Round 302

Date Released: March 12th, 2009.
Date Rewarded: March 17th, 2009.

271 - 482 = 436

148 + 591 = 866

4881 + 8914 = 6562

8827 * 1178 = ?

Do NOT show your work; submit only a single number, with no punctuation, as your answer.

Prize


Rotawheel

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To be easy to work on this problem, we convert all the digits into alphabetical letters in the following scheme: a = 1, b = 2, c = 3, d = 4, e = 5, f = 6, g = 7, h = 8, i = 9. Then we have three equations:

bga = dhb + dcf (1)
adh + eia = hff (2)
dhha + hiad = fefb (3)

Before we solve the problem, we have to assume that each letter represent a unique digit (distinct property.) For example, if a is 4, then b must not be 4. Also, all the digits in the hundreds' places in (1) and (2) and thousands' places in (3) must not be 0.

If we look at the equation (1) at the hundreds' place, we have b = d + d. So, we know that 1 <= d <= 4 because d can't be 0 and if d >= 5 then b is not a digit. There are four cases:

• d = 1: b can be either 2 or 3 (in the case of carry from the tens' place.)
• d = 2: b can be either 4 or 5.
• d = 3: b can be either 6 or 7.
• d = 4: b can be either 8 or 9.

If we look at the equation (3) at the unit's place, we have a + d = b. Then, the four cases will be following:

• d = 1: b = 2 => a = 1 (violate the distinct property), b = 3 => a = 2
  
• d = 2: b = 4 => a = 2 (violate the distinct property), b = 5 => a = 3
• d = 3: b = 6 => a = 3 (violate the distinct property), b = 7 => a = 4
• d = 4: b = 8 => a = 4 (violate the distinct property), b = 9 => a = 5

If we look at the equation (1) at the unit's place, we have a = b + f. Then, the four cases will be following:

• d = 1: a = 2, b = 3 => f = 9
• d = 2: a = 3, b = 5 => f = 8
• d = 3: a = 4, b = 7 => f = 7 (violate the distinct property)
• d = 4: a = 5, b = 9 => f = 6

There are only three cases left after the last analysis. If we look at the equation (2) at the unit's place, we have h + a = f. Then, the three cases will be following:

• d = 1: b = 3, f = 9, a = 2 => h = 7
  
• d = 2: b = 5, f = 8, a = 3 => h = 5 (violate the distinct property)
• d = 4: b = 9, f = 6, a = 5 => h = 1

There are only two cases left now. However, if we look at the equation (2) at the hundreds' place, we have a + e = h. In the last case, a = 5 and h = 1. It is impossible to get since h > a. Therefore, only the first case is correct. Then, we have: a = 2, b = 3, d = 1, f = 9, h = 7

Now, we look at the equation (2) at the tens' place. Since there is no carry on the unit's place addition, we can write: d + i = f <=> i = f - d = 9 - 1 = 8 (no carry either). Then, we look at the same equation at the hundreds' place, we can write: a + e = h <=> e = h - a = 7 - 2 = 5.

Then, we look at the equation (1) at the tens' place. Since there is a carry on the unit's place (a = 2 = 3 + 9 = b + f), we can write: g = h + c + 1 <=> g = c + 8. Since 1, 2, 3, 5, 7, 8, 9 are already taken. c and g can only be 0, 4 or 6. Only the pair c = 6 and g = 4 are satisfied this equation. Therefore, we have: a = 2, b = 3, c = 6, d = 1, e = 5, f = 9, g = 4, h = 7, i = 8.