Is there a simple formula for this?

Question

Is there a simple formula that works to square numbers up to 25 besides multiplying it by itself and the distributive property? Preferably something quick, easy and that you can do in your head.

Thanks!

Answer 1

You can add successive odd numbers:

1: 1 = 1

2: 1 + 3 = 4

3: 1 + 3 + 5 = 9

4: 1 + 3 + 5 + 7 = 16

5: 1 + 3 + 5 + 7 + 9 = 25

and so forth:

The sum of the first n odd number is n^2.

Answer 2

lols. i just remember it somehow O.o

1 - 1

2 - 4

3 - 9

4 - 16

5 - 25

6 - 36

7 - 49

8 - 64

9 - 81

10 - 100

11 - 121

12 - 144

13 - 169

14 - 196

15 - 225

16 - 256

17 - 289

18 - 324

19 - 361

20 - 400

21 - 441

22 - 484

23 - 529

24 - 576

25 - 625

(i dont think this really answers your question. and btw i didnt use a calculator pretty good huh? ^^ )

Answer 3

11*11 = (10*12) + 1*1 = 121

12*12 = (10*14) + 2*2 = 144

13*13 = (10*16) + 3*3 = 169

14*14 = (10*18) + 4*4 = 180 + 16 = 196

15*15 = (10*20) + 5*5 = 225

16*16 = (10*22) + 6*6 = 220 + 36 = 256

17*17 = (20*14) + 3*3 = 289

18*18 = (20*16) + 2*2 = 324

19*19 = (20*18) + 1*1 = 361

20*20 = 2*2*100 = 400

21*21 = (20*22) + 1*1 = 441

22*22 = (20*24) + 2*2 = 482

23*23 = (20*26) + 3*3 = 529

24*24 = (20*28) + 4*4 = 560 + 16 = 576

25*25 = (20*30) + 5*5 = 625

the formula would be:

Lets say the number you want to square it 'n'

And the base is either 10 or 20... we shall call it 'b'

n*n = (n-(b-n))(n+(b-n)) + (b-n)(b-n)

But (b-n) is the difference between the number and the base so we could say (b-n)=d

n*n = (n-d)(n+d) + (d*d)

Answer 4

I doubt it. Try memorizing, it's only 25 things. And you should already know at least a few of them.

And really how much easier can a formula get? multiplying two two digit numbers (I will skip the single ones since you SHOULD know all those) isn't exactly a mind numbing task.