@rkshtsurana Hmm... thats was very basic .... And if we want to find the number at ten's place ,then what should be the approach (step by step )?
@busar005 @fedbite @krum
first the concept
|x-a| + |x-b| + |x-c| has minimum value at x = b ( median value)
so our ques
|x-1| + |2x-1| + |3x-1| .........|119x-1|
rewiting as
|x-1| + |x - 1/2| + |x- 1/2| + |x - 1/3| + |x - 1/3| + |x - 1/3| + ......| x - 1/119|
total terms = 119*120/2 = 7140
middle term = 3571 / 3570
so, median term will be n(n+1)/2 = 3570
n = 84
so it will be min at x = 1/84
|1/84 -1| + | 2/84-1| + ......| 84/84-1| + |85/84 - 1| + .......| 119/84 -1|
83/84 + 82/84 + ....0 + 1/84 + 2/84 ......35/84
=49
@Boson said:@rkshtsurana Hmm... thats was very basic .... And if we want to find the number at ten's place ,then what should be the approach (step by step )?
There is no general set method, you have to adapt as per question.
First is to learn few basics:
Know Eulers Toitent function.
and Fermat and wilson theroam :)
@Boson said:@Brooklyn Mera Sawaal bilkul general hai yaani kisi bhi place ke liye ...Abhi ke liye maan lo units place par..
@Boson said:Can somebody help me with Q's like something raise to paower something has which digit in some given place ............ for eg 3^2001 has which digit at ones place ........Please give a detailed step by step approach ..... @krum@Brooklyn@chandrakant.k@anytomdickandhary
Method 1: Cyclicity as @rkshtsurana explained
Method 2:
when you have to find out unit digit, divide the number by 10 and the remainder.
when you have to find out unit digit, divide the number by 100 and the remainder.
(holds good when problem is complex)
using this method, it becomes quite easy.
3^2001 mod 10
=3^1 mod 10 (using Eular)
=3 mod 10
A two-digit number x, when increased by x% becomes a perfect square 'y'. Find y.
@rkshtsurana : Aate hi teer chhod diye apne :splat:
Abhi naya naya thread start hua h, bachhon ko darao mat 😛 :D
@Boson said:@Brooklyn Mera Sawaal bilkul general hai yaani kisi bhi place ke liye ...Abhi ke liye maan lo units place par..
for tens digit use mod 100 😃
similarly for higher.
Of course there are other methods. But Eulers is easiest to grasp when starting of with remainder
@invincible2910 said:hi all!please post an explanation for this one-the constt term in the expansion of ( x^2 + y +1/y +1/x^2)^8 is4900;5000;5100;8000
8c4(x^2+y)^4*(1/y+1/x^2)^4
=>8c4*(4c0x^8+4c1x^6y+4c2x^4y^2+4c3x^2y^3+4c4y)
(4c01/x^8+4c1*1/x^6*1/y+4c2*1/x^4*1/y^2+4c3*1/x^2*1/y^3+4c4*1/y)
const term
=>8c4(4c0^2+4c1^2+4c2^2+4c3^2+4c4^2)
=>8*7*6*5/24*(1+16+36+16+1)
=>4900
=>8c4*(4c0x^8+4c1x^6y+4c2x^4y^2+4c3x^2y^3+4c4y)
(4c01/x^8+4c1*1/x^6*1/y+4c2*1/x^4*1/y^2+4c3*1/x^2*1/y^3+4c4*1/y)
const term
=>8c4(4c0^2+4c1^2+4c2^2+4c3^2+4c4^2)
=>8*7*6*5/24*(1+16+36+16+1)
=>4900
@vijay_chandola said:A two-digit number x, when increased by x% becomes a perfect square 'y'. Find y.@rkshtsurana : Aate hi teer chhod diye apne Abhi naya naya thread start hua h, bachhon ko darao mat
80 + 0.8*(80) = 144
y = 144 😃 ..
Set S = { a1 , a2 , a3.....} where elements of set S are only the combination of powers of 2 and 3 like a1 = 2 a2 = 3 a3= 4 a4 = 6 a5 = 9 a6 = 12 and so on..
Find 1/a1 + 1/a2 + 1/a3 ............
a) 1
b) 2
c) 3
d) 4
@vijay_chandola said:A two-digit number x, when increased by x% becomes a perfect square 'y'. Find y.@rkshtsurana : Aate hi teer chhod diye apne Abhi naya naya thread start hua h, bachhon ko darao mat
x + x^2/100 = y
x^2 + 100x = 100y
x^2 + 100x + 2500 = 100y + 2500
(x+50)^2 = 100(y +25)
x = 10 * rt( y + 25) - 50
now since x is an integer
5 12 13 is a pytagorus triplet
y = 144
x = 10*13 - 50 = 80
bhai darana jaruri he..taaki pata chale ki aagge ka way is not easy
@Brooklyn said:ye to strike kaise kiya be???
Maximum two digit number = 99
99 + 0.99*(99) = approx 198..
-> Consider the perfect squares below that.. 13^2 = 169 , 12^2 = 144 and so on..
Bas dikh gaya ki 144 = 80 + 64 .
@rkshtsurana said:Set S = { a1 , a2 , a3.....} where elements of set S are only the combination of powers of 2 and 3 like a1 = 2 a2 = 3 a3= 4 a4 = 6 a5 = 9 a6 = 12 and so on..Find 1/a1 + 1/a2 + 1/a3 ............a) 1b) 2c) 3d) 4
1/2+1/2^2+1/2^3+....=1
1/3+1/3^2+1/3^3+....=1/2
1/3(1/2+1/2^2+1/2^3....)+1/3^2(1/2+1/2^2+1/2^3..)....=1/2*1=1/2
1+1/2+1/2=2
1/3+1/3^2+1/3^3+....=1/2
1/3(1/2+1/2^2+1/2^3....)+1/3^2(1/2+1/2^2+1/2^3..)....=1/2*1=1/2
1+1/2+1/2=2
@rkshtsurana said:Set S = { a1 , a2 , a3.....} where elements of set S are only the combination of powers of 2 and 3 like a1 = 2 a2 = 3 a3= 4 a4 = 6 a5 = 9 a6 = 12 and so on..Find 1/a1 + 1/a2 + 1/a3 ............a) 1b) 2c) 3d) 4
2?
@rkshtsurana said:Set S = { a1 , a2 , a3.....} where elements of set S are only the combination of powers of 2 and 3 like a1 = 2 a2 = 3 a3= 4 a4 = 6 a5 = 9 a6 = 12 and so on..Find 1/a1 + 1/a2 + 1/a3 ............a) 1b) 2c) 3d) 4
1/2+1/4+....+1/3+1/9+.....+1/6+1/12+1/18+.....
1/2/1/2+1/3/2/3+1/6(1+1/2+1/3+..)~~2
ans. 2
@Brooklyn said:ye to strike kaise kiya be???
x*(x+100)/100= perfect square
means 100 se divisible hoga
so check only multiples of 10
80
means 100 se divisible hoga
so check only multiples of 10
80
@krum said:1/2+1/2^2+1/2^3+....=11/3+1/3^2+1/3^3+....=1/22 hai kya?
s1 = 1/2 + 1/2^2 + 1/ 2^3 ........ = 1
s2 = 1/3 + 1/3^2 .......= 1/2
combination = ( 1+ s1)(1+ s2) - 1
2 * 3/2 - 1 = 2