prove that sqrt(20 ones - 10 twos) = 10 threes?

Answer 1

See the inductive process here :
11 -- 2 = 9 and sqrt(9) = 3
1111 -- 22 = 1089 and sqrt(1089) = 33
111111 -- 222 = 110889 and sqrt110889 = 333 and so on up to
20 ones -- ten twos
= 11111111108888888889 and
sqrt(11111111108888888889)
= 3333333333
=> sqrt(20ones -- 10twos) = 10threes.

Answer 2

The 20 ones, 10 ones, and 10 threes can be expressed as a finite geometric series
sum = (a -a*r^n)/(1-r)

20 ones = (1-1*10^20)/(1-10) =(10^20 -1)/9

10 twos = (2-2*10^10)/(1-10) =(2*10^10- 2)/9

10 threes = (3-3*10^10)/(1-10) =(3*10^10 -3)/9

sqrt(20 ones - 10 twos) = 10 threes
20 ones - 10 twos = (10 threes)^2
(10^20 -1)/9 -(2*10^10- 2)/9 = ((3*10^10 -3)/9 )^2

combine the first two parts on the left side
(10^20 -2*10^10 +1)/9 = ((3*10^10 -3)/9 )^2

Divide the top and bottom of the right side by 3
(10^20 -2*10^10 +1)/9 = ((10^10 -1)/3 )^2

expand the square on the right side
(10^20 -2*10^10 +1)/9 = (10^20 -2*10^10 +1)/9

multiply both sides by 9
(10^20 -2*10^10 +1) = (10^20 -2*10^10 +1)

The left and right side are equal

Answer 3

Let's say, Y = 10^9 + 10^8 + ... + 10^0 (10 ones)

So

20 ones
= 10^19 + .. +10^10 + 10^9 + ... + 10^0
= 10^10 * (10^9 + 10^8 + ... + 10^0) + (10^9 + 10^8 + ... + 10^0)
= 10^10 * Y + Y

10 twos
= 2 * 10^9 + 2 * 10^8 + ... + 2 * 10^0
= 2 * (10^9 + 10^8 + ... + 10^0)
= 2Y

and,

10^10 (10000000000 = 9999999999+1)
= 9*10^9 + 9*10^8 + ... + 9 * 10^0 + 1
= 9*(10^9 + ... +10^0)+1
= 9Y+1

Therefore,

sqrt(20 ones - 10 twos)
= sqrt(10^10 * Y + Y - 2Y)
= sqrt((9Y+1) Y - Y)
= sqrt(9Y^2 + Y - Y)
= sqrt(9Y^2)
= 3Y
= 3 * (10^9 + 10^8 + ... + 10^0)
= 3333333333 ( 10 threes )

Answer 4

sqrt(11111111111111111111 - 2222222222)=
sqrt(11111111108888888889) = 3333333333

Answer 5

Please mouse over the dots to see complete answer.

sqrt(11111111111111111111-2222222222)=3333333333