What Is 40 2/5 Divided By 2 1/50

How is 40 divided in 1/2?

Follow the steps below.40÷1/2It is always considered that any whole number is assumed or has 1 as its denominator.40/1 ÷ 1/2We will get the reciprocal of 1/2 and do multiplication for both numerators and denominators so it becomes:40/1 * 2/140*2=801*1= 1Now we have80/1Let’s simplify and we’ll get80 as the answer.Therefore, there are 80 one halves in 40.

What is 5 divided by 1/8?

When dividing a number by a fraction, it is much easier to think of it as the opposite of the multiplication process. So, instead of dividing by 1/8, multiply by the fraction’s reciprocal—-in other words, turn the fraction upside down. Thus, we have a much easier problem to deal with: 5 times 8/1 or just 5 times 8 which is 40.To divide 6 by 2/3, think 6 times 3/2. Multiply numerators 6 times 3 and put that product over the product of the denominators 1 times 2. Now just divide 18 by 2, and your answer is 9.

What is 2 divided 5/6?

Fractions are fun![math]2 \div \frac{5}{6}[/math][math]= \frac{2}{1} \div \frac{5}{6}[/math][math]= \frac{2}{1} \times \frac{6}{5}[/math][math]= \frac{2 \times 6}{1 \times 5}[/math][math]= \frac{12}{5}[/math][math]= \frac{24}{10}[/math][math]= 2.4[/math]

How do you show that 41 divides 2^(20)-1?

This is essentially asking us to show that [math]2^{20} = 1 \pmod{41}[/math]. Because 41 is an odd prime, we know by Fermat’s Little Theorem that [math]x^{(41 - 1)/2} = 1 \pmod{41}[/math] whenever [math]x[/math] is a nonzero square modulo 41. (In fact, by counting roots, this is furthermore an “if and only if”; see below). Since [math](41 - 1)/2 = 20[/math], this applies precisely to what we want to know. We now just have to show that 2 is a square modulo 41.Here we invoke another great number theoretic tool, quadratic reciprocity, which lets us quickly determine what values are squares modulo what values. In particular, it tells us 2 is a square modulo an odd prime just in case that odd prime is next to a multiple of 8. 41 is indeed next to a multiple of 8 (in this case, 40), so 2 is indeed a square modulo 41, and so we are done.Although I’ve invoked quadratic reciprocity here, we can actually inline the relevant argument for those who are unfamiliar with quadratic reciprocity, like so: Consider all [math]40[/math] nonzero values modulo [math]41[/math]. By Fermat’s Little Theorem, we know that they are the [math]40[/math] roots of [math]x^{40} - 1 = (x^{20} + 1)(x^{20} - 1)[/math]. Since neither factor can have more than 20 roots, we find that each factor gets 20 roots; in particular, there are 20 solutions to [math]x^{20} = -1[/math]. Pick any such solution [math]x[/math] and let [math]y = x^{10} + x^{-10}[/math] while [math]z = x^5 + x^{-5}[/math]. Observe that [math]y^2 = x^{20} + x^{-20} + 2 = -1 + -1 + 2 = 0[/math]; thus [math]y = 0[/math]. Next observe that [math]z^2 = x^{10} + x^{-10} + 2 = y + 2 = 0 + 2 = 2[/math]. Thus, [math]2[/math] is a square modulo [math]41[/math]. And as to our original problem, [math]2^{20} = (z^2)^{20} = z^{40} = 1[/math]. (All equations here are in modulo 41 land, of course)

What is the remainder if 2^81 is divided by 5?

Let us understand a simple concept first.2^1=22^2=42^3=82^4=162^5=32And so on . . .Now note that the last digit on the RHS starts repeating itself after operating the power of two 4 times. Now in 2^81 the power is 81, let us see how many 4s are contained in 81.81÷4=20 with a remainder 1.So we have to look at 2^1=2. So the answer of 2^81 is going to have its last digit as 2. Multiples of 5 will have their last digit as either 5 or 0. Since the last digit in our case is 2, the remainder will also be 2.Hence the remainder when 2^81 is divided by 5 the remainder will be 2.This technique can be applied to similar such problems.Hope this clears your doubt.