**PUZZLE**

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?

**ANSWER**

**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.**

**CHALLENGE YOURSELF**

Without the odd number conddition for the question, what other possible number(s) would satisfy the question?