for i in range(25,100,10):
print(str(i)+"*"+str(i)+"=",i*i)
for j in range(10,80,10):
print(str(i)+"*"+str(i+j)+"=",i*(i+j))
print""
"""
Results
There is a sequence observe the first digits they are (6,8,11,13,16,18...)(2,3,2,3,2,3...)(6+2,8+3,11+2,13+3...)
I know 25*25 = 625 (25*25=(2*3)25)==>625
To get 25*35 all i need to do do is add 2 to 6 = 8
and change 25 to 75
so final answer would be 875
('25*25=', 625)
('25*35=', 875)
('25*45=', 1125)
('25*55=', 1375)
('25*65=', 1625)
('25*75=', 1875)
('25*85=', 2125)
('25*95=', 2375)
There is a sequence (12,12+3,15+4,19+3,22+4,26+3).
Addition of (3,4,3,4,3,4)...
('35*35=', 1225)
('35*45=', 1575)
('35*55=', 1925)
('35*65=', 2275)
('35*75=', 2625)
('35*85=', 2975)
('35*95=', 3325)
('35*105=', 3675)
Addition of (4,5,4,5,4,5)...
('45*45=', 2025)
('45*55=', 2475)
('45*65=', 2925)
('45*75=', 3375)
('45*85=', 3825)
('45*95=', 4275)
('45*105=', 4725)
('45*115=', 5175)
Addition of (5,6,5,6,5,6)
('55*55=', 3025)
('55*65=', 3575)
('55*75=', 4125)
('55*85=', 4675)
('55*95=', 5225)
('55*105=', 5775)
('55*115=', 6325)
('55*125=', 6875)
('65*65=', 4225)
('65*75=', 4875)
('65*85=', 5525)
('65*95=', 6175)
('65*105=', 6825)
('65*115=', 7475)
('65*125=', 8125)
('65*135=', 8775)
('75*75=', 5625)
('75*85=', 6375)
('75*95=', 7125)
('75*105=', 7875)
('75*115=', 8625)
('75*125=', 9375)
('75*135=', 10125)
('75*145=', 10875)
('85*85=', 7225)
('85*95=', 8075)
('85*105=', 8925)
('85*115=', 9775)
('85*125=', 10625)
('85*135=', 11475)
('85*145=', 12325)
('85*155=', 13175)
('95*95=', 9025)
('95*105=', 9975)
('95*115=', 10925)
('95*125=', 11875)
('95*135=', 12825)
('95*145=', 13775)
('95*155=', 14725)
('95*165
Learn python for fun.The popular blog with questions and answers to the python.Solutions to facebookhackercup,codejam,codechef.The fun way to learn python with me.Building some cool apps.
Showing posts with label finding patterns. Show all posts
Showing posts with label finding patterns. Show all posts
Monday, March 30, 2015
Finding Patterns Multiplication of 2 digit number part 2
Sunday, March 29, 2015
Finding Patterns two digit multiplication part 1
21 * 21 step 1 20 * 20 =400 step 2 20*1 1*20 ----- 40 Step 3 1*1 step 4 441 Base Method base is 2 21 *21 = 2*(21+1)|(1*1) =44|1=441 21*22 =2*(21+2)|2=462 21*23=2*(24)6 = 486 21*24=2*(25)|4=504 visualise 21*26 considering base 20 21 1(1 away from base 20) 26 6(6 away from base 20) ---------- 21+6|1*6 21+6 or 26+1 followed by 1*6 27|6 27*2|6 546 So within 20 or 30 we can do these caluclations easily Base 30 33*34 37*3|12 111|12 1122 or 33*34 I know 35*35=1225 (35-2)*(35-1) 1227-105 1122Learn python for fun.The popular blog with questions and answers to the python.Solutions to facebookhackercup,codejam,codechef.The fun way to learn python with me.Building some cool apps.
Finding Patterns Multiplication by 19 the harder way
for i in range(1,30):
print (str(19)+"*"+str(i)+"=",19*i)
print "difference=",((19*(i+1))/10 - (19*i)/10)
print("=======================================")
"""
Multiplication by 19 ==> The harder way
Many would suggest ==> 1| 3|5| 7|9|11| 13| 15| 17|19 ==> odd numbers(1,3,5..)
9| 8|7| 6|5|4 | 3 | 2 | 1 | 0 ==> numbers from 9,8,7....
------------------------------
19,38,57,76.......
It's easy but you should write it down
This is for kids ==> Rather than by heart just know that
19*n=(2*n)-1 ==> This will be first digit
From the observations
19*36=??
Once you know 19 table it's not a big deal but sometime for starters it's a mess to remember numbers.
I assume you know 2 table;i assume you know compliment of a number;
for compliment pairs ==> (1,9),(2,8),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2),(9,1)
Ex : 19* x = (2x-1)[compliment of x]
19*36==> (2*3 - 1)7 [compliment of 3 is 7] and (2*6 -1)4 [compliment of 6 is 4]==> 57 114 ==> 11 carry over ==> 57+11 =68==>684
"""
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Wednesday, March 25, 2015
Patterns-Finding squares in the hard way
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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