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Worked Examples

Example-1

Solution

Here, each letter represents a unique number and also we know that ‘three + three + two + two + one = eleven’! It is hard to construct such a beautiful alphametics!

 

The answer is,

 

 

Example-2

A much more elementary example is the following problem:

Solution

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

 

Here is how we do the analysis:

  • Since 3 × ON is a two-digit number, 3 × ON ≤ 99; therefore, ON ≤ 33 and O ≤ 3. Thus, O = 1, 2 or 3.
  • Since 3N ≡ O, we examine the multiplication table for 3. We deduce the following:
     
    O = 1 ⇒ N = 7
    O = 2 ⇒ N = 4
    O = 3 ⇒ N = 1
  • Each of these possibilities gives rise to a solution. The solutions are:
     
    17 × 3 = 51 24 × 3 = 72 31 × 3 = 93
     
     
    It is clear that this list of solution is complete.

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.
 

 

Example-3

Decode the following problem.

 

TWO2 = THREE

Solution

We reason as follows:

  • Since TWO2 has five digits, it follows that TWO ≤ 316 (because Description: Description: 103654.png lies between 316 and 317). Therefore, T is 1, 2 or 3.
  • Since the left-most digit of TWO2 is the same as that of TWO, namely T, we must have T = 1. (If T were 2, then TWO2 would exceed 40,000 and we would have T ≥ 4; and if T were 3, then TWO2 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: 1232 = 15129 and 1372 = 18769, 2432 = 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:
     
    1WO2 = 1HR44
     
    Since the number on the right is less than 20,000, it follows that 1WO ≤ 141 (since Description: Description: 103667.png ≈ 141.42). As O is 2 or 8, it follows that W < 4; therefore, W is either 2 or 3. (Since T = 1, the possibility that W = 1 is ruled out.)
  • 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, 1282 = 16,384; 1322 = 17,424; 1382 = 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:
     
    TWO = 138, THREE = 19044.

 

Example-4

Solve this alphametic.

 

Solution

Here what could A be? If A is greater than 1, then we would have ABCD ≥ 2000; but 9 × 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 × D is either A or an integer with units digit A. Since, A = 1, it follows that 9 × D is a number with units digit 1. This forces D to be 9 (because 9 × 9 = 81).

 

Now, we have;

 

 

 

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

 

Now, we have;

 

 

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





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