# What is the positive three-digit integer that is 4 times the product of its digits? Thank you! :-)

## Question:

What is the positive three-digit integer that is {eq}4 {/eq} times the product of its digits?

## Expressing number in terms of digits

A number can be expressed as the sum of many numbers varying by the order of multiple of 10 in value. It can be dividing by segregating the digit by dividing it by the appropriate power of 10 and expressing it in the method mentioned in the following examples.

For example, 25 can be expressed as $$25 = 5\times 10^0 + 2\times 10^1 + 0\times 10^3 + 0\times 10^4.... .. . . . . .. 0\times10^n$$

Similarly an unknown number, say {eq}abcd {/eq}

$$abcd = d\times 10^0 + b\times 10^1 + c\times 10^3 + d\times 10^4$$

Let us take a three-digit number {eq}XYZ {/eq} which can be expressed as {eq}100x + 10y +z. {/eq}

According to the question, the number is 4 times the product of its digits.

$$100x + 10y +z = 4\times x\times y\times z$$

$$100x + 10y +z = 4xyz$$

$$100x + 10y = 4xyz -z$$

$$10(10x + y) = z(4xy -1)$$

$$z=\dfrac{10(10x + y)}{4xy -1}$$

The value of {eq}z,\ x,\ and\ y {/eq} is an integer and between 1 to 9.

To get that the denominator should be able to divide the 10. For that, the numerator should be ending by either 10 or 5.

If the denominator were to end with 10 the value of {eq}4xy {/eq} will end in 11, which is not possible. The denominator has to end with the digit 5, the value of {eq}4xy {/eq} will end in 6, which is possible.

If {eq}4xy {/eq} were to end in 6, the possible pair of values of {eq}(x,y) {/eq} are (3,3),(7,2), (3,8),(2,7) and (8,3).

After substituting all the values, the ones that satisfies all the constraints is {eq}x=3,\ and\ y=8 {/eq}

Substituting the values of {eq}x \ and\ y {/eq}:$$z=\dfrac{10(10x + y)}{4xy -1}$$

$$z=\dfrac{10(10*3 + 8)}{4*3*8 -1}$$

$$z=\dfrac{10(30 + 8)}{96 -1}$$

$$z=\dfrac{10*38}{95}$$

$$z=4$$

The three-digit number is {eq}384. {/eq}