51 of 100: Some Simple Fractions

$\frac{1110}{1111}=\frac{6660}{6666}$

$\frac{2221}{2223}=\frac{6663}{6669}$

$\frac{3331}{3334}=\frac{6662}{6668}$

For each fraction, the difference between numerator and denominator is the same (6), so the fraction with the largest numerator or denominator will be the closest to 1, hence the answer is $\boxed{\frac{2221}{2223}}$ .

Rewrite problem as: which is largest: $1 - \frac{1}{1111}, 1 - \frac{2}{2223},$ or $1-\frac{3}{3334}$ . This is clearly equivalent to: which is smallest, $\frac{1}{1111}, \frac{2}{2223},$ or $\frac{3}{3334}$ ? Since $\frac{1}{1111} = \frac{2}{2222} > \frac{2}{2223},$ the first is clearly not the smallest. That leaves us with the task of comparing $\frac{2}{2223}$ and $\frac{3}{3334}$ . Quick cross-multiplication (compare $3*2223$ with $2*3334$ ) shows that $\frac{3}{3334}$ is larger, leaving $\frac{2}{2223}$ as the smallest. Thus $\frac{2221}{2223}$ is the largest of the three original fractions.

$\frac{a}{b} < \frac{a+c}{b+c}$ given that $b>a$

$\frac{1110}{1111}=\frac{2220}{2222}<\frac{2221}{2223}=\frac{6663}{6669}>\frac{6662}{6668}=\frac{3331}{3334}$

$\begin{aligned} \frac {1110}{1111} & = 1 - \frac 1{1111} \\ \frac {2221}{2223} & = 1 - \frac 2{2223} = 1 - \frac 1{1111\frac 12} \\ \frac {3331}{3334} & = 1 - \frac 3{3334} = 1 - \frac 1{1111\frac 13} \end{aligned}$

Since $\dfrac 1{1111\frac 12} < \dfrac 1{1111\frac 13} < \dfrac 1{1111}$ , $\implies \boxed{\dfrac {2221}{2223}} > \dfrac {3331}{3334} > \dfrac {1110}{1111}$ .

Each of these is close to but not greater than one, so you may rephrase the question as "Which has the smallest difference from 1?" Subtracting and making a common numerator gives:

1- $\frac{1110}{1111}$ = $\frac{1}{1111}$ = $\frac{6}{6666}$

1- $\frac{2221}{2223}$ = $\frac{2}{2223}$ = $\frac{6}{6669}$

1- $\frac{3331}{3334}$ = $\frac{3}{3334}$ = $\frac{6}{6668}$

Since the numerators are all the same, the largest denominator is the smallest number, and therefore $\frac{2221}{2223}$ is the correct answer to the problem

if $x = 1110$

$A =$ $\frac{1110}{1111}$ = $\frac{x}{x+1}$

$B =$ $\frac{2221}{2223}$ = $\frac{2x+1}{2x+3}$

$C =$ $\frac{3331}{3334}$ = $\frac{3x+1}{3x+4}$

Cross multiply the A and B, you end up with

$2x^2+3x < 2x^2+3x+1$

in the numerator, so B is larger.

Now cross multiply B and C, which gives you

$6x^2+11x+4 > 6x^2+11x+3$

in the numerator, so B or $\frac{2221}{2223}$ is larger.

$\frac { 1110 }{ 1111 } =1-\frac { 1 }{ 1111 } \\ \frac { 2221 }{ 2223 } =1-\frac { 2 }{ 2223 } =1-2\times \frac { 1 }{ 2223 } =1-2\times \frac { 1 }{ 1111\times 2+1 } =1-2\times \frac { 1 }{ 2\times (1111+\frac { 1 }{ 2 } ) } =1-\frac { 1 }{ 1111+\frac { 1 }{ 2 } } \\ \frac { 3331 }{ 3334 } =1-\frac { 3 }{ 3334 } =1-3\times \frac { 1 }{ 3334 } =1-3\times \frac { 1 }{ 1111\times 3+1 } =1-3\times \frac { 1 }{ 3\times (1111+\frac { 1 }{ 3 } ) } =1-\frac { 1 }{ 1111+\frac { 1 }{ 3 } }$

The largest fraction is the one in which the least quantity is subtracted from 1.

So, the answer is $\boxed{\frac { 2221 }{ 2223 } }$ ,

In fact this fraction has the same numerator of the others and a greater denominator.

if we multiply these 3 fractions with big integers, the difference between numerator and denominator will be seen. and the fraction with the largest numerator or denominator will be close to 1 than other numbers. so, the answer will be = $\frac {2221}{2223}$

again, their is another way, if we subtract these all fractions from 1 (as they all are less than 1), than these fractions will be, $\frac {1}{1111}$ < $\frac {2}{2223}$ > $\frac {3}{3334}$

LaTex: $\Huge \text {Let :} \; A = 1111$

LaTex: $\Huge \text {Let : } \; e = \frac{A-1}{A}, \; f = \frac{2A-1}{2A+1}, \; g = \frac{3A-2}{3A+1}$

LaTex: $\Large \text {Let's suppose : } \; \boxed {e<f}$

LaTex: $\large e<f \iff \frac{A-1}{A}< \frac{2A-1}{2A+1} \iff (A-1)(2A+1)<A(2A-1) \iff -1<0 \Rightarrow \; \text {True}$

LaTex: $\Large \text {Let's suppose : } \; \boxed {e<g}$

LaTex: $\large e<g \iff \frac{A-1}{A}< \frac{3A-2}{3A+1} \iff (A-1)(3A+1)<A(3A-2) \iff -1<0 \Rightarrow \; \text {True}$

LaTex: $\Large \text {Let's suppose : } \; \boxed {f<g}$

LaTex: $\large f<g \iff \frac{2A-1}{2A+1}< \frac{3A-2}{3A+1} \iff (2A-1)(3A+1)<(2A+1)(3A-2) \iff -1<-2 \Rightarrow \; \text {False}$

LaTex: $\Large \text {Thus :} \; \boxed {f>g}$

LaTeX: $\Large \text {Finally } \; (e<f), (e<g) \; \text {and}\; (f>g) \Rightarrow \color{#D61F06}{ e<g<f \Rightarrow \boxed{ f = \frac{2A-1}{2A+1}= \frac{2221}{2223} }}$

LaTeX: $\Large \text {That is to say : "\$f\$" is the largest number between the 3 }$

LaTex $\large \text {Another way to find the largest number :}$

LaTex: $\huge \text {Let :} \; a = 1110$

LaTex: $\huge \text {Let : } \; A(a) = \frac{a+0}{a+\frac{1}{1}}, \; B(a) = \frac{a+\frac{1}{2}}{a+\frac{3}{2}}, \; C(a) = \frac{a+\frac{1}{3}}{a+\frac{4}{3}}$

LaTex: $\Large lcm(1;2;3) = 6 \; \text {then let's multiply every numerator and denominator :}$

LaTex: $\Large \text {Let : } \; A(a) = \frac{6a+0}{6a+6}, \; B(a) = \frac{6a+3}{6a+9}, \; C(a) = \frac{6a+2}{6a+8}$

LaTex: $\Large [ \lim\limits_{a \rightarrow 0} A(a) = \frac{0}{6} = 0], \; [\lim\limits_{a \rightarrow 0} B(a) = \frac{3}{9}= \frac{1}{3}\approx 0.33], \; [\lim\limits_{a \rightarrow 0} C(a) = \frac{2}{8}= \frac{1}{4}= 0.25]$

LaTex: $\Large \text {We can now sort the 3 numbers :} \; A(a)<C(a)<B(a) \iff 0<\frac{1}{4}<\frac{1}{3}$

LaTeX: $\Large \text {Thus, the largest number between the 3 is : } \; \color{#D61F06}{ \boxed {B(a) = \frac{a+\frac{1}{2}}{a+\frac{3}{2}} = \frac{2221}{2223}}}$

For positive values, if $\frac{a}{b}>\frac{c}{d}$ then $ad>bc$ .

Let $x=1110$ , then the fractions are:

$\frac{1110}{1111} = \frac{x}{x+1}$ , $\frac{2221}{2223} = \frac{2x+1}{2x+3}$ , and $\frac{3331}{3334} = \frac{3x+1}{3x+4}$

For the first two,

$x(2x+3) < (2x+1)(x+1)$

$2x^2+3x < 2x^2+3x+1$

$0 < 1$

So the second is greater than the first. Then for the second vs. the third,

$(2x+1)(3x+4) > (3x+1)(2x+3)$

$6x^2+11x+4 > 6x^2 + 11x + 3$

$4 > 3$

So the second is the highest.

Just do some changes 1-1/1111 1-2/2223 3-3/3334 2/2223=1/2223/2=1/1111.5<1/1111

So by this u get ur answer morons

All these fractions are clearly a little less than 1. Write them initially as "1 minus something":

$\frac{1110}{1111}$ =1- $\frac{1}{1111}$

$\frac{2221}{2223}$ =1- $\frac{2}{2223}$

$\frac{3331}{3334}$ =1- $\frac{3}{3334}$

Now look at the amounts that have been deducted from 1; the smallest deduction will lead to the largest original fraction.

Make the numerators the same; lowest common multiple is 6 ...

$\frac{1}{1111}$ is $\frac{6}{6666}$ , $\frac{2}{2223}$ is $\frac{6}{6669}$ , and $\frac{3}{3334}$ is $\frac{6}{6668}$

Looking at the denominators, 6669 is the largest, so $\frac{6}{6669}$ is the smallest deduction from 1, so we can conclude that $\frac{2221}{2223}$ was the largest of the original fractions.

a/(a+1)< (2a+1)/(2a+3)>(3a+1)/(3a+4)

$\frac{1110}{1111}$ - $\frac{2221}{2223}$ = $\frac{1110*2223-1111*2221}{1111*2223}$ = $\frac{1110*(2223-2221)-2221}{1111*2223}$ = $\frac{-1}{1111*2223}$ <0; $\frac{2221}{2223}$ > $\frac{1110}{1111}$

$\frac{2221}{2223}$ - $\frac{3331}{3334}$ = $\frac{2221*3334-2223*3331}{2223*3334}$ = $\frac{2221*(3334-3331)-6662}{2223*3334}$ = $\frac{1}{2223*3334}$ >0; $\frac{2221}{2223}$ > $\frac{3331}{3334}$

The largest is $\frac{2221}{2223}$ .

Let 1110 = a. The numbers now are a/(a+1), (2a+1)/(2a+3), and (3a+1)/3a+4). Now let a=1, so now the fractions are 1/2, 3/5, and 4/7. The second, 3/5, is the largest.

1110/1111 = 1 – 1/1111

2221/2223 = 1 – 2/2223 = 1 – 1/1111.5

3331/3334 = 1 – 3/3334 = 1 – 1/1111.3333......

Since obviously,

1/1111 > 1/1111.33…. > 1/1111.5

So,

1110/1111 < 3331/3334 < 2221/2223.

Therefore 2221/2223 is largest.

Let $a = 1110$ . Then the three fractions become $\frac{a}{a+1}$ , $\frac{2a+1}{2a+3}$ and $\frac{3a+1}{3a+4}$ . We can systematically cross multiply the three pairs: $A$ and $B$ ; $A$ and $C$ ; $B$ and $C$ and compare the result.

$A$ and $B$ : $(a)(2a+3) \ ? \ (a+1)(2a+1) \Rightarrow (2a)(a) + 3a \ ? \ (2a)(a) + a + 2a + 1$ . Since $0 < 1, A < B.$

$B$ and $C$ : $(2a+1)(3a+4) \ ? \ (2a+3)(3a+1) \Rightarrow (2a)(3a) + 8a + 3a + 4 \ ? \ (2a)(3a) + 2a + 9a + 3$ . Since $4 > 3, B > C.$

Since $B$ is greater than both $A$ and $C$ , we can conclude $B$ is the greatest out of the three fractions.