Math Is Fun Forum

Lets take the length of sides Mathsy has given.
The Hero's formula for area of a triangle is
Area = √(s(s-a)(s-b)(s-c)
where a, b, c are the sides and s = (a+b+c)/2
Therefore, Area = √[(11/2)(11/2 - 2)(11/2 - 3) (11/2 - 6)]
11/2 - 6 is a negative number,
So the Area of the triangle is an imaginary number

Thanks, MathsisFun.
I'll continue from where you left.
a(a+1) is certainly even.
b(b+2) is odd if b is odd.
If b is even, b(b+2) is even, and is always a multiple of 4.
a(a+1) is a multiple of 4 only if a is multiple of 4 or it is a number of the form 4n+3.
Lets assume a is a number of the form 4n+3.
a(a+1) = (4n+3)(4n+4) = 16n² + 16n + 12n + 12 = 16n² + 28n + 12
= 4(4n²+7n+3)
Let this number be equal to b(b+2)
b(b+2) = 4(4n²+7n+3)
b² + 2b - 4(4n²+7n+3) = 0
b = [-2 ± √{4 +16(4n²+7n+3)}]/4
Let n=0, we get an irrational number as b.
Let n=1, we get an irrational number as b.
Let n=2, again, we get an irrational number as b.

Can {4 +16(4n²+7n+3)} never be a perfect square?

Is it because the last digit of {4 +16(4n²+7n+3)} is 2 or 8 and not 4 or 6?

The Triangle inequality is
AB + BC > AC
AC + BC > AB
AB + AC > BC

You have given BC=4 and AC = 8 - AB
Therefore, the required inequality is
AB < BC + AC
AB < BC + 8 - AB
2 AB < BC + 8
AB < (BC + 8)/2

Problems #n+6
There are 5 boys and 5 girls and they are made to sit in a row such that no two boys and no two girls sit next to each other. How many arrangements are possible?

I thought of it in a different way. I used the summation of a Geometric Progression and this is the result I got:-
0.9999999999999999999999999999.....................
= 0.9(1 + 1/10 + 1/100 + 1/1000 + 1/10,000 + 1/100,000 + 1/1,000,000.......)
The ∑ of the series within the brackets would be 1/(1-1/10) = 1/(9/10)
= 10/9
10/9 x (0.9) = 9/9 = 1
I know this is an approximation for an infinite series.
But 0.999999999999999999999999999999999.................... is a devilish number. You never encounter this number in any mathematical problem. I have never come across recurring 9s after the decimal in any division. It can be proved that there can be no recurring 9s in any division.
Let me prove this by a counter-solution.
Let there be a recurring decimal 0.2999999999999999999.......
When we try to convert this into a fraction, this is what happens.
x = 0.2999999999999999999............
10x = 2.999999999999999999999......
100x = 29.9999999999999999999999......
90x = 27
x = 27/90 = 3/10, which is 0.3 !
Therefore, I think 0.9999999999999..........exists only theoretically. Only in the imagination.

I get a different answer.
There are 9 numbers ending with zero up to 99. 10,20,30,40,50,60,70,80,90.
100 has two zeros. Then, in each set of ten consecutive numbers (1-10, 11-20, 21-30 etc.),
one ends in 2 and one ends in 5, which gives us another 0. There are ten tens in a hundred.
Therefore, I thought 100! would end with 21 consecuitive zeros.
But when I calculated 100! using the full precision calculator on this website, I got
93,326,215,443,944,152,681,699,238,856,
266,700,490,715,968,264,381,621,468,592,
963,895,217,599,993,229,915,608,941,463,
976,156,518,286,253,697,920,827,223,758,
251,185,210,916,864,000,000,000,000,000,
000,000,000
That is 24 zeros! I wonder where those three extra zeros came from
Okay, one additional zero 40, 50 since 40x50 = 2000, the other by using 4 (instead of 2)for 25 and 8 for 75?

Did I miss something, Mathsy?

You have shaken the foundations of Mathematics.
If someone asxked me to do it,
I'd simply say take cube roots on both sides.
If the cube roots are equal, the numbers are equal

Problme #n+5
How many zeros would 1,000! end in?

4n-1 = a² + b²
4n    = a² + b² + 1
  n    = (a² + b² + 1)/4
Since 4n-1 is an odd number, one of a and b is odd and the other is even.
When an even number is squared, the resultant is always a multiple of 4.
An odd number when squared is always of the form 2n+1.
Any number of the form 2n+2 would never be divisible by 4 when n is even! When n is odd, it is of the form(2n+1). When this is squared, we get
4n² +4n+1. i.e. 4(n² +n)+1. When 1 is added to this number, we get
4(n² +n)+2. This number is NOT divisible by 4.
Therefore,
a² + b² ≠ 4n-1
q.e.d

Problem # n+4
(1) How many diagonals would a polygon of 'n' sides contain?
(2) What would be the sum of the internal angles of a polygon of 'n' sides?

Well done, tt! You got it right!

Here are the squares and square roots of all squares up to 31.
1² = 1
2² = 4
3² = 9
4² = 16
5² = 25
6² = 36
7² = 49
8² = 64
9² = 81
10² = 100
11² = 121
12² = 144
13² = 169
14² = 256
15² = 225
16² = 256
17² = 289
18² = 324
19² = 361
20² = 400
21² = 441
22² = 484
23² = 529
24² = 576
25² = 625
26² = 676
27² = 729
28² = 784
29² = 841
30² = 900
31² = 961

Remember,
(1) The last digit of a perfect square is always 0, 1, 5, or 6.
(2) Finding square roots is easier by prime factorisation.
eg. 784 = 2 x 2 x 2 x 7 x 7 x 7
Therefore, the square root of 784 is 2 x 2 x 7, i.e. 28