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

Hint: 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.