Math Is Fun Forum

Outstanding! You are really supersmart!
Try this one....But don't post your reply immediately.
Let others too try.

(2) A mixture of 40 liters of milk and water contains 10% water. How much water must be added to make water 20% in the new mixture?

Thanks, Jenilia. Visit the 'Games and Puzzles' topic regulalry. I shall post mathematical problems and their solutions which could be of some help to you, for the Olympiad.:)

Excellent! You are correct! Tell us how you did it!

23+18+12+22 is already greater than 30! So, it should be 'people' as pointed out by Mathsy!:)

1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + .............1/ ∞ = 2
I cannot think beyond that.

First, the formula or equation is checked for values like 1,2,3 etc.Only if this test is pased, we asume it is true for an arbitrary number k. Thereafter, if it is proved true for k+1, it becomes a 'proof beyond doubt'.

Mathematical Induction is a brilliant way of proving.
For example, the sum of the first n Natural Numbers is
n(n+1)/2.
For n=1, we get 1.
For n=2, we get 3.
Therefore, the summation formula works for 1 and 2.

Let's assume it is true for an arbitrary Natural Number k.
Therefore, the sum of the first k Natural Numbers would be k(k+1)/2.

For k+1, the sum would be k(k+1)/2 + (k+1),
that is [k(k+1) + 2(k+1)]/2, or [(k+1)(k+2)]/2

And, according to the formula we had at the beginning of the proof,
the sum of the first k+1 Natural Numbers would be [(k+1)(k+2)]/2

Since the two are equal, this can be said to be true for sum up to any natural number.

Similarly, it can be proved that:-
the sum 1² + 2² + 3² + 4² +... n²  = [n(n+1)(2n+1)]/6
and also
the sum 1³ + 2³  + 3³ + 4 ³  + ......... n³  = [n(n+1)/2]²

Hannah,
Mathematics is fun when you try to understand the subject.
Trying to memorize the tables and the formulae may be difficult for some.
Yet, it is easier to get a good score in Mathematics than any other subject.
For example, When you forget what 7x7 is,
try to remember what 7x6 is, then add 7 to the result.
You can get doubts clarified here.

Maybe, I should have waited for some more days
before giving the solution.

I am unable to find any serious flaw, yet,
I am not fully convinced.
I shall try with a much simpler number, viz. y76.
y76 is (y*10^2)+(7*10^1)+(6*10^0).
When this number is squared,
[This is of the form (a+b+c)²
which is equal to a² + b²  + c² + 2ab + 2bc + 2ac]
we get
10000y²  + 4900 + 36 + 14000y + 840 + 1200y
which is the same as
10000y² + 15200y + 5776
From this, it is clear, for any y belonging to Natural Numbers,
the last two digits are not affected. They continue to be 76.
But, does this conclusively prove there exists a value y such that
the last three digits of y76² are y76?

Mathsy, I am confused.