 + 1. So x + 1. Formula a^3 + b^3 (a cube + b cube) = (a+b) (a^2 - a.b + b^2). So take x as a, 1 as b. So (x+1) (x^2 - x.1 + 1^2). 1^2 (1 square = 1) and x.1 = x...... So they asked divisible by x+1. So (x + 1)(x^2 - x + 1)/(x + 1). So we get (x^2 - x + 1). Because (x + 1) in numerator and denominator gets cancelled. |

Tarun said: (Feb 3, 2016) | |

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). |

Aman said: (Feb 4, 2016) | |

How we can solve 2^64, 3^432 means? What is the best way to solve high exponential/powers of a number? |

Saurabh Gupta said: (Feb 19, 2016) | |

Dear sir I can't understand your explanation. Please explain clearly from basics. |

Siprarani Tripathy said: (Mar 29, 2016) | |

Dear sir, I can't understand. Kindly explain clearly. |

Savitri said: (Mar 31, 2016) | |

How to solve 2^64, 3^432? |

Madhu said: (Apr 11, 2016) | |

How you take 2^96 +1 exactly? |

Antony Mary said: (Jul 3, 2016) | |

Confused with this problem. Let me know the shortcut method? |

Rayanna said: (Jul 28, 2016) | |

I can't understand please explain me easily. |

Sravanthi said: (Aug 16, 2016) | |

I can't understand. Please explain it. |

Sravanthi said: (Aug 16, 2016) | |

I don't understand this problem. Please explain it in a simple way. |

Nishu Kumari said: (Aug 19, 2016) | |

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. |

Ambika said: (Sep 9, 2016) | |

Was it possible to use power cycle of 2? Can anyone please explain how to use this? |

Karthika said: (Sep 19, 2016) | |

I could not understand the problem please explain me easily. |

Owais Majeed said: (Sep 23, 2016) | |

It is written in question no should be completely divisible. For that number should be greater than given number i.e. 2^32+1. Ex 1/2 = 0.5 (not completely divisible). 4/2 = 2 (completely divisible). So option A, B, C all three numbers are less than given number 2^32 +1. Now come on option D = 2^96+1. Clearly shows number is greater than 2^32+1. This is what up to which some logic can b made. Rest how it is possible to refer to given explanation of the question. Thanks. |

Mahalakshmi said: (Sep 28, 2016) | |

I can't understand please explain again in a simple way. |

Shirisha said: (Oct 21, 2016) | |

I understand this. But I want simple method. |

Pranshu said: (Dec 22, 2016) | |

Why not option A? Explain. |

Sajal said: (Jan 4, 2017) | |

I didn't understand the last step. Please explain it. |

Apsana said: (Jan 17, 2017) | |

I can't understand this please explain this in a simply way. |

Sajid said: (Jan 18, 2017) | |

I don't understand. Please, someone explain me. |

Saruu said: (Jan 31, 2017) | |

Can anyone give a simple method to explain this? |

Deva said: (Mar 9, 2017) | |

I don't understand the solution. Please, someone explain me. |

Kranthi said: (Mar 27, 2017) | |

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. |

Aravinth Kumar said: (Apr 5, 2017) | |

@Kishore Sulthana. Good explanation. I can understand it now. |

Lupu Gogoi said: (Apr 23, 2017) | |

We are finding the answer. We don't know what is the answer above this problem. Then how to you directly put the answer 2^96+1. How you know the answer is that? If the problem given you another then how to you solve that. Please explain. |

Shyam said: (May 16, 2017) | |

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. |

Karna said: (Aug 18, 2017) | |

Here, We know that (x^n+y^n) is divisible by (x+a) for natural numbers. |

Yuvraj said: (Aug 20, 2017) | |

I can't understand to which way solve this question please explain in simple way. |

Yuvraj said: (Aug 20, 2017) | |

I Can't understand, how to solve this? Someone help me. |

Priya Dp said: (Jan 23, 2018) | |

By trial and error method only option A&D are applicable |

Akriti said: (Mar 11, 2018) | |

I can't understand last step how it will completely divisible? Please explain again in simple step. |

Shreya said: (May 19, 2018) | |

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. |

Habib said: (Jul 2, 2018) | |

I can't understand properly, please anyone explain me easily. |

Siri said: (Jul 9, 2018) | |

I can't understand properly, please anyone explain me easily. |

Arjhun said: (Jul 13, 2018) | |

@All. Please refer the following. Formula : (a^3+b^3)=(a+b)(a^2-ab+b^2). Substitute : (x^3+1^3)=(x+1)(x^2-x*(1)+1^2). |

Kannan said: (Jul 17, 2018) | |

Here, 2^96 / 2^32 = 0 is completely divided. |

Binod Barai said: (Aug 26, 2018) | |

2^32+1=2^33. 2^96+1=2^97. So, it is completely divided. |

Supriya Pawar said: (Feb 9, 2019) | |

96/3 = 32 so the divisible no is 3. |

Kamal said: (Mar 14, 2019) | |

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. |

Renu said: (May 21, 2019) | |

I can't understand the last step. Can anyone help me to get it? |

Kavitha said: (May 31, 2019) | |

Here, the options a,b,c are wrong because all are the smallest value than 2^32 +1. When you divide a number by largest number, the answer must be decimal so it not whole and not completely divisible. Then final option 2^96 is larger than 2^32. It may be divisible. Let x= 2^32. We can write, Power 96=32 *3. Then 2^96+1=(2^32)^3+1, We know that (a^3+b^3)=(a+b)(a^2-ab+b^2). Let a=x i.e,x=2^32. b=1. Then X^3+1=(x+1)(x^2-x+1). ÷by x+1i.e x+1=2^32+1. Then we get (x^3+1)/(x+1)=x^2-x+1. And the answer that must be the whole number. Yes, it is completely divisible by 2^32+1. |

Abarna said: (Jun 18, 2019) | |

I can't understand it, please explain me. |

Tej said: (Jul 2, 2019) | |

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. |

Diya said: (Jul 16, 2019) | |

When it comes to 2^16+1, how can we find its not divisible by n? (2^8)2+1 = x^2+1=? |

Nayudu said: (Aug 16, 2019) | |

Agree @Diya. Can anyone please explain it? |

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