How Do You Calculate the Cardinality of a Special Set?

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The cardinality of the set A, which includes all natural numbers from 1 to 6000 that are divisible by 3 or 7 but not by 105, is calculated to be 2548. This is derived from counting 2000 multiples of 3 and 857 multiples of 7, then subtracting the 24 multiples of 105 that overlap. For the second part of the problem, it is established that among the multiples of 7, approximately 190 will yield a remainder of 2 when divided by 3, based on the identified pattern of remainders.

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[Resolved][Sets] Cardinality problem

Homework Statement



let A be a Set of all natural numbers from 1 to 6000 that are divsible by 3 or 7 but not 105.

1.What is the cardinality of A?

2.How many numbers in A give 2 as the remained of division by 3.

Homework Equations


The Attempt at a Solution



1. My thinking was like this

How many multiples of 3 are from 1 to 6000. Well 6000/3 = 2000. How many multiples of 7? Well 6000/7 = 857. How many multiples of 7 are also multiples of 3? I would say 2000 / 7 = 285 that is in that 2000 multiples of 3 , 285 multiples are multiples of 7 aswell. That gives

+2000 multiples of 3
+857 multiples of 7 -285 multiples of 7 that are multiples of 3 aswell

that gives a total of 2572 numbers

now to put !divison by 105 into picture. Soo they must not be multiples of 105. We just substract multiples of 105, right? that is 2572 / 105 = 24, that is

the cardinality of A is 2572 - 24 = 2548.

Is any of the above correct or even close to a solution?

Now for the second part my thinking is like this, what do we have in those 2548 numbers, well we have something like this multiples of 3 and 7 but not 105. Now multiples of 3 will ofcorse not give the proper remainder, now for 7,

7 // remainders 3*2 + 1 rem 1
14 // rem 2
21 // 0
28 // 1
35 // 2
42 // 0
49 // 1
56 // 2
63 // 0

Now I noticed a pattern that multiples of 7 that give remainder 2 have a pattern, meanin starting from 2nd multiple which gives 2 as a remainder then every 2*k + 1 multiple gives 2 as the remainder. Meaning if I take 3 successive multiples of 7 one of them is guarantted to give 2 as the remainder. Meaning if there are 572 multiples of 7 (not multiples of 3) then there are 572 / 3 = 190 that will give 2 as the remainer.

Any good? At least close?
 
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I think I got it. Thanks anyway.
 

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