How Do You Divide 2n-5/4n^2-16n 15

When n is divided by 4, the remainder is 3. What is the remainder when 2n is divided by 4?

N/4 = x + 3Multiplying both sides with 2, which gives( 2×N )/ 4 = 2X + 6N/2 = 2X + 6Since, 6 is divisible by 4 and gives a remainder of 2, above statement can be written asN/2 = 2X + 1X + 2N/2 = 3X + 2From here we will reach to the solution of the above problem as… 2 is remainder when 2n is divided with 4.ORSay N = 11Dividing N by 4 gives you remainder of 3.Dividing 2N i.e. 22 by 4 gives you remainder of 2.Similarly, dividing 3N i.e. 33 by 4 gives you remainder of 1 only.And so on so forth.

How do you find the remainder of 54^124 divided by 17?

4Rem [54^124 / 17]= Rem[(3)^124 / 17]= Rem[81^31 / 17]= Rem[(-4)^31 / 17]= Rem[(-4)^30 * (-4) / 17]= Rem[(16)^15 * (-4) / 17]= Rem[(-1)^15 * (-4) / 17]= Rem[(-1) * (-4) / 17]= 4I have answered a bunch of very similar questions on remainders. You can get the complete list here: Remainder Theorem and related concepts for CAT Preparation by Ravi Handa on CAT Preparation

What is the number of divisors which are in the form of 4n+2 of a number 240?

Well, let us first find out the number of factors of the given number[math]240 = 2^4.3^1.5^1[/math]Therefore, the number of factors as we know is [math](4+1)(1+1)(1+1)[/math][math]= 20.[/math]Now, we are to find the factors of the form [math]4n+2.[/math][math]4n+2 = 2(2n+1)[/math]Therefore, the term [math]4n+2[/math] is always even.So, let us eliminate the odd factors of [math]240[/math].No. of odd factors = [math](1+1)(1+1) = 4[/math]Subtracting the number of odd factors from the total number of factors, we get[math]20-4 = 16[/math] factors.Now, if we give a closer look to the set of even natural numbers, we can see, that every non-multiple of 4 has a difference of two with its consequtive multiples of 4.Therefore, the term [math]4n+2[/math] is every even natural number, except the multiples of [math]4[/math], in the set.Therefore from the [math]16[/math] factors, if we eliminate the multiples of [math]4,[/math] we will get our desired result.So,[math]2^4.3.5 = 2^2(2^2.3.5)[/math]Therefore the number of multiples of [math]4[/math] in its set is [math](2+1)(1+1)(1+1)[/math] [math]= 12[/math]Now, it's time to subtract the multiples of 4 i.e [math]12 [/math]from [math]16.[/math]Thus the number of factors of the number [math]240[/math], of the form [math]4n+2[/math] is [math](16-12) = 4.(Ans.)[/math]

How can you prove by induction that 7^n+ 4^n + 1  is divisible by 6, for n = 1, 2, 3, …?

The question can be presented as:Prove that  6|7^n+4^n+1   when n is positive intergerLet P(n) be the proposition that asserts 7^n+4^n+1 is divisible by 6.Basic step:P(1) is true, since  7+4+1 = 12, 6|12Inductive step:Assume that P(k) is true, with k>=1.That is:6|7^k+4^k+1 We need to prove P(k+1) is true:That is 7^(k+1)+4^(k+1)+1 = 7*7^(k) + 4*4^(k)+1=(6+1)*7^(k) + (3+1)*4^(k)+1=(7^k+4^k+1 ) + (6*7^k + 3*4^k)two terms are divisible by 6, so P(k+1) is true.This completes the proof