Easy Trick

EDIT (Friday, November 13, 2015): $20142014 = 20,142,014$ . $20142014$ is not the same as $2014\times 2014$ . I've also changed the colors.

Simple.

Let $a=20142014$ . Then the given expression is equivalent to $a^2 - (a-1)(a+1)$ $=a^2-(a^2-1)$ $=\boxed{1}$ and we are done.

Disregard the first seven digits. Use only the last digits. (4x4)-(3x5). 16-15=1.

$2014^2-(2014-1)(2014+1)$

$=2014^2-(2014^2-1)$

$=2014^2-2014^2+1$

$=1$

Let X=20142014 then According to Question, x^2-(x-1)(x+1)=x^2-(x^2-1) =x^2-x^2+1 = 1 ans.

Let $x=20142014$

Then the problem becomes $\begin{aligned} x^2-(x-1)(x+1)&=x^2-(x^2-1^2)\\&=x^2-x^2+1\\&=1 \end{aligned}$

=20142014^2 -( 20142014-1) x (20142014+1) ; // As (a-b)x(a+b) = a^2-b^2 ; = 20142014^2 - (20142014^2 - 1^2 ) ; = 1 ;

Let x =20142014, Then The above eq. is: $=x\times x-(x-1)\times (x+1)\\As (a+b)(a-b)=({ a }^{ 2 }-{ b }^{ 2 })\\ \therefore (x-1)(x+1)=({ x }^{ 2 }-1)\\{ =x }^{ 2 }-({ x }^{ 2 }-1)\\ ={ x }^{ 2 }-{ x }^{ 2 }+1\\ =1$

It's really really easy ^^ nearly the same as Victor Loh but maybe some don't understand so let's make it clear *but a bit longer :( * .

   20142014 x 20142014 -  20142013 x 20142015

= (20142014)^ 2  -  (20142014 - 1) x (20142014 + 1) 

= (20142014) ^ 2 -  [ (20142014) ^ 2 - 1]

= 0 - (-1) = 1

We also have another way to solve this, no expanding or factoring formulas required :)

    20142014 x 20142014 - 20142013 x 20142015

= 20142014 x 20142014 - 20142013 x (20142014 + 1) 

= 20142014 x 20142014 - 20142013 x 20142014 - 20142013 

= 20142014 x ( 20142014 - 20142013) - 20142013 

= 20142014 - 20142013 = 1

That's all :) Have a nice weekend :)

I defined x=2014

X(2014) * X(2014) which would be X^4

X^4 - x(X-1) * x(X+1) X^4 - X^4 = 1

The problem is invariably wrong. It should have a^2-(a-1)(a+1) in which case the figures should have appeared as (20142014 20142014) - (20142013 20142015). Here we are given the problem as a^2-(a-1)a in which the answer comes as "a" itself always. The answer to (20142014 20142014)-(20142013 20142014) evaluates to 20142014 itself.

Thanks, Siddhartha

its simple take 20142013 on the right side then take anyone of the 20142014 and divide it u will be left with jst 20142014 and 20142013 on both sides subtact them and theres ur answer

we have given 20142014^2= x^2, and 20142013*20142015= (x-1)(x+1) so that why, Let's considered x^2-(x-1)(x+1) therefore x^2-1^2=(x-1)(x+1) =x^2-(x^2-1^2) =x^2-x^2+1 =1

Since every time 2 of the same numbers are multipled together, for example 4x4 which is 16. And every time the numbers is one less and one more, the answer is one less, like 5x3 which is 15 which is one less than 16. Another example could be 7x7 makes 49, and 6x8 makes 48. Or 10x10 makes 100 and 11x9 makes 99. So for this question it is 2014x2014-2015x2013 (this is when you simplify it) which we know equals one, even when we dont know the answer for 2014x2014 or 2015x2013.

x=20142014

x²-{(x-1)(x+1)} =x²-(x²-1²) =1

Consider $20142014 = a$ Then, $20142013 = a-1$ and $20142015 = a+1$ So, we have $20142014\times 20142014 = a^2$ And $20142013\times 20142015 = (a-1)(a+1) = a^2-1$ So, we have $a^2-(a^2-1) = (a^2-a^2)+1 = 1$

I did the easier way: the numbers are almost the same: 20142014-20142014-20142013 -20142015 just the last digit has changed so... I forgot the whole number and keep just the last digit and did it 2014201(4)x2014201(4)-2014201(3)x2014201(5) then.. 4x4-3x5 16-15 = 1

hahhahahah

Let 20142014 = x - 1

Thus 20142013 = x - 2

And 20142015 = x

So,

=(x-1)^2 - ((x-2)*x)

=x^2 -2x +1 -x^2 + 2x

=1

Alright, let's see what we've got!: $20142014\times 20142014-20142013\times 20142015$

Let us first re-write 2013 & 2015 as $\pm 1$ of 2014, as such: $20142014\times 20142014-(20142014-1)\times (20142014+1) 20142014\times 20142014-(20142014-1)\times (20142014+1)$

Now, we distribute the negative sign, to the one, as so: $20142014\times 20142014-20142014+1\times 20142014+1$

Now, we just simplify, as such: $20142014\times 20142014-20142014\times 20142014+1$

Now, we merely factor out the 2014, as so: $(2(20142014)-2(20142014))+1$

Now, we merely cancel out the 2014s, as both of them are $2(20142014)$ , therefore leading us to: $\boxed{1}$

$Q\bullet E\bullet D$

TIP: Oh, and one more thing, start using current dates, like 2016. 2014 was 2 years ago!

I must say that the answer is not right. Its simply 0. The last digit is useless as you take one and move to the other. You did there (a x a) - (b x c) but the fact is that (a x a) = (b x c).

Two close numbers multiplied have a trick, Average^2 - difference^2

If we apply this here we get: 20142014 * 20142014 - 20142013 * 20142015

20142014^2 - 20142014^2 -1^2 -1^2=1

20142014(20142014) - (20142014 + 1 )(20142013) = a(a) - (a+1)(b) = a(a) - a(b) - b =a(a-b) - b = (20142014)(20142014 - 20142013) - (20142013) =20142014 - 20142013 = 1

Simple. x^2 -(x^2 -1) = 1

Why would a single number ever be displayed as two different colors? The years to me have no significance in solving this equation other than if they were being multiplied by each other, seems like most other people commenting have the same thought process.

It's a trap

Just regards the last digits of every term which is 4x4-3x5=>16-15=1

20142014 x 20142014 - 20142013 x 20142015 =

20142014 x 20142014 - (20142014-1) x (20142014+1) =

20142014 x 20142014 - (20142014 x 20142014 - 1 x 1) =

20142014^2 - 20142014^2 +1=

=1

Color formatting makes question unclear. I'd prefer if all 8 digits of a number one color instead of splitting it pink and blue . It implies multiplication at a glance. Had to work the problem both ways to check myself.

20142014 X 20142014 - 20142013 X 20142015 =(20142014)^2 - (20142014 - 1)(20142014 + 1) =(20142014)^2 - (20142014^2 - 1^2) =20142014^2 - 20142014^2 + 1^2 =1

I just used other numbers as an example.

3 * 3 - (3 - 1) * (3 + 1)

3² - 2 * 4

9 - 8

1

20142014 x 20142014 - (20142014-1) x (20142014+1)

20142014^2 - ( 20142014^2 - 1 )

20142014 - 20142014 + 1 = 1

=(20142014)^2 -[(20142014-1)(20142014+1) =(20142014)^2-[(20142014)^2 -1] =(20142014)^2 - (20142014)^2 +1 = 1

a²-(a-1)(a+1) =a²-a²+1 =1.

The solution by Victor Loh is very clear, but for people who want to deal with the numbers only, here is another form of the solution:

20142014 * 20141014 - 20142013 * 20142015 =20142014 * 20141014 + 20142015 - 20142014 * 20142015 (we add 20142015 to both terms) =20142014 * 20141014 + 20142015 - 20142014 - 20142014 * 20142014 (subtracting 20142014 from both sides) = 20142015 - 20142014 =1

Its very easy. You just need to multiply the last numbers i.e. 4×4 - 3×5=16-15=1.

20142013 = 20142014 - 1 and 20142015 = 20142014 + 1. Then: (20142014-1)×(20142014+1) = 20142014² - 1. 20142014² - (20142014²-1), will be 20142014² - 20142014² + 1; resulting in 1.

Yeah, I realised it while giving my last try

= ((2014201)^(2) * 4 * 4) - ((2014201)^(2) * 3 * 5) = 16 - 15 = 1

I looked at it and said screw that. Set 20142014 = 2

20132013 = 1

20152015 = 3

Then... 4-(1*3) = 1

We can simply minus LHS AND RHS. 2014-2014=0 2014-2014=0 2014-2013=1 2015-2014=1 and at last we will multiply answer 1×1=1

let 20142014 be x

now you will get the expression like this

a^2-(a-1)(a+1) a^2-(a^2-1)

a^2-a^2+1

1

The problem can be easily solved by breaking down numbers into simple variables: For example, we can assign 20142014 to be x. Rewriting the equation:

(x*x)-(x+1)(x-1). Notice how 20142013 is really just 20142014-1 or x-1. Same follows for 20142015. Simplifying the difference of squares gives us x^2-(x^2-1). The x^2's cancel out leaving -(-1), which is just 1.

I did not use a formula that I was aware of, I noticed that the numbers on each side of the minus were the same, and then looked closer to see that there was only 1 number difference ib the two numbers, and 1 times one is one. it was strictly observational intuition.

SUBTRACTING EACH NUMBER WITH 20142013... (20142014-20142013)X (20142014-20142013)-(20142013-20142013)X(20142015-20142013) = 1X1-0X2 = 1

${2014}^{2}-(20142014-1)(20142014+1)$

Since we already know the formula for difference in squares, which:

${a}^{2}-{b}^{2}=(a-b)(a+b)$

We can simplify the question even further:

${2014}^{2}-({2014}^{2}-{1}^{2})$

Remove the bracket and you will have:

${2014}^{2}-{2014}^{2}+1$

Evaluate and you will end up in the answer as:

$\boxed{1}$

Let x = 20142014

Then, 20142013 = ( x - 1 )

And , 20142015 = ( x + 1 )

So the given conditions become,

x^2 - ( x - 1)( x + 1 )

= x^2 - ( x^2 - 1 )

= x^2 - x^2 + 1

= 1

I simply worked with the differentials - the end digits - because, in any normal operation, the common factors would cancel each other. Hence: 4×4 -3×5 = 16-15 = 1.

very simple

just take x = 20142014

then x X x -(x-1)(x+1) =x^2 - (x^2-1^2) =x^2 - x^2 + 1 =1 . :) simple

The correct answer is 1

Write a solution. x^2 -(x-1)(x+1) =x^2-((x^2)-1) =(x^2)-(x^2)+1 =1

20142014×20142014 - 20142013×20142015= 2014 (2014^2)- 2014 (2013x2015)= 2014^2 - 2013×2015= 1

Let 20142014 be x x^2-(x-1)(x+1) x^2-x^2-(-1) 1

only the claw knows

Ur method is wrong and ans I'd also not one..... a=2014x2014 ...doesn't mean a-1=2014x2013......rather it's 2014x2014-1

it took me a while with the colors and all, but in the end I realized it was incredibly easy. by the way i never gnu that color meant factor, I'll have to keep that in mind

use the last digit 4x4-3x5

$20142014 \times 20142014 - 20142013 \times 20142015$ $= (20142010 + 4)(20142010 + 4) - (20142010 + 3)(20142010 + 5)$ $= (20142010^{2} + 8(20142010) + 16) - (20142010^{2} + 8(20142010) + 15)$ $= 1$

Simple trick of squared numbers n^2 -1 = (N-1)(N+1) example 8x8=64, 7x9 =63, try with any square numbers

A^2-[(A-B)(A+B)]=A^2-[A^2-B^2]=A^2-A^2+B^2=B^2; here A=20142014 and B=1;

This problem seems difficult at first, though thinking of it differently can help.

For the equation:

20142014 * 20142014 - 20142013 * 20142015 = ?

think of the number 20142014 as a variable, in this case x therefore we get the equation:

X * X - ( X - 1 ) * ( X + 1 ) = ?

Solving we get:

X ^ 2 - (X ^ 2 - 1)

Distributing the -1 from in front of the parentheses:

X ^ 2 - X ^ 2 + 1

Subtracting the X ^ 2:

1

Easy just guess

Use the last digit of all numbers, multiply and subtract

Ignore the redundant digits (2014201) to leave 4x4-3x5=1.

20142014^2 - (20142014-1)(20142014+1) = 20142014^2 - 20142014^2 - 20142014+20142014 +1 = 1

I disagree with the solution given of 1. Here is my reasoning. 2014x2014x2014x2014-2014x2013x2014x2015

Take out a common factor of 2014^2. 2014^2x(2014^2-2013x2015)

Now you have to do the calculations. Using algebra will not simplify the equation any more. (I've tried) The answer comes out as 4,056,196.

I've checked some of the other solutions mentioned and the algebra is flawed. If you still don't believe me then use a calculator to solve the equation from the start and you will see that the answer isn't 1 but 4,056,196. it is impossible to use algebra to find a different solution to an equation.

Let 20142015 be y^3 Let 20142014 be y^2 let 20142013 be y Then 20142014 × 201402014 =20142013 × 20142015

Y^2 × y^2 = y × y^3 Y^4 = y^4

Then y^4/y^4 = 1

Is it right or wrong ?

What the hell! I had no idea how to do it. I casually put 1in the answer. Turned out right!!

X^2 - (X+1) (x-1)= X^2 - X^2-1= X^2-X^2 - - 1= 0- -1= +1

Multiplying the last digits we get 16 - 15 = 1 Ans

4 X 4=16 and 3 X 5=15 and other digits are same. So 1 is the answer.

let 20142014 be x. So the problem boils down to x^{2}-(x-1)*)(x+1) = x^{2}-[ x^{2}-1 ] = x^{2}-x^{2}+1 = 1

I found that any number times itself minus the product of the number minus one and the number plus one is always one. For example, (4X4)-(5X6) =1. Also, (6X6)-(5X7) is also 1. (100X100)-(99X101) is also 1. Therefore, the answer to this problem must be 1.

a squared - (a-1)(a+1)= always 1

The numbers are all the same except the last digit so I if you just pretend the rest of the numbers aren't there the problem looks like "4x4 - 3x5." 4x4=16 and 3x5=15 so I just did 16 -15= 1.

calculator is giving answer 0. what's wrong?

I just did this

$4*4$ - $3*5$ =16-15=1

Your answer is wrong.

let 20142014 =a then the expression would look like

(a x a) - (a-1)(a+1) = a^{2} - a^{2} + a - a - 1 = -1

20142000 is same . I just took 14x14 - 13x15 = 196-195 =1

if you notice that its basically just the same multiplication problem on each side except for the one at the end. the answer is one.

=[20142014 +(20142013 20142014)]-[(20142013 20142014)+20142013] =20142014-20142013

in my opinion if we replace those big numbers by small ones as if we put 2 instead of 20142014 and put 1 instead of 20142013 and put 3 instead of 20142015 because it's the biggest number so if we normally calculate 2 2-1 3 it will be simply equal 1

All you have to do is 4 4-3 5=1!

i just multiplied last one digits and subtracted

20142014 x 20142014 - 20142013 x 20142015

(20142013 + 1)(20142013 + 1) - 20142013(20142013 + 2)

20142013^2 + 2(20142013) + 1 - 20142013^2 - 2(20142013)

= 1

(14 14-15 13)=1

I just guessed :)

This is what I did 2x2-1x3

When I first looked at this question I thought of this:

20142014 x 20142014 - 20142013 x 20142015

let 2014 be 1

(11 x 11) - (10 x 12) = 121 - 120 = 1

I know this is strange, but for a situation like this it works! Where (a*a) - (a-1)(a+1).

I started with easier example like 4* 4= 16 , 3 5=15 6 6 =36 , 5*7 = 35 the difference each time is 1

simple matter, you do remove every 5 number.than last 1 number (4\times 4-3\times 5=1)={(4\times 4=16)-(3\times 5=15)=1}.....

= (20142013+1) (20142015-1) - (20142013* 20142015) = (20142013* 20142015)-20142013+ 20142015-1- (20142013* 20142015) =20142013+20142015-1 =1 Write a solution.

You could look at this the fast way and take the last digits of each of the numbers. You end up with

(4x4)-(5x3)

16-15

1

Simple :)

Write a solution.

= (2014 * 2014) - (2013 * 2015)

= 4,056,196 - 4056195

= 1

you see, we can see a pattern if we start from the whole numbers starting from 0. Like 1 times 3 is 3 and the square of 2 (the number in the middle) is one plus 3, equals to 4. Like 2 times 4 is 8, and the square of 3 (the number in the middle) is one plus 8, equals 9. 3x5 is 15 and square of 4 is 16. 4x6 is 24 and 5x5 is 25. 5x7=35, 6x6=36. . . . . . 99x101=9999, 100x100=10000

And you can see in our problem we are asked to substruct 20142013x20142015 from the square of the number in middle, 20142014! And if we take that our pattern will work, the answer would be 1. This is my simple solution. (Sorry if any linguistic mistakes have been done during the solution, I'm not a native.)

$20142014^2$ -- (20142014-1) (20142014+1)

$a^2$ -- $a^2$ +1

1 : ans

what I have done is I disregarded the first six digit (201420), then used the calculator. So, 14 14 - 13 15 = 1

20142014 x 20142014 - 20142013 x 20142015 = 20142014^2 - (20142014 - 1)(20142014 + 1) = 20142014^2 - (20142014^2 - 1) = 20142014^2 - 20142014^2 +1 = 1

let 20142014 x (x*x)-(x-1)(x+1) then x^2 - (x^2+x-x-1)=1