1. Each of the digits 1,2,3,4,5,6 is used exactly once in forming the 3digit integers X and Y. How many possible values of X+Y are there if XY=111?
2. The product of all digits of positive integer M is 105 How many such M's are there with distinct digits?
I tried 6 for the problem above but it wasn't right... Any thoughts on why?
3. Two boys and three girls are going to sit around a table with 5 different chairs. If the two boys want to sit together, in how many possible ways can they be seated?
4. What fraction of all the 10digit numbers with distinct digits have the property that the sum of every pair of neighboring digits is odd?
5.Mort walks each day for exercise and likes to vary his path. He starts each day at position 1 on the circle shown, then walks in a straight line to another numbered point on the circle, and continues, before returning to his starting point. If everyday Mort visits at least one point numbered 2  6 but doesn't visit the same point twice (except for the starting and ending point), how many different possible paths can he take?
Thanks!
2. The product of all digits of positive integer M is 105 How many such M's are there with distinct digits?
I tried 6 for the problem above but it wasn't right... Any thoughts on why?

I think the answer is not 6 because 1 can also be a digit of M.
105 = 3 * 5 * 7
So 5, 3, and 7 must be digits of M, while 1 might be a digit of M.
So the possiblites are:
357
375
537
573
735
753
In addition to:
1357 1375 1537 1573 1735 1753  ____  3157 3175 3517 3571 3715 3751  ____  5137 5173 5317 5371 5713 5731  ____  7135 7153 7315 7351 7513 7531 
the total number of possibilities = 3! + 4! = 6 + 6*4 = 6 + 24 = 30
3. Two boys and three girls are going to sit around a table with 5 different chairs. If the two boys want to sit together, in how many possible ways can they be seated?
We can anchor the boys in any two adjacent chairs.....and they can be seated in 2! = 2 ways in each of these chairs
And for each of these arrangements, the 3 girls can be seated in 3! = 6 ways
So....the number of possible seating arrangements is 2! * 3! = 2 * 6 = 12 ways
Hi CPHill,
For some reason when I checked 12 as the answer it came up as wrong... do you have any thoughts on why?
Thanks,
sudsw12
Cphill, the answer is 60. i believe you forgot the fact that the table is round.
the solution would be that There are 5 different neighboring pairs of chairs that the boys could choose to sit in. Once they do, there are 2! ways for the boys to arrange themselves in those chairs and 3! ways for the girls to take their seats. In total, there are 5*2!*3!=60 ways for them all to sit.
Mmmm....I think it might have to do with the fact that we have " different" chairs
Maybe this is the approach, but I'm not sure
Call the chair colors
Red Blue Green Yellow White
The boys can occupy the chairs in the following ways
(R B)
(B G)
(G Y)
(Y W)
(W R)
And for each of these the boys can be seated in two ways = 5 * 2 = 10 ways
And for each of these ways, the girls can be seated in 3! = 6 ways
So.....maybe 10 * 3! = 10 * 6 = 60 is your answer ????
5.Mort walks each day for exercise and likes to vary his path. He starts each day at position 1 on the circle shown, then walks in a straight line to another numbered point on the circle, and continues, before returning to his starting point. If everyday Mort visits at least one point numbered 2  6 but doesn't visit the same point twice (except for the starting and ending point), how many different possible paths can he take?
Since we can draw a (straightline) chord between any two points on a circle.....the number of possible paths is just al lthe permutations of the set { 1,2,3,4,5,6,1 } with 1 as the first and last elements
So....this is just 5! = 120 possible paths
1. Each of the digits 1,2,3,4,5,6 is used exactly once in forming the 3digit integers X and Y. How many possible values of X+Y are there if XY=111?

I'm not as sure of this one but here's my guess:  
We can let Y < X without changing the number of possible values of X + Y 

For every value of Y such that X  Y = 111 there is one unique corresponding value of X  
So if we just count the possible values of Y, that will be the number of possible values of X + Y 

And since Y < X , the possible digits of Y are only: 1, 2, 3, 4, 5  
(If 6 was a digit of Y then there is not a possible corresponding X which makes X  Y = 111) 

the number of possibilities of Y = 5 * 4 * 3 = 60  
So there are 60 different pairs each with a different sum. 

And so there are 60 different possibilities of X + Y 
To help see this...here is the beginning of the list of pairs:
123, 234
124, 235
125, 236
132, 243
134, 245
135, 246
142, 253
143, 254
.
.
.
4. What fraction of all the 10digit numbers with distinct digits have the property that the sum of every pair of neighboring digits is odd?
Here's my best attempt :
We either have
Even Odd Even Odd Even Odd Even Odd Even Odd or
Odd Even Odd Even Odd Even Odd Even Odd Even
In the first case we have the following choices for each position....note that the first position cannot be 0, so we only have 4 evens to choose from
4 * 5 * 4 * 4 * 3 * 3 * 2 * 2 * 1 * 1 = 5! * 4 * 4! = 11520 integers
In the second case we have
5 * 5 * 4 * 5 * 3 * 3 * 2 * 2 * 1 * 1 = 5! * 5! = 14400 integers
So 11520 + 14400 = 25920
And we have 10^10  10^9 = 9000000000 ten digit integers
So....the fraction is 25920 / 9000000000