The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = n² Use the diagram to show that any odd number is the difference of two squares.
How many zeros are there at the end of the number which is the
product of first hundred positive integers?
Is it true that $99^n$ has 2n digits and $999^n$ has 3n digits? Investigate!
Oops, there was a typographic error in the formula at the start
of the month. It is right now. Thank you to Andrei Lazanu of school
205 Bucharest for what follows.
First, I tried to solve the problem in the "conventional
mathematical way", and I listed the powers of 1, 2, 3, 4, 9 from 0
to 9, looking in each case to eliminate those greater than 5 digit
numbers, attributed to both $r^n$ and $c^h$.
Because it seemed very long, I tried the problem from a
"calculatoristic" point of view:
First I observed that N must be greater than 0, and each other
number must have the values from 0 to 9. I wrote a program in
MATLAB to calculate the numbers.
The idea of the program is to test for all digits the values
from 0 (1) to 9. For this I wrote a cycle allowing $n$ to have
values between 1 and 9.For each value of $n$, I give an inside
cycle, to $r$ values from 0 to 9. So, this works as follows: $n$
has the value 1, and for this $r$ has in turn values from 0 to 9.
This is the idea of a cycle inside another one. Each cycle ends
When I arrived at this cycle, I test if the condition of the
problem: $r^n - i + c^h = 1000\times r+100\times i+10\times c+h$is
fulfilled. If yes, the number $1000\times r+100\times i+10\times
c+h$is written on the screen.
Because this way I'll find all solutions, even with repeated
digits, I must in a further step eliminate them.
Here is the program:
if $r^n - i + c^h - 10000 n - 1000 \times r - 100 \times i - 10
\times c - h == 0$
a =[n r i c h]
The results given were the following:
a = 5 9 0 5 0
a = 5 9 2 6 3
As in the first one the digits are not distinct, this is not a
The second is the unique solution of the problem:
n = 5
r = 9
i = 2
c = 6
h = 3