Junior College Math Questions

[Andy Nguyen]

This requires high-school or first year college math level. Each question has a maximum of 3 minute time.

1) A multiplicative magic square (MMS) is a square array of positive integers in which the product of each row, column, and long diagonal is the same. The 16 positive factors of 2010 can be formed into a 4x4 MMS. What is the common product of every row, column, and diagonal?

2) If you roll three fair dice, what is the probability that the product of the three numbers rolled is prime

3) A palindrome is a positive integer like 11, 313, and 5445 which reads the same left to right and right to left. If a number is chosen at random from all four-digit palindromes, find the probability that it is divisible by 7

4) You have 4 red, 4 white, and 4 blue identical dinner plates. In how many different ways can you set a square table with one plate on each side if two settings are different only if you cannot rotate the table to make the settings match?

5) A 100 m long railroad rail lies flat along level ground, fastened at both ends. Heat causes the rail to expand by 1% and rise into a circular arc. To the nearest meter, how far above the ground is the midpoint of the rail

 

[bigbadwolf]

For #2, two of the dice would have to show 1, and the third could be 2,3 or 5. If the first was 1, the second was 1, and the third 2,3, or 5, then you would have 1/6*1/6*3/6 = 3/216. Multriply by 3 (because first could be 1 and third could be 1, or second and third could be 1) to get 9/216.

For #4, take out one of the plates to be a reference point. It can be white, blue, or red: 3 possibilities. You are left with 11. If they were 11 different plates, you would have 11! different arrangements, but some are the same, so you should get 11!/4!4!3!. Mustn't forget to multiply by 3 to get 3*(11!/4!4!3!)

Correction: This strikes me as wrong. Start again. If you arrange them in a straight line, then you have 12!/4!4!4!. But now, in a circle, this is overcounting, because by rotation, some will be equivalent. You are overcounting by a factor of 12, so the answer should be 11!/4!4!4!

 

[euroazn]

3) is probably an AMC question... I recall seeing something like it somewhere...

Anyway: a four digit palindrom is in the form ABBA, or 1001A+110B. Note that 1001 is a multiple of 7. Thus 110B must be a multiple of 7. Hence, B is either 7 or 0, for odds of 2/10=1/5

Time to tackle 1 and 5

 

[pruse]

For #4, take out one of the plates to be a reference point. It can be white, blue, or red: 3 possibilities. You are left with 11. If they were 11 different plates, you would have 11! different arrangements, but some are the same, so you should get 11!/4!4!3!. Mustn't forget to multiply by 3 to get 3*(11!/4!4!3!)

Why? 11!/4!4!3! is the number of ways to permute 11 objects some of which are identical. But here we're not arranging 11 objects, we're only arranging 4.

Here's another approach...

If we only use 1 color, there are clearly 3 ways to arrange the table.

If there are 2 colors, that's either a) 1 of one kind, 3 of another, b) 2 of each of two kinds. In case a), we have 6 (=(\binom{3}{2}\cdot 2)) ways ((\binom{3}{2}) ways to pick the two colors, 2 ways to pick the one that occurs once). In case b), we have 6 (=(\binom{3}{2}\cdot 2)) ways ((\binom{3}{2}) ways to pick the two colors, 2 ways to arrange them around the table -- like colors sitting next to each other, or opposite each other). So a total of 12 ways with 2 colors.

If there are 3 colors, then we must have 2 of one kind and one of each of the other 2 kinds. There are 3 ways to pick the one that appears twice. Once the selection of colors is picked, say AABC, there are (3-1)! = 2 ways to permute A, B, C around the table, and then 2 ways to insert the extra A (either next to the first A -- either side of that A gives the same arrangement -- or between B and C). So we get 2*3 = 6 ways in this case.

Total is 3+12+6=21.

 

[euroazn]

1) Upon half a minute of thinking, I realized that 1 is very easy.

Suppose the magic square is

A B C D
E F G H
_I J K _L
M N O P

The product ABCD=EFGH=IJKL=MNOP. ABCDEFGHIJKLMNOP=2010^8
Thus the product is 2010^8^1/4=2010^2=4040100

 

[bigbadwolf]

Why? 11!/4!4!3! is the number of ways to permute 11 objects some of which are identical. But here we're not arranging 11 objects, we're only arranging 4.

I didn't read the question properly. I shall now commit seppuku.

 

[pruse]

I didn't read the question properly. I shall now commit seppuku.

haha...

 

[euroazn]

I didn't read the question properly. I shall now commit seppuku.

Noooooooooo....... for all I know you may be a commie but you're still one of us... don't do it! :O


:P

 

[koupparis]

Got 6 meters for 5, but I used Taylor series expansions. Are those taught in high school? I had an AP student that was taught Taylor series.

 

[pruse]

I got 6m also. Yeah, Taylor series are covered in AP BC Calculus. The types of students these questions are geared towards would most probably know Taylor.

 

[szhong]

question 4

this is my first post, and I want to say that this forum has been immensely helpful for me, so plz be gentle with me.

for number 4, i believe if you apply burnside's lemma, you should get 24 possibilities.

specifically, you have 1/4(3^4 + 3 + 3^2 + 3) = 24

peterruse solution is correct except for the last part, in which the combination AABC has 3 possibilites of arrangement, not 2.

specifically, you can have ( following letters represent "top right bottom left" positions ) : A A C B, A A B C, C A B A

The first 2 types are not equal since they are not the same by rotation, but are the same by reflection only.

Please correct me if you see anything wrong with my argument

 

[pruse]

szhong, you're right, it is 24.

the way to look at my last case is, (\frac{4!}{4\cdot 2}=3), the 4 to account for cyclic permutations, the 2 to account for repetition of A's.

Burnside's Lemma is indeed a powerful combinatorial tool. good call!

 

[nburk]

1) Upon half a minute of thinking, I realized that 1 is very easy.

ABCDEFGHIJKLMNOP=2010^8

I am not sure where this step came from....

 

[euroazn]

Nburk, good question!

The 16 positive factors of 2010 can be formed into a 4x4 MMS.

2010 isn't a perfect square.... so there are exactly 8 pairs of factors that multiply into 2010... so 2010^8. Do you get it now?

 

[koupparis]

The 16 factors of 2010 come from the 4 prime factors of 2010, 2, 3, 5, and 67. To get the other 12 factors we group the prime factors. 2 can be grouped 8 different ways, ie it shows up 8 times within the 16 factors, hence so do all the other prime factors. So when we take the product of all 16 factors we get all prime factors to the 8th power, hence 2010^8.

 

[euroazn]

The 16 factors of 2010 come from the 4 prime factors of 2010, 2, 3, 5, and 67. To get the other 12 factors we group the prime factors. 2 can be grouped 8 different ways, ie it shows up 8 times within the 16 factors, hence so do all the other prime factors.

That's correct, but unnecessary for this problem since they give you the number of factors

 

[koupparis]

Is there a better way to get 2010^8? Without my logic? Too lazy to think ...

 

[euroazn]

All I meant that your first part was unnecessary since they stated that 2010 has 16 factors. The 16/2 part is absolutely right though.