# Worked Examples

Here, each letter represents a unique number and also we know that â€˜three + three + two + two + one = elevenâ€™! It is hard to construct such a beautiful alphametics!

The answer is,

A much more elementary example is the following problem:

The problem may also be written as ON + ON + ON = GO.

Here is how we do the analysis:

- Since 3 Ã— ON is a two-digit number, 3 Ã— ON â‰¤ 99; therefore, ON â‰¤ 33 and O â‰¤ 3. Thus, O = 1, 2 or 3.
- Since 3N â‰¡ O, we examine the multiplication table for 3. We deduce the following:
O = 2 â‡’ N = 4O = 3 â‡’ N = 1
- Each of these possibilities gives rise to a solution. The solutions are:
17 Ã— 3 = 51 24 Ã— 3 = 72 31 Ã— 3 = 93

Here, we have three possible values to each of the letters in the given problem. But, generally, such problems with multiple answers are not entertained.

Decode the following problem.

TWO^{2} = THREE

We reason as follows:

- Since TWO
^{2 }has five digits, it follows that TWO â‰¤ 316 (because lies between 316 and 317). Therefore, T is 1, 2 or 3. - Since the left-most digit of TWO
^{2}is the same as that of TWO, namely T, we must have T = 1. (If T were 2, then TWO^{2 }would exceed 40,000 and we would have T â‰¥ 4; and if T were 3, then TWO^{2}would exceed 90,000 and we would have T = 9.) - We know that no perfect square ends with 2, 3, 7 or 8. So, the digit in the unit place of a square must be one of the following: 0, 1, 4, 5, 6 and 9. Since O and E are distinct digits, they can be neither 0 nor 5. Therefore, E is 1, 4, 6 or 9.
- We now use the property of perfect squares. The digit in the tens place of the square of an odd number can never be odd. (For example, consider the squares of odd numbers: 123
^{2 }= 15129 and 137^{2 }= 18769, 243^{2 }= 59049.) - Since the last two digits in THREE are both E, it follows that E cannot be odd. Therefore, E is either 4 or 6.
- Since E is even, it follows that THREE is even. So, TWO itself is even. Therefore, THREE is divisible by 4. We now use the test for divisibility by 4. A number is divisible by 4 if the two-digit number formed by its last two digits is divisible by 4. This implies that the number EE is divisible by 4. Since EE is either 44 or 66, and of these only 44 is divisible by 4, it follows that E = 4. This implies, in turn, that O is either 2 or 8.
- The partial reconstruction now has the following appearance:
^{2 }= 1HR44 - Since O = 2 or 8 and W = 2 or 3, it follows that TWO is one of the following: 128, 132, 138 (Only three possibilities are left!). The finish is now rather quick. We have, 128
^{2 }= 16,384; 132^{2 }= 17,424; 138^{2 }= 19,044. Only the third possibility fits (it is the only one where both the last two digits are 4). So we have our answer:

Solve this alphametic.

Here what could A be? If A is greater than 1, then we would have ABCD â‰¥ 2000; but 9 Ã— 2000 = 18,000, a five-digit number, whereas the product DCBA is a four-digit number. So, A cannot exceed 1. Further, as A is the first digit, it cannot be 0. So, A = 1.

What would D be? Arguing from the units end, 9 Ã— D is either A or an integer with units digit A. Since, A = 1, it follows that 9 Ã— D is a number with units digit 1. This forces D to be 9 (because 9 Ã— 9 = 81).

Now, we have;

We observe that 1200 Ã— 9 = 10,800 a five-digit number. This implies that B â‰¤ 1. Since we already know that A = 1, B cannot be again 1. So, B = 0.

Now, we have;

Now check: 1089 Ã— 9 = 9801. Here, we have not only obtained an answer but also shown that the solution is unique!Since (9 Ã— C) + 8 is a number with units digit 0, 9 Ã— C is a number with units digit 2; therefore, C = 8 and ABCD = 1089.