There is a three digit odd number. The second digit is four times as big as the third, while the first digit is three less than the second. What is the number?
Preambles: 1. The number is three digits. 2. The number is odd.
Notes: For any number to be odd, the last digit of the number must also be odd. Thus 17 is odd because the last digit, ‘7’ is odd; 12 is not odd because the last digit, ‘2’ is not odd. This holds when we are working in base 10, which in this case we are.
Therefore, the last digit of the number we seek can not be even. 0, 2, 4, 6, 8 are thus not possible values of the third digit. Now let us find our three digit number following the steps below:
Step 1: From the preamble and notes above, the only possible values for the third digit are 1, 3, 5, 7 and 9.
Step 2: The second digit is four times as big as the third. If we picked either 3, 5, 7 or 9 as the third digit, four times any of those numbers will give us a two digit number. Example: 4*3=12; 4*5=20; 4*7=28; 4*9 = 36. But since the number we seek is a three digit number, the second digit must be a single digit. Thus the third digit of the number can not be 3, 5, 7 or 9.
If we picked 1 as the third digit, the second digit will be = 4*1 = 4. ‘4’ satisfies the preamble because 4 is four times bigger than 1 and four is a single digit number. Thus the third digit is ‘1’ and the second digit is ‘4’.
Step 3: The first digit is three less than the second means the first digit must be 1, because 1 is 3 less than 4.
Therefore, the number is 141.
Without the odd number conddition for the question, what other possible number(s) would satisfy the question?