# Math Challenge - There's A Three Digit Number Math Puzzle

Math puzzles are a fun and engaging way to test your problem-solving skills and logical reasoning abilities. Exercise your brain and make you think deeply. Try to solve the below math challenge and find the answer and explanation.

## Math Challenge

I am a three-digit number my second digit is four times bigger than the third digit my first digit 3 less than my second digit who am I?

Try to solve it. It's little tricky

**Hint:**You have to use an algebraic equation

**Answer:** 141

** **

**Explanation: **

Let the three numbers be **100s** number **a**, **10s** number **b**, and **1s** number **c**.

therefore number will be **100a+10b+1c**

Let the third number c be x

The alternate number is four times as big as the third number,

therefore **2nd** number **b = 4x**

the first number is three lower than the alternate number, therefore **1st** number **a=(4x-3)**

** **

Now there are the following possibilities,

Since **x** is a number, it must lie between a single number of **1** and **9**.

**x** cannot be **0** or further than **3** because If** x** is considered **0**,

also the number a will be **0** and the resulting number will be only a two-digit number.

**x** cannot be **3** because

However, the alternate number b which is **4x**, won't be between** 1** and **9** if **x = 3**.

So, **x** can only take the value of c as **1** and **2**.

However, **2nd** number,

If **x = 1** also the third number c will be 1.

**b** will be **4x = 4** and

the first number a will be **4x- 3 = 4- 3 = 1**

** **

Thus, the three-digit number will be** 141**.

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