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?

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.

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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

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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

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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.

AFOKE88 AFOKE88 · Answer 2 · 2021-10-08T15:29:14+02:00

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

141 and 582

We are told that we have a three digit number.

Let this 3 digit number by xyz.

Second digit is four times as big as the third digit. Thus; y = 4z

First digit is three less than the second digit. Thus; x = y - 3

Let's try values for z.

If z = 0, then y = 4 × 0

y = 4 × 0

y = 0.

This means that; x = 0 - 3

x = -3

x cannot be negative and so z cannot be 0.

Let's try z = 1;

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

Let us try z = 2;

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

Let us try z = 3;

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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