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1. Each of the digits 1,2,3,4,5,6 is used exactly once in forming the 3-digit integers X and Y. How many possible values of X+Y are there if |X-Y|=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 10-digit 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!

Jun 26, 2019

#1
+4

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?

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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

 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

Jun 26, 2019
#4
+1

Thanks, that works!!

sudsw12  Jun 26, 2019
#2
+1

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   Jun 26, 2019
edited by CPhill  Jun 26, 2019
#3
+1

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

sudsw12  Jun 26, 2019
#5
+1

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.

sudsw12  Jun 26, 2019
#6
+2

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 ????   CPhill  Jun 26, 2019
#7
+1

OK....we agree  !!!!   CPhill  Jun 26, 2019
#8
+1

yeah that's right.

sudsw12  Jun 26, 2019
#9
+2

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 (straight-line) 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   Jun 26, 2019
#10
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This answer is incorrect however, do you have any ideas why? Is it maybe because it is a circle like the previous question answered?

sudsw12  Jun 26, 2019
#11
0

Don't know..   CPhill  Jun 26, 2019
#12
+4

1. Each of the digits 1,2,3,4,5,6 is used exactly once in forming the 3-digit integers X and Y. How many possible values of X+Y are there if |X-Y|=111?

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 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

.

.

.

Jun 26, 2019
edited by hectictar  Jun 26, 2019
#13
+1

Nice job, hectictar   !!!   CPhill  Jun 26, 2019
#14
+2

4. What fraction of all the 10-digit 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   Jun 26, 2019
#15
+1

Hmm.. that doesn't work... I simplified that fraction to 9/3125000 and it was wrong. A hint is How can you build a number in which every pair of neighboring digits has an odd sum?

sudsw12  Jun 26, 2019
#16
0

Why don't you do your own homework?

Jun 27, 2019