Sunday, September 28, 2008

Skeleton in the Problem Solving Closet

How many of you often wonder why you can’t ‘see through’ a problem? Why is it that you know all the formulas and have gone through a lot of problems associated with a particular chapter, and yet you are unable to solve a new problem when it comes across to you? What is it that those math geniuses possess that you don’t?

There are two differences between a mathematical mind and an ordinary mind:

· A mathematical mind keeps its armory of formulas and theorems very handy and in an assorted manner, ready to be recalled at an instant notice.

· A mathematical mind identifies the uniqueness to a problem and subconsciously searches through its repository of formulas to find one that fits to that uniqueness.

Which brings me to the most essential part of problem solving-

Identifying that uniqueness can often help us solving the problem. Here is a specimen:

Where will you start?

Would this problem make sense if it was something like ?

Yes. As the unique part of this problem is the number DDD, we start with this number itself.

· The number DDD = D × 111 = D × 3 × 37 = 3D × 37.

· 37 is a two digit number and since it cannot be reduced further it can be one of the numbers. Let it be AB.

· Then CB is a number ending in 7 because both AB and CB have the same unit digit.

· CB = 3D, i.e. CB is a multiple of 3.

Now let’s check our solution- Multilying 37 and 27, we get 37 × 27 = 999. Therefore, our logic is correct and A = 3, B = 7, C = 2, and D = 9. And A + B + C + D = 21.

Broadly speaking, here are the steps that you should try in solving a problem, regardless of the topic that problem came from:

Here is another simple problem:

Where will you start?

There are two unique points about this problem- first that the digits of the number are getting reversed and second that the number is being multiplied by 4. Notice that ABCD × 4 = DCBA is different from ABCD × 4 = EFGH or ABCD × 7 = DCBA.

Right now you can handle that the number is multiplied by 4 because you know something about properties associated with number 4. Let’s start with that.

· Any number multiplied by 4 will give us an even number. Hence, the digit D when multiplied by 4 will give us an even number. Since A is the unit digit of the product it is even. Hence, A = 2, 4, 6 or 8 (It cannot be 0. Why?)

· A is also the first digit of the multiplicand and if A = 4, 6 or 8 the product will become a 5 digit number. Hence A = 2. Writing the value of A we get

· What can be the value of D? looking at the first and last digits of the multiplicand, we can see that 4 × D gives the unit digit of 2 and 4 × 2 gives the first digit of D. Yes, you got it right. D = 8. Writing the multiplication again with the value of D we get:

· What can be the value of B? From your repository of formulas associated with 4 recall the one about divisibility of 4. A number is divisible by 4 if the number formed by the last two digits is divisible by 4. Since the number 8CB2 is a multiple of 4, the number B2 should be divisible by 4. Or, the number B2 = 12, 32, 52, 72 or 92. Hence the original number ABCD is 21C8, 23C8, 25C8, 27C8 or 29C8. But the last 4 numbers when multiplied by 4 will not give you the first digit of 8 in the product! Therefore B = 1 and the original number is 21C8. We write the multiplication again:

· Can you identify C now? Notice that when you multiply 8, the unit digit of 21C8, by 4 you write 2 in the unit digit of the product and carry 3. The tenths digit of the product is 1. Therefore, 4 × C + 3 (carryover) gives a unit digit of 1. Hence, C is 2 or 7. You can easily check by the hundreds digit in the product (which is C again) that C = 7.