## 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.

Notice how we grabbed a uniqueness of the problem, started unraveling it, and slowly discovered the complete solution to the problem.

The beauty of the above two problems is that they require elementary knowledge of numbers but they will confuse many of you because you do not know where to start looking. Now you have your answer: