Aptitude - Numbers - Discussion

Discussion :: 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?

 [A]. (216 + 1) [B]. (216 - 1) [C]. (7 x 223) [D]. (296 + 1)

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.

 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.

 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.

 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.

 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?

 Jyoshna said: (Sep 19, 2019) I am not understanding please anyone help me to get it.

 Saad said: (Sep 30, 2019) I can't understand. Please, anyone, explain it briefly.

 Sasikumar Dpi said: (Dec 1, 2019) 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)

 Sonu said: (Dec 27, 2019) Your explanation is very complicated.

 Dilip said: (Mar 2, 2020) I'm not understanding this problem.

 Deva.Harshitha Patel said: (Apr 2, 2020) WKT,(x^n+a^n) is divisible by (x+a),if n is odd. Here,if we take x=2^32 and a=1,x+a=2^32+1 divides (x^n+a^n),if n is odd i.e,i may be 1 or 3 or 5 or 7.. If we take n=1,(2^32)^1+1^1=2^32+1 is not in the option, if n=3,(2^32)^3+1^3=2^96+1 is the correct answer.

 Octavia said: (May 28, 2020) I don't understand. Please explain in detail.

 Anuj Yadav said: (Aug 5, 2020) I think the formula will be 2^(n.3)+1.

 Yaduvanshi said: (Oct 30, 2020) @Deva Harshitha Patel. You are absolutely right, thanks.

 Vilaz said: (Jan 21, 2021) Let 2^32 = x. Lets take, (2^96 + 1) = [(2^32)3 + 1] = (x^3 + 1) { we took 2^32=x} =(x^3 + 1) { In the form of [a^3+b^3 ]=[a^2- ab+b^2 ] = (x + 1)(x2 - x + 1). Which is completely divisible by N, since (x + 1) is divisible by N.

 Vivek Singh said: (Feb 23, 2021) This question is to solve using the formula:- a^3+b^3 = (a+b)(a^2-ab+b^2). Only option D satisfies the above formula. We can write (2^96+1) as ((2^32)^3 + 1). So if we solve ((2^32)^3 + 1) using the above formula then we get: ((2^32)^3 + 1) = (2^32+1)(2^64-2^32+1). Now, divide (2^32+1)(2^64-2^32+1) by (2^32+1). If we divide these two then (2^32+1) gets cancelled and we get the result as (2^64-2^32+1). Hence option D is correct.