17 September
2013
Puzzle time
Returning home from New York. Ergo puzzle time. Today’s is a simple yet challenging problem.
Think of the various pairs of two numbers you can make using the digits 1,2,3,4,5,6,7,8,9 (each used only once) e.g. (2345, 16789) or (743, 821569) and so on.
Which is the pair (of two numbers) has the maximum product value (compared to other pairs)?
Ugh. Word puzzles please
No limits for the number of pairs? It can be too many combinations!
Yes. But which combination gives the highest product value (of those two numbers)
I see… Lemme think…
An intuitive guess..I have to chose between 9876 and 54321 and 98765 and 4321..Or may be 987654 and 321 too.
You would be wrong in all those and also no posting answers in Comments section
94321 and 8765
Rajibda, I have seen that some of your friends don’t follow your instruction about giving answers and i thought they were not from bengali medium schools
Baishali .No… we have posted the wrong answers and misled others
According to Rajibda’s rule no answer (right/wrong) should be in comment section.
students can be naughty..
The answer to the puzzle is 9642 and 87531. While many came close, the correct answers were cracked by Bob and Subbu
If you want the methodical way to do it, I think the following should work:
Principle 1: For obvious reasons, the numbers will have descending order of digits (else, you can make the number itself bigger by rearranging the digits to make them descending and thus the product will be biggest).
So, we have one number starting with 9 and the other with 8.
Now let’s look at what to do with 7 and 6. Whether we arrive at 97 and 86 or 96 and 87, the sum is the same (183).
Principle 2: If you break a number into two parts, the product is higher when the difference between the parts is lower. So, if you break 10 into two parts, the product is highest when the two parts are 5 and 5 (difference 0). Product of 9 and 1 (with difference of 8) is less than product of 6 and 4 (with difference of 2). This can be proven algebraically too. (See below #1)
So, to break up 183, we are better off with 96 and 87 (lesser difference).
Now, take 5 and 4 and you come up with 964 and 875 (the other option will have greater difference).
Next take 3 and 2 and you come up with 9642 and 8753.
Now what to do with the 1? Go with 96421 and 8753 or 9642 and 87531? A quick multiplication – which can be proven algebraically (see below #2) will give you the answer.
Algebraic Proof #2
Given X and Y (X > Y) and having to add 1 to either gives you either 10X+1 and Y or X and 10Y+1
The products are 10XY+Y or 10XY+X.
Since X>Y, the second one is bigger. So add the 1 at the end of the lesser number.
Algebraic Proof #1
Let’s say the number is N and you break up it into x and (N-x).
The product is Nx – x(squared)
which is – ( x(squared) -Nx)
which is – (x(squared) – Nx + (N/2)(squared)) + (N/2)(squared)
which is (N/2)(squared) – (x – (N/2))(whole squared)
(N/2)(squared) is always positive.
the total becomes lower and lower as the (x-(N/2))(whole squared) becomes bigger and bigger.
That is true when x-(N/2) becomes bigger and bigger
That is true when you pick x further and further away from the midpoint.
(realize that the distance of x from midpoint is twice the difference between x and the other number N-x)
Ergo….
My whole body was hurting after playing two sets with Anindya and now my head is starting to hurt, reading your solution to the puzzle. Need some Aspirin. đŸ™‚
Classic Dynamic Programming from Algo 101 đŸ™‚
Ore baba!eto boro answer amar meyera je ki kore kare!! ora bollo oder ans naki almost close chilo, apni message diye chen. jak,apner puzzle solve kore jodi buddhi bare.