The four numbers A, B, A+B and A-B are all prime. The sum of these four numbers is

A) Even

B) Divisible by 3

C) Divisible by 5

D) Divisible by 7

E) Prime

Source: 2002 AMC 10/12B #15

The four numbers A, B, A+B and A-B are all prime. The sum of these four numbers is

A) Even

B) Divisible by 3

C) Divisible by 5

D) Divisible by 7

E) Prime

Source: 2002 AMC 10/12B #15

I love these kinds of puzzles because there is only one solution but there are several ways to get there. It’s at least as interesting to hear the different approaches as it is to solve it.

Here was the puzzle posted last week:

Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with 4-digit number in which each of two digits appears two times.

“Look, daddy!” she exclaims. “That number is evenly divisible by the age of each of us kids!”

“That’s right,” replies Mr. Jones, “and the last two digits just happen to be my age.”

What is the four digit number in the license plate?

The first thing I noted when I saw the problem is that the solution must be in one of the following forms: aabb, abba, or baba (where a and b are distinct digits). Also, I noticed that since the oldest child is 9, the sum of digits must be a divisible by 9. So that 2a + 2b = 9k for some k. Because 2a+2b is even, k must be even. We know that a and b are distinct digits between 0 and 9, inclusive, the largest 2a+2b could be is 34, so k must be 2. In other words, a+b=9.

Next thing I noticed is that Mr. Jones will have either 4 year old or an 8 year old or both. Either way, the 4-digit number is divisible by 4. This means the last two digits must be divisible by 4. We thus have the following possibilities: 9900, 5544, 1188, 3636, 6336, 2772, 7272. I arrived at these by thinking first of numbers of the form bb that are divisible by 4 (00, 44, 88) and providing the appropriate a to get aabb that is divisible by 9. Then I considered the numbers of the form ba, that are divisible by 9 and 4 (36, 72) and listed both abba and baba.

Notice that in this list only 9900 is divisible by 5 and 00 can’t possibly be the age of Mr. Jones so he must not have a child age 5. Only 5544 is divisible by the digits 1, 2, 3, 4, 6, 7, 8, and 9.

SOLUTION: 5544

A few folks emailed me their solutions and were all correct, but each had an approach that was not quite the same as mine. Well done to those!

Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with 4-digit number in which each of two digits appears two times.

“Look, daddy!” she exclaims. “That number is evenly divisible by the age of each of us kids!”

“That’s right,” replies Mr. Jones, “and the last two digits just happen to be my age.”

What is the four digit number in the license plate?

Source: 2006 AMC 10/12B #25 (American Mathematics Contest)

Two days ago, I posted this simple little number puzzle. Quite a few folks came up with the answer below. One of the interesting questions you can ask is whether that solution is unique.

Clearly there are two lines of symmetry in the original problem so by reflection alone we come up with a total of four solutions: [tex] \{ I, F_x, F_y, F_x \circ F_y \} [/tex] where [tex]F_x [/tex] and [tex] F_y [/tex] represent “flips” across the lines of symmetry and [tex] I [/tex] represents the identity, or the solution above. By [tex]F_x \circ F_y[/tex], I mean the composition of the flipping operations or just consecutive flipping.

There also exists radial symmetry at [tex]180^{\circ}[/tex], but this is equivalent to [tex]F_x \circ F_y[/tex]. So for this arrangement above, there are four solutions of the same “type”.

Are any other arrangements possible besides these four?

I’m not keeping very good track of how many of these kind of puzzles I’m posting, so we’ll just say this is the 6th.

**Problem**: Place the digits 1 through 8 in the circles below such that no two adjacent circles contain consecutive digits.

*Update: I was missing a couple of lines. The picture is now correct.*

I know some of you who read this blog also read 360 but the sequence posted there today was too good to not pass along:

Puzzle: Identify the next term in the sequence

1, 11, 21, 1211, 111221, 312211, 13112221, . . .

[HINT: Read the digits aloud. The sequence is aural/visual and not numeric.]

I’ll post the answer tomorrow or the next day.

I’m embarrassed by the fact that I have 3 weeks of Twitter updates cluttering the first page of this blog. That means, that I’ve gone without blogging for at least two weeks. I was trying to get on a roll with the number puzzles. I have a backlog of some interesting links and topics in my bookmarks list so.

Back on December 8, I posted the following problem:

Form a number from the digits 0 to 9 such that the first digit is divisible by one, the first two digits form a number that is divisible by two, the first three digits form a number that is divisible by three, and so forth.

As was pointed out by one of the commenters, this was ambiguous because I wasn’t clear if I meant first from the left or first from the right. I actually meant from the left but the problem can be solved either way.

For the intended question, the solution is 3816547290.

Here’s a solution given by Jenning Y. Seto.:

Let each of the digits in the number be represented by a letter

so that the number is: abcdefghijFirst of all, we can see that ab/2, abcd/4, abcdef/6, abcdefgh/8,

and abcdefghij/10 are integers, therefore, b, d, f, h, and j are all

even. Since we only have five even digits (0, 2, 4, 6, and 8), a, c,

e, g, and i are all odd digits. Furthermore, knowing that certain

parts of the number are divisible by certain numbers gives us the

following constraints:1. a/1 is an integer tells us nothing.

2. ab/2 is an integer tells us that b is even.

3. abc/3 is an integer tells us that (a+b+c)/3 is an integer.

4. abcd/4 is an integer tells us that cd/4 is also an integer and d is even. Since c is odd, d must be 2 or 6.

5. abcde/5 is an integer tells us that e MUST be 5.

6. abcdef/6 is an integer tells us that (a+b+c+d+e+f)/3 is an integer and that f is even.

7. abcdefg/7 is an integer can be used later.

8. abcdefgh/8 is an integer tells us that fgh/8 is an integer and h must be 2 or 6 since g is odd.

9. abcdefghi/9 is an integer tells us that (a+b+c+d+e+f+g+h+i)/9 is an integer.

10. abcdefghij/10 is an integer tells us that j MUST be 0.Constraint 9 tells us nothing since constraint 10 states that j is 0,

and the sum of the remaining digits is divisible by nine.From constraints 4 and 5, the only possible values for the digits

d and e arede = 25 or 65.

Combining constraints 3 and 6 we see that (d+e+f)/3 must be

an integer and that f is even. Thus,de = 25 implies f = 8.

de = 65 implies f = 4.

This gives us two possible cases:

def = 258 or 654.

Using constraint 8, we can see that

def = 258 implies gh = 16 or 96.

def = 654 implies gh = 32 or 72.

Now we have four possibilities that satisfy constraints 4, 5, 6,

and 8:defgh = 25816, 25896, 65432, or 65472.

Notice that in all 4 cases, 2 and 6 are used. Thus b = 4 or 8.

Now using constraint 3 with the above constraints leaves us with

a number of possibilities for abc:abc = 147, 183, 189, 381, 741, 789, 981, or 987.

Even with the above constraints, this gives us ten possibilities

for the first eight digits:abcdefgh = 14725896, 18365472, 18965432, 18965472,

38165472, 74125696, 78965432, 98165432, 98165472, or 98765432.However, we can use constraint 7, and see that only

abcdefg = 3816547 is divisible by 7.

Thus the final answer is 3816547290. Furthermore, this is a unique

answer.

Sally has five red cards numbered 1 through 5 and four blue cards numbered 3 through 6. She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?

From: MAA Minute Math

Form a number from the digits 0 to 9 such that the first digit is divisible by one, the first two digits form a number that is divisible by two, the first three digits form a number that is divisible by three, and so forth.

I’ll post my solution in a day or two.

HT: QYV

A couple of days ago, I posted the following puzzle:

There is a ten digit number where the leftmost digit is also the number of zeros in the number, the second leftmost digit is the number of ones and so forth until the last digit (or rightmost digit) is the number of nines in that number. What is this number and is it unique?

Congratulations to the folks that figured it out, including Trae (of TraeBlain.com)who posted the solution to the comments section. I’d award him a prize if I had one to give. Instead he’ll have to settle for a virtual pat on the back.

The answer: **6210001000**

Now the real question is how do you come up with the answer and is it unique? Additionally, you might ask how this works for a smaller number of digits, that is, do the same problem but start with only 3 digits, or 4 digits, etc.?

I came up with a very similar explanation as the original source for the problem so I’ll give credit where credit is due. You can read the original post here: http://begghilos2.ath.cx/~jyseto/Academia/Math-Problem-1.php