Saturday

March 28, 2015

March 28, 2015

Posted by **Francesca** on Tuesday, April 12, 2011 at 10:55am.

• Using the Principle of Inclusion-Exclusion, find the number of integers between 1 and 2000 (inclusive) that are divisible by at least one of 2, 3, 5, 7.

A = {n| 1 ≤ n ≤ 2000, 2 |n}

B = {n| 1 ≤ n ≤ 2000, 3 |n}

C = {n| 1 ≤ n ≤ 2000, 5 |n}

D = {n| 1 ≤ n ≤ 2000, 7 |n}

|A| = [2000/2] = 1000

|B| = [2000/3] = 666

|C| = [2000/5] = 400

|D| = [2000/7] = 285

Also, I'm kind of confused with how to approach this problem:

• John Sununus was once the governor of New Hampshire, and his name reminds one of the authors of a palindrome (a words which is spelt the same way forwards as backwards, such as SUNUNUS).

How many seven-letter palindromes (not necessarily real words) begin with the letter S and contain at most three letter?

Thanks for any helpful replies!

- Discrete Math -
**Count Iblis**, Tuesday, April 12, 2011 at 1:30pmYou can add up the |A| , |B| etc. and then say that you have double counted the elements that are in both A and B, A and C, etc. etc. So, you subtract these. But then you have counted the elements that are in the intersection of A, B and C a total of zero times, because they are counted in

|A| one time

|B| one time

|C| one time

|Intersection of A and B| minus 1 time

|Intersection of A and c| minus 1 time

|Intersection of B and C| minus 1 time

So, you need to add the number of elements in the intersection of A, B and C, and thus also all other pairs of intersections of 3 sets.

Proceeding in this way, you will find the usual inclusion-exclusion rule. You can derive this more formally, by first computing the number of elements that are not divisible by any of the numbers.

You can then directly apply the inclusion-exclusion formula in its usual formulation, so you then find that this is the total number of elements minus |A|, minus |B|,.. plus |A intersection B|, etc. etc. etc.

This is then also N minus the number of elements divisible by either of the numbers, you get the same result.

- Discrete Math -
**Francesca**, Tuesday, April 12, 2011 at 3:08pmOkay I continued the first problem:

|A ∩ B| = [2000/6] = 333

|B ∩ C| = [2000/15] = 133

|C ∩ D| = [2000/35] = 57

|A ∩ D| = [2000/14] = 142

|A ∩ B ∩ C ∩ D| = [2000/210] = 9

1000 + 666 + 400 + 285 - 333 - 133 - 57 - 142 + 9 = 1695 <--Answer

- Discrete Math -
**James**, Tuesday, April 12, 2011 at 4:52pmI too am having trouble with that SUNUNUS problem, any idea how to approach it?

- Discrete Math -
**Francesca**, Tuesday, April 12, 2011 at 7:15pmAny suggestions?

- Discrete Math -
**James**, Tuesday, April 12, 2011 at 8:16pmThe only thing I can think of is starting with the method from the first problem and plugging in the numbers? How have you started it?

- Discrete Math -
**Francesca**, Tuesday, April 12, 2011 at 8:33pmTo be honest I haven't started yet, but your method sounds like a step in the right direction. . .I'll play around with it for a little and see what I get. . .If you figure out anything post. Hopefully someone who knows something will post cuz I'm lost

- Discrete Math -
**James**, Tuesday, April 12, 2011 at 9:01pmYeah same here and it's due tomorrow lol. So far I got this for #2 but I think it's wrong.

A.) 1 x 26 x 25 x 24 = 15,600

B.) 26 x 25 x 24 x 23 = 14,398

C.) 25 x 24 x 23 = 13,800

- Discrete Math -
**MathMate**, Wednesday, April 13, 2011 at 7:41pmFrancesca,

INCLUSION/EXCLUSION PRINCIPLE:

Your post: 2011-04-12T15:08

I believe the present problem can be solved using the principle to calculate the count of numbers NOT divisible by ANY of the 4 factors (2,3,5,7). Subtract that from 2000 to get the count of numbers divisible by at least one of the four factors.

The inclusion/exclusion principle works as follows:

For a two set case, and using u to denote the cardinality of the universal set (2000),

a=count of numbers divisible by 2,

b=count of numbers divisible by 3, then

ab=count of numbers not divisible by 2*3 (i.e. |A∩B|)

Count of numbers NOT divisible by either factor (2 or 3) is:

N̅=u-(a+b)+ab

=2000-(1000+666)+333

=667

For the case of 3 factors,

N̅=u-(a+b+c)+(ab+bc+ca)-(abc)

For the case of 4 factors:

N̅=u - (a+b+c+d) + (ab+ac+ad+bc+bd+cd) - (abc+abd+bcd+acd) +abcd

where

a=count of numbers divisible by 2

ab=count of numbers divisible by 2*3

abc=count of numbers divisible by 2*3*5

abcd=count of numbers divisible by 2*3*5*7

If you proceed this way, you should get the count of numbers NOT divisible by any of the 4 factors as:

N̅=2000-2351+960-160+9=458

So the count of numbers divisible by AT LEAST one of the four factors 2,3,5,7 is 2000-458=1542.

If you have questions, please post.

**Answer this Question**

**Related Questions**

Discrete Math - 1.Using the principle of inclusion-exclusion find the numbers of...

discrete math - Find how many positive integers with exactly four decimal digits...

Math - What is the smallest natural number divisible by each of the integers 1-...

DISCRETE MATH - HOW MANY POSITIVE INTEGERS LESS THAN 1000 A.are divisible by ...

Easy Math - The number of positive integers that are less than 500 and that are ...

Maths - How many integer solutions of x1 + x2 + … x6 = 20 satisfies 1 ≤ xi...

Math - What are the positive integers between 1 and 100 that are only divisible ...

math - Using the numbers 1 through 9 with no repeats, find a 9 number such that...

algebra - how would i find least common multiple of 20, 50 and 3 ? okay well you...

Math - The question is this: You know that a number is divisible by 6 if it is ...