Explain the Pattern - Part 2

Consider the perfect square,

$\begin{aligned} N_n & = (50+n)^2 & \small \color{#3D99F6} \text{where }n \in \mathbb N \\ & = 2500 + 100n + n^2 \\ & = (25+n)\times 100 + n^2 \\ \implies N_7 & = (25 + 7)\times 100 + 49 \\ & = \boxed{\color{#3D99F6}32\color{#D61F06}49} \end{aligned}$

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In response to Challenge Master's note:

The formula $N_n = 100(25+n) + n^2$ is arithmetically true for all $n$ . When $n = 10$ , then $25+n=\color{#3D99F6}35$ and $n^2=\color{#D61F06}100$ , which exceed two digits and $N_n = 3600$ and the pattern breaks.

When $n$ is negative, $|n|$ is in the reverse decreasing order. Then $\color{#3D99F6}25+n = 25-|n|$ follows the increasing order. But $\color{#D61F06} n^2 = |n|^2$ follows the decreasing order. The pattern breaks again when $n=-10$ .

$\large \begin{aligned} 51 \times 51 &= {\color{#3D99F6}26}{\color{#D61F06}01} \\ 52 \times 52 &= {\color{#3D99F6}27}{\color{#D61F06}04} \\ 53 \times 53 &= {\color{#3D99F6}28}{\color{#D61F06}09} \\ 54 \times 54 &= {\color{#3D99F6}29}{\color{#D61F06}16} \\ 55 \times 55 &= {\color{#3D99F6}30}{\color{#D61F06}25} \\ 56 \times 56 &= {\color{#3D99F6}31}{\color{#D61F06}36} \\ & \vdots \end{aligned}$

The pattern is:

• In $R.H.S,$ the two digits on the $\color{#3D99F6}\text{left}$ are consecutive positive integers.

• The two digits on the $\color{#D61F06}\text{right}$ are consecutive perfect squares.

In this way, $57 \times 57 = \boxed{\color{#3D99F6}32 \color{#D61F06}49}$

Using what has preceded .... (56+1)(56+1) = 56 x 56 +56 x 2 + 1 = 3136 + 113 = 3249

57×57 = 3.249 This pattern appears when we first square the number 50, which equals 2.500. As we keep incrementing the base by 1, we can see this pattern appear until the value of the base equals 60, in which case, 60 squared is 3.600, breaking the pattern (since 3.600 doesn't follow 59²'s 3.481) The reason as to why this pattern appears in the first llace is because of the nature of the number 5. Whenever we multiply a number times 5, we are essentially getting the half of 10 times that number: since 5×3=15 and 10×3=30. Thus when we double a number multiplied by 5, we get the original number times 10. This general rule is the reason for why this pattern appears. If we break up these problems' numbers' units and tens into the form (tens+unit)×(tens+unit), we can showcase what is happening by using the distributive property. For example, let's use 57×57 (or 57²) and break it into: (50+7)×(50×7), and perform the operation using the distributive property; the integers we'll get are: 2.500 + 350 + 350 + 49 = 2.500 + 700 + 49 = 3.200 + 49 = 3.249 What's evident here is the nature of 5 at play. When we multiply 50 times 7 twice, it will be the same as simply multiplying 100 times 7 just once, thus getting 700 + 2.500, leaving the last two digits of the number untainted for us to place our squared unit without it affecting the rest of our number. This happpens to the integer bases from 50 to 59. Since we'll start with 2.500, add the unit x 100 (incrementing the hundreds place by the value of that unit) , and then add the squarr of the unit. Making a pattern of increment in the first half and of exponentiation in the second as the base's value increments by 1. (I thought I had to write this as part of my answer, thus I wrote this on my phone without using images or anything.)

5n * 5n=(25+n)n*n

Simply square the first digit, add the tens digits to the product, multiply by 100, and add the square of the tens digit

Broken down:

Thus, 51 x 51 is 2601

Do the same with 52 x 52

Thus, 52 x 52 is 2704

Now, try 57 x 57

Therefore, the answer to this question is 3249

31+1=32, the next number increases by odd numbers, 3,5,7,9,1,13....

Square of same number is ab ab=(a a+b)(b*b)

5*5=25 25+1 = 26 1 * 1 = 1 —> 2601

25+2 = 27 2*2 = 4 —> 2704

25+3 =28 3*3 = 9 —> 2809

....

25+ 7 = 32 7*7 = 49 —> 3249

57*57=3249

(31+1)=32

36+13=49 (sum next prime Number)

It happens for two reasons, the process of squaring multi-digit numbers and how adding two 5s together equals 10, with a 0 in the ones place.

57 x57 = first multiply units place 7 ×7 =49 place 9 in the answer in units place and hold '4' .,then cross multiply 5×7 5×7 add both the answers(35+35) =70. Then add'4' to the 70 =70+4 =74 place 4 in the tens place in the answer and hold 7 .then multiply 5 ×5 =25 add 7 that gives 32 ..so total 3249

I look at it as being ultimately based on (A + B)^2 = A^2 + 2AB + B^2. It COULD be used indefinitely, but after you hit 60^2 it would become somewhat less user-friendly

5 * 5= 25, 25 +7=32 -->32

7*7=49 -->49

3249

(7 7)=49 (5 5)+7=32 3249

I can't explain why it appears, but i can tell you that it breaks at 59X59, because 9X9 is 108.

Then what's 50 squared?

Simple Trick , use (5*5)+Unit digit number- for first two digits & and square of unit digit for last two digits

Cpnsider number in series The difference between the numbers in red are odd numbers 3,5,7,9,11. So add 13 to 36 to get 49 57×57=3249

5 5=25 +7 = 32 ,7 7=49 Multiply first digit of both the numbers then add the corresponding number(second number) number which will give your hundreds and thousands place number then multiply the corresponding number the answer of which will be your units and tens place. E.g.: 5 5=25 +7 = 32 ,7 7=49 3249,eureka

4-1=3 9-4=5 16-9=7 25-16=9 36-35=11 incremento di 2 nelle ultime 2 cifre di 1 nelle prime 2

57×57

Solution : 5×5 》: 25 + 7 : 32 》: 7×7 : 49 Answer : 3249

57 X 57 = 64 X 50 (add and subtract 7) + 7 X 7, 64 X 50 = 3200 then add 7 X 7 , 3249 easy way to square any number !

57*57 is obviously 3249 if you include the pattern. You don't really have to do all of that crazy stuff! It is very simple

Find a value of k such that n-k=10p or n+k=10p where we consider n^2. In this case, n=57, we have k=3 so that n+k =60. Now, n^2 = (n+k)(n-k)+k^2. Thus, 57^2 = 60×54+9=10×6×6×9+9=10×36×(10-1)+9=10×324+9=3249.

57x57= 5x5+7=32(7x7)

Since we are square we can write it as (50+n)^2 = (50+n)(50+n) = 2500 + 100n + n^2. so 2500 + 700 + 49 = 3249. Since the number each time is 100n + n^2 to 2500, this means that each time the thousand + hundred value increases by 1 (because of the increase in the hundreds (if it is taken as one 2 digit number)) and the units and tens would be the next square number. This patter will stop at 60 * 60, where the thousands + hundred value will increase by 2 not 1 because the square of 10 is 100 which changes the hundreds value.

This is actually a question:does this apply to other no. Sets?

57 × 57 ; 5 × 5 + 7 = 32 ; 7 × 7 = 49 = 3249 !

I just looked at the first digit of each. Being 5, I multiplied them which makes 25. Since they have a second digit that isn't zero, I added the value of that single digit, then multiplied both of the second digits together while remembering that there are 4 total digits. So, 5x5 = 25, 25 + 1 = 26, 26 + (1 x 1) to the ones place = 2601. It's a bit like solving binomials in your head.

The pattern breaks at 60*60

Think of the first two blue digits as hundreds,since 55 x 55 = 3025, then 56 x 56 = ${\color{#3D99F6}31}$ 00 + ${\color{#D61F06}25}$ + 11 = ${\color{#3D99F6}31}$ ${\color{#D61F06}36}$ , so 66 x 66 = ${\color{#3D99F6}41}$ 00+ something The second two (red) digits have been increasing by adding the progression of ten odd numbers after 11 , that is ${\color{#D61F06}36}$ +13+15+17+19+21+23+25+ 27+29+31 = 256 + 4100 = 4356 ......66 x 66 = 4356 Q.E.D. Try 67 x 67; 4200 + 256 + 33 = 4489 and 67^2 = 4489

Going with the original sequence.... ${\color{#D61F06}01}$ + 3 = ${\color{#D61F06}04}$ + 5 = ${\color{#D61F06}09}$ + 7 = ${\color{#D61F06}16}$ + 9 = ${\color{#D61F06}25}$ + 11 = ${\color{#D61F06}36}$ .. and so on

Of course.when it reaches 60^2. the "pattern" breaks because 3600 is (3500 + 10^2), so it skips the 35 XX entry. If we go downwards from 49^2, 48^2, 47^2....we get 24 01 , 23 04 , 22 09 , 21 16 , 20 25 , 19 36 , 18 49 , 17 64 , 16 81 ,16 00 (1500+ 100 ), 1521 = 1400 + 121

If we use that arithmetic rule, when it gets down to 24^2 we are in partial negative number, i.e., 576 = -100 + 676,( 676 is the sum of all odd positive integers from 1 to 49).

The following is the pattern. ;-

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✓ 46*46 = 21 16

✓ 47*47 = 22 09

✓ 48*48 = 23 04

✓ 49*49 = 24 01

✓ 50*50 = 25 00

✓ 51*51 = 26 01

✓ 52*52 = 27 04

✓ 53*53 = 28 09

✓ 54*54 = 29 16

✓ 55*55 = 30 25

✓ 56*56 = 31 36

✓ 57*57 = 32 49

✓ 58*58 = 33 64

✓ 59*59 = 34 81

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To solve this question you should calculate , go with the logic, or go through the logic. If you followed any other metjod pls write in comments

Any doubt can be asked Please share your metho in comments

(50+h)^2 =50^2 +2 50 h +h^2

= 50^2 + 100h+ h^2

= 2500+ 100h+ h^2

= (25+h)100 + h^ 2

Thus the pattern 's logic is studied in a glance

We could now find the ways to solve it

Here we can solve in 2 ways

✓ On by seeing the pattern. Which would just be a prediction . We dont know if it is a deliberate one or omtenrional one. Even he deliberate ones have logics

    (Or)

✓. (50+7)^2 = 2500+700+ 49. = 3249

Multiply the tens column togethet =5x5=25 then add the units digit eg 56x56 = 5x5 +6=31 (blue digits) then square the units column, as in this example 6x6=36 (red digits). Hence answer to 56x56=3136. I think the pattern falls down when you nit 60x60.

To explain why you have to put the number as a sum of fifty plus n. The formula of the square: $(50+n)^2 = 50^2 + 2*50*n + n^2$ . This gives $2500 + 100*n + n^2$ You can see the hundreds rise as $n$ rises. You can see the two first digits correspond to $n^2$ .

It is always the previous number Plus the root of that number plus the next number. Like 3136+56+57.

The difference between two consecutive perfect squares is always two greater than the difference between the set of two perfect squares directly previous. If $n \times n = x_n$ then $(n + 1)(n + 1) = n^2 + n + n + 1 = n^2 + 2n + 1 = y_n$ . $2n + 1$ is the difference. $2(n-1) + 1 = 2n + 1 - 2$ proves my statement. The answer is 3249.

The numbers on the right are perfect squares (in order) and the numbers on the left are simply ordered integers starting at 26. The color separation was not necessary to solve this problem, but it would have been much harder to find a solution with ill - defined groupings.

All I did was take the last number from the current and last multiplicands (56 & 57 -> 6 & 7) and added them together (6+7=13) then I took the previous result into two parts 31 & 36. Each new step you just add one to the frist part and add the sum you calculated to the second part and then you just reattach the numbers. So 31+1=32 and 36+13=49; 32 & 49 => 3249. Not a ‘real’ mathematical solution but a solution nonetheless. :)

All squares are equal to the previous square plus the next consecutive odd number after the difference from the previous squares 1x1=1 2x2=1+3=4 3x3=4+5=9 4x4=9+7=16, etc.

The values in red increase by consecutive odd numbers starting from 03 i.e. 01+03=04, 04+05=09, 09+07=16, etc.

The numbers in blue increase by a factor of 01.