Can you answer these 4 hard questions (It's for math homework)?

Question

Ok, I have a couple of questions i do not know how to solve and i've tried for like forever....i'll give each question and if needed will put extra information,
1. Here is one way to use six 9's to obtain 100
9+9+9x9+*9/9
* in fraction form
Now find another way to use six 9's to obtain 100,
such answers can have 9 unsquared and other things as long as its all 9's

The next set of questions is basically guess the number,
A. The number is the square of it's units digits
B. The number exceeds its reversal by 20%
C. The number and its reversal add to a perfect square...
Any help would be greatley appreciated

Answer 1

Nice set of Diophantine questions!
There were some errors in my previous post.
Here's the corrected version.
For no. 1, I came up with 99 + 9/9 +(9-9).
For the next set let's work them out mathematically!
I will assume all the wanted numbers are positive.
If the number has 1 digit, 1 is the only solution.
Suppose our number has 2 digits, i.e., it is 10x + y.
Then 10x + y = y².
So y² - y -10x = 0,
Now use the quadratic formula.
The discriminant must be a perfect square to
have integer roots.
So 1 + 40x = t² (*)
The smallest solution occurs for x = 2, t = 9
So y² -y -20 = 0.
(y-5)(y+4)= 0.
Since y is positive, we get y = 5.
Thus one such number is 25.
Another solution of (*) is x = 3, t = 11.
So our quadratic is now
y² -y -30 = 0.
(y-6)(y+5) = 0
Since y is positive, we get y = 6.
So another answer is 36.
The final solution is x = 9, t = 19.
Now we get y² - y -90 = 0
and y = 10, but this does not give a solution.
Since no number bigger than 99 can be
the square of a single digit(the largest
such square is 81) we are done!
The answers to A. are 1, 25 and 36.

B). Let's find a number that exceeds
its reversal by 20%.
No one digit number works, so
again, let's assume that the number is a 2 digit number
of the form 10x+ y. Here x is any number between
1 and 9 and y any
number between 0 and 9.
Then
10x + y = (10y +x) + 0.2(10y+x)= 1.2(10y + x)
8.8x - 11y = 0.
.8x-y = 0
4x - 5y = 0.
Now just try all the possible values of y.
Since y must be a multiple of 4, these are 0, 4 and 8.
Now 0 is not a solution, since then x = 0 also.
If y = 4, x = 5, so 54 is a solution.
Note that 45 + .2(45) = 45 + 9 = 54.
If y = 8 x = 10.
But this doesn't yield a solution.
So 54 is an answer to B).
I haven't checked whether any number
with more than 2 digits can be a solution.

C). The only 1 digit numbers that work and 2 and 8.
So let's look for numbers of 2 digits that satisfy
the conditions of the problem.
Again, let's let our number be 10x + y, with
the same conditions as in B).
Now
10x + y + 10y + x = z²
11x + 11y = z².
Now z must be a multiple of 11, so put z = 11t
and cancel. Then
x + y = 11t².
Suppose t = 1.
Then we get the solutions 29, 38, 47, 56, 65, 74, 83 and 92, all adding to 121. Note that
t can't be larger than 1 because then x
or y would exceed 9.
Now let's get the answers for 3 digit numbers.
Suppose our 3 digit number is 100x + 10y + z.
Its reversal is 100z + 10y + x,
and we must have 1 <= x <=9, 0 <= y <= 9, 0 <= z <= 9.
Thus we must have
101x + 20y + 101z = t².
I wrote a PARI program to get the solutions
for this equation and found 19 solutions:
110, 143, 164, 198, 242,
263, 297, 341, 362, 396,
440, 461, 495, 560, 594,
693, 792, 891, 990.
The only squares obtained were
121, 484, 625 and 1089.
I haven't had time to pursue this for
numbers of more than 3 digits.

I hope that helps!!