There is a three digit number. The second digit is four times as big as the third digit, while the first digit is three less than the second digit. What is the number?

The possible three digit numbers that suits the question description are;

141 and 582

Let this 3 digit number by xyz.

Let's try values for z.

y = 4 × 0

y = 0.

This means that; x = 0 - 3

x = -3

x cannot be negative and so z cannot be 0.

y = 4 × 1

y = 4

This means that; x = 4 - 3

x = 1

Thus, we will have a possible 3 digit number which is 141

y = 4 × 2

y = 8

This means that; x = 8 - 3

x = 5

Thus, we will have a possible 3 digit number which is 582

y = 4 × 3

y = 12

y cannot be double digits and so z cannot be equal to 3.

Since any value of z ≥ 3 gives a y value that is not single digit, we can conclude that our possible 3 digit numbers are; 141 and 582.

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There are 2 possible answers and they are

Explanation:

a = first digit

b = second digit

c = third digit

the three-digit number is abc

b = 4c because the second digit is 4 times that of the third digit

a = b-3 because the first digit is 3 less than the second digit

If c = 0, then b = 4*c = 4*0 = 0. This leads to a = b-3 = 0-3 = -3, but 'a' must be some positive whole number between 0 and 9. So c = 0 is not possible.

If c = 1, then b = 4*c = 4*1 = 4. This leads to a = b-3 = 4-3 = 1.

So (a,b,c) = (1,4,1) is one solution

If c = 2, then b = 4*c = 4*2 = 8 and a = b-3 = 8-3 = 5.

This means (a,b,c) = (5,8,2) is another solution

We can't have c = 3 because b = 4*c = 4*3 = 12 is not a single digit number. The same can be said for any c value from 4 onto 9.

In other words, only c = 1 and c = 2 are possible.

This means the two possibles are 141 and 582. We do not have enough information to narrow it down to just one answer only.

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