The Big Riddle of Four

After a long dry period, I'm going to publish just one post here, and it contains a riddle for you, without my solution. Today is a very special day for me (well, it's my birthday), so I'm giving you the opportunity to solve something truely difficult.

Use the number 4 twice to get 64.

That's it. Not complicated, yet not very simple. You can use any mathematical operation as long as the final results contains only two 4's. For example, 4² x 4 = 64, but it contains the digit 2, which is unacceptable.

Let's see how long it takes people to solve this, and how old I'll be when it's solved.

The answer is purely mathematic. No word tricks or anything involved.

Post your answers in the comments. Can't wait to see them.

Nadav

nadavs

Reversing a Number

After a pretty boring question yesterday, I found a good one today. Not just good - great. It's a general mathematics question with almost no hidden hints. All it takes to answer that question is pure thinking. Here it is:

There is a number abcde (each letter represents a different digit). When multiplied by 4, this number becomes edcba. What is the number?

Try thinking for a minute. What is the boldest clue in here? Of course, the number of digits. This information alone gives us two digits.

Since there are five digits on both numbers, the first digit, a, must be 1 or 2 (if it's 3 and above, the big number would have six digits). Now, because digit a appears at the units digit of a number divisible by 4, it must be a possible digit of a multiplication by 4. No integer will end in 1 when multiplied by 4, but ending with 2 is possible (for example: 3*4 = 12), so a = 2 and the number is:
2bcde

Now, because this number is multiplied by 4 and starts with 2, it must end with 8 or 9. 4 * 9 = 36, ends with 6, not good. 4 * 8 = 32 - perfect for us, since it ends with 2. So now the number is:
2bcd8

Notice how the bigger number starts with 8 and not 9. This means that b * 4 < 10, so b must be 1 or 2. Since 2 is taken, b = 1:
21cd8

In the bigger number, 8dc12, the tens digit is 1, but 8 * 4 = 32. This means we need a number that when multiplied by 4 gives 8 (to add to the 3). This means d is either 2 or 7. 2 is taken, so d = 7:
21c78

Now two things can be done: one, guessing all possible values of c. Two: calculating c. Since I don't think guessing is a legitimate reason, let's calculate:
c is in the hundreds place, so the number is 21078 + 100c, and the bigger number is 87012 + 100c. That gives the equation:
4(21078 + 100c) = 87012 + 100c
84312 + 400c = 87012 + 100c
300c = 2700
c = 9

Finally, the number is: 21978. Multiply that by 4 and you'll get 87912. Job complete.

Want to see more of these tricks? Join super math tips, there are some really nice things over there.

Hope you liked it,
Nadav

nadavs

Magic C

For today's question, I took another Yahoo Answers great question to give you the opportunity to solve a nice question. Here it comes:

The digits from 1 to 5 should be placed on a C shaped form like this:

x-x
x
x-x

(Each x represents a different digit)
The two horizontal rows (x-x) and the vertical row with three x's should have the same sum of digits.
In how many ways can this be done (reflections don't count as different), and why there are no other possibilities?

We can see at first that in order to fill in the digits, we need to find two pairs of digits that add up to the same number (the two x-x). There are three numbers that work like that: 5 (1+4, 2+3), 6 (2+4, 1+5), and 7 (3+4, 2+5).

However, 5 is impossible as a sum, since it leaves the digit 5 outside, which means it'll be in the middle x, creating a sum definitely greater than 5. We are left with only 6 and 7 as sums. After some playing with arrangements, we can see that the two solutions are:

For 6:
1-5
3
2-4

For 7:
2-5
1
4-3

There are no other possibilities because all other sums (except 5) have a maximum of one pair that works for them, and we already showed why 5 doesn't work.

Now it's your turn. Send in the riddles!

Have a great weekend,
Nadav

nadavs

The Riddle of Four

Yesterday I saw an amazing riddle on Yahoo Answers, and I had to share it, but it was after I published the calculus question, and I didn't want to post too much. Even though I answered it with detail, the asker promised to give the best answer to the first one who answers, and my long answer took away my 10 points.

Without further explaining, here is the riddle:

A number has six digits, and the last one (the units digit) is 4. When the 4 is removed from the units place and placed on the beginning of the number (the hundred thousands), the number is multiplied by 4. What is the number?

It sounds difficult and tempting to solve with complex algebra and calculus methods, but the solution is much simpler: use your head.

First, we know that when the 4 is moved to the beginning of the number, it's multiplied by 4. Since it was and remains 6 digits, the first digit must be 1. Now we have two digits:
1 _ _ _ _ 4

When we multiply 4 by 4, we get 16. Since the tens digit becomes the units digit and the bigger number must end in 6, the tens digit is 6. Now we're half way there!
1 _ _ _ 6 4

Now look at the end of this number: 64. When multiplied by 4, we get 4 * 64 = 256. Notice it ends in 6, which is perfect for us, because it'll fit right in the bigger number. That means we have two more digits, and only one is missing!
1 _ 2 5 6 4

To find the last digit, we need some more thinking. The bigger number looks like 4 1 _ 2 5 6. Notice that the 1 is not changed, and so does the 2. That means we need a digit that does not change when multiplied by 4, and the winner is zero.

The number is 102564. When you multiply it by 4, you get 410246. Riddle solved!

Nadav

nadavs