Question

If we have an expression as $500!={{2}^{m}}\times \text{an integer}$ then
1. $\text{m=494}$
2. $\text{m=496}$
3. It is equivalent to no. of $\text{n}$ is $\text{400!=}{{\text{2}}^{n}}\times \text{an integer}$
4. $\text{m=}{}^{500}{{C}_{2}}$

Answer 1

As we know that when an integer is less than $[\dfrac{k}{2}]$ or equal to $(k)$ then even numbers of those half are divisible by $\text{4}$ and those half are divisible by $8$ etc. Hence we will define the sum of values and we will find the power of $\text{2}$ that is less than $500$ so that we could choose the correct answer in the given options.

Complete step-by-step solution:
Factor, in mathematics a number or algebraic expression that divides another number or expression evenly that is with no remainder.
There are no shortcuts to getting better at identifying factors and multiples. Factoring a polynomial means writing it as a product of other polynomials.
In algebraic expressions, terms are formed as products of factors. For example, in the algebraic expression $5xy+3x$ the term $5xy$ has been formed by the factors $5,x$ and $y$.
Observe that the factors $5,x$ and $y$ of $5xy$ cannot further be expressed as a product of factors. We may say that $5,x$ and $y$ are prime factors of $5xy$ . In algebraic expressions, we use the word irreducible in the place of prime. We say that $5\times x\times y$ is the irreducible form of $5xy$
We must note that $5\times (xy)$ is not an irreducible form of $5xy$ , since the factor $xy$ can be further expressed as a product of $x$ and $y$ i.e, $xy=x\times y$
When we factorize an algebraic expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions.
Expressions like $3xy,5{{x}^{2}}y,2x(y+2),5(y+1)(x+2)$ are already in factor form. Their factors can be just read off from them, as we already know.
On the other hand expressions like $2x+4,3x+3y,{{x}^{2}}+5x,{{x}^{2}}+5x+6$ . It is not obvious what their factors are. We need to develop a systematic method to factorize these expressions, i.e, to find their factors.
Now according to the question:
We know that when a integer is less than $[\dfrac{k}{2}]$ or equal to $(k)$ then even numbers of those half are divisible by $\text{4}$ and those half are divisible by $8$ and so on ….
Hence the sum of the values can be written as:
$\Rightarrow [\dfrac{k}{2}]+[\dfrac{k}{4}]+[\dfrac{k}{8}]+.............$
Now for $500$ we have to go till ${{\text{2}}^{8}}$ as ${{\text{2}}^{8}}=256$ which is less than $500$ and if we take ${{\text{2}}^{9}}$then that will be greater than $500$ as ${{\text{2}}^{9}}=512$
$\Rightarrow [\dfrac{500}{2}]+[\dfrac{500}{4}]+[\dfrac{500}{8}]+[\dfrac{500}{16}]+[\dfrac{500}{32}]+[\dfrac{500}{64}]+[\dfrac{500}{128}]+[\dfrac{500}{256}]$
$\Rightarrow 250+125+62+31+15+7+3+1$
$\Rightarrow 494$
Hence option $(1)$ is correct as $\text{m=494}$.

Note: We must remember that factors are always whole numbers or integers and never decimal or fractions and all the even numbers will have $2$ in their factors. The factor of a number is always less than or equal to the given number. Division and multiplication are the operations that are used in finding the factors.