Question

Three different non-zero digits can be arranged in six different ways to

form six three-digit numbers. If the smallest three of these numbers add
to 540, what is the sum of the largest three numbers?

Answer 1

1134

Explanation:

We have 3 digits:

a, b, c

a 3 digit number can be written as:

a*100 + b*10 + c*1

Such that these numbers can be:

{1, 2, 3, 4, 5, 6, 7, 8, 9}

Let's assume that:

a < b < c

Then the 3 smaller numbers are:

a*100 + b*10 + c

a*100 + c*10 + b

b*100 + a*10 + c

The 3 larger numbers are:

b*100 + c*10 + a

c*100 + a*10 + b

c*100 + b*10 + a

We know that the sum of the 3 smaller numbers is equal to 540, then:

(a*100 + b*10 + c) + (a*100 + c*10 + b) + (b*100 + a*10 + c) = 540

Let's simplify this:

(a + a + b)*100 + (b + c + a)*10 + (c + b + c) = 540

(2a + b)*100 + (b + c + a)*10 + (2c + b) = 540

The sum of the 3 larger numbers is equal to X, we want to find the value of X:

(b*100 + c*10 + a) + (c*100 + a*10 + b) + (c*100 + b*10 + a) = X

Now let's simplify the left side:

(b + c + c)*100 + (c + a + b)*10 + (a + b + a)*1 = X

(b + 2*c)*100 + (c + a + b)*10 + (2a + b) = X

Then we have two equations:

(2a + b)*100 + (b + c + a)*10 + (2c + b) = 540

(b + 2*c)*100 + (c + a + b)*10 + (2a + b) = X

Notice that the terms are inverted.

By looking at the first equation, we can see that:

(2c + b) = 10 (because the units digit of 540 is 0)

Then, we can see that:

(b + c + a + 1 ) = 14 (the one comes from the previous 10)

finally:

(2a + b + 1) = 5 (the one comes from the previous 14)

Then we can rewrite:

(2*c + b) = 10

(b + c + a) = 14 -1 = 13

(2a + b) = 5 - 1 = 4

Now we can replace these 3 in the equation:

(b + 2*c)*100 + (c + a + b)*10 + (2a + b) = X

(10)*100 + (13)*10 + 4 = X

1000 + 130 + 4 = X

1134 = X

The sum of the 3 largest numbers is 1134.