Round 304

we don't have time to post the whole solution but this problem need to be solved using differential equation to get the exact answer... we probably post during the weekend...

Reviving the blog

Today, we have decided to revive the blog. We are still busy with school and work. However, we still want to have some fun and archive old puzzles.

However, we do not guarantee to post solutions at the time of the puzzle. Instead, we create an empty page and allow discussing through comments for each puzzle (Thanks to the advance posting feature from Blogger). We will post solutions when we have free time (mostly during weekend).

In this way, some people can't complain that they can't get into first 250 because of our site. However, we still provide solutions/hints for those who struggle. So I hope everyone will be happy.

LC Solver

Round 303

Date Released: March 19th, 2009.
Date Rewarded: March 26th, 2009. (estimated)

Rearrange the letters in the following phrase to come up with a Neopian word or phrase:

ROTATE SOON VIA RAIL

Submit only the answer with no other information, otherwise your answer will be marked as incorrect!

Prize

Steam Train Gummies

Click to show/hide hint

It is related to a character in the current Daily Dare event: Aristotle A Avinroo.

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

Click to show/hide solution

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.

Round 301

I should be back here soon but can't spend too much time on it now... We will do a survey to see what answer you submitted last week. It may not be accurate but it's worth a shot...