The methods 1-5 are available in most aptitude books.
Basic Assumptions:
1.You know how to caluclate squares ending with number 5
Many of us know how to caluclate square of numbers ending in 5
Ex: x5*x5=(x*x+1) 25
1. 45*45 = 4*(4+1) 25 = 2025
2. 75*75=7*8 25=5625
3. 85*85 = 8*9 25=7225
2.You know how to caluclate squares of numbers near to 100.
Ex1:
102 * 102 = 10404
Number By how much it is away from 100
102 +2 Line 1
102 +2 Line 2
------- -----
04(2*2) ==>last two digits
Take 102 from first line and add 2 from 2nd line(cross)
102+2=104 ==> First first 3 digits
final answer: 10404
Ex:2
112*112
Number By how much it is away from 100
112 +12 Line 1
112 +12 Line 2
------- -----
144(12*12) ==>last two digits
Take 112 from first line and add 12 from 2nd line(cross)
112+12=124 ==> First first 3 digits
final answer:12 (4+1)44 ==> 12544
3.You know how to caluclate squares of numbers from 31-50
47*47=??
- 47 = 50-3
- (-3)*(-3) = 09 ==> Last two digits
- for first two digits ==>[25 is used as standard]
- So use 25-3 = 22
- final answer is 2209
4.You know how to caluclate squares of numbers from 51-80
64*64 =??
- 64 = 50+14 (for 66 write as 50+16)
- 14*14=196 take 96 for last two digits
- First two digits ==> 25+14+1(this 1 is from 196) = 40
[25 is used as standard] - final answer 4096
5. You know how to caluclate squares of numbers from 81-100
88*88=??
88=100-12- Last two digits: (-12)*(-12)=144 ==>44(1 carried)
- First two digits:88-12+1=77
- final answer = 7744
The above methods 4 and 5 involves two steps.If you can remember some squares by heart then Instead of two steps we can acheive in one step.The above methods are easy,recommended.The methods below are for mnemonic guys.
So for academic purposes use the above methods (4,5)
1,2 methods are common and the only easy methods.
I've seen some aptitude books,finding squares,it's fine but i want to find out a hard way.
I was in my class way back in 2012 in mining lecture.I wanted to play with numbers .So after many caluclations i thought i found a pattern.But i didn't go beyond some huge numbers.Just confined to 3 digit caluclations.
This is a good method to find squares upto 100.You can extend upto 200 though.
for i in range(10,127):
print (str(i)+"*"+str(i)+"=",i*i)
print(((i*i)/100)%10)
print "difference=",(((i+1)*(i+1))/100 - (i*i)/100)
print("=======================================")
This piece of got bought to me to the above conclusions"""
Values to Memorise
base Number Square Squares you can find
4 17 289 [17 to 23] (base for these numbers is 4)
6 27 729 [27-33] (base for these numbers is 6)
8 37 1369 [37-43] (base for these numbers is 8)
10 47 2209 [47-53] (base for these numbers is 10)
12 57 3249 [57-63] (base for these numbers is 12)
14 67 4489 [67-73] (base for these numbers is 14)
16 77 5929 [77-83] (base for these numbers is 16)
18 87 7569 [87-93] (base for these numbers is 18)
"""
"""
45*45 =(4*5)(5*5) = 2025 i guess everyone know for numbers ending
Type 3 Example-1
Type 2 Example-1
31*31=?
base of 31 is 6
first two digits
Now I know 30*30=900;i will take first two digits 90
31 is one number away from 30;So i need to add 6 to first two digits;
90+6=96
last digit
31*31 = 1
final answer
961
Type 1 Example-1
28*28 =
first two digits
I know 27*27 = 729,it's base is 6;Now to get the answer i will take 72 from 729 and add 6 to it
72+6=78 ==> first two digits
last digit
8*8=4(last digit)
final answer
784
PS:Post incomplete...will update in time
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