Question

let u be set with numbers of element in it is 2009 a subset of u with n(a) equals to 1681 and out of this 1681 elements exactly 1075 element belong to a subset b of u if n(a-b)=m^2 + P1 P2 P3 for some positive integer m and distinct prime P1 P2 P3 then for least m. FindP1 + p2 + P3

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

The least value of m is 2.

The sum of  $P_{1} +P_{2} +P_{3}$ is 52.

Step-by-step explanation:

Using the concept of sets and subsets,

Given,U be set with numbers of element in it is 2009 a subset of U with n(A) equals to 1681 and out of this 1681 elements exactly 1075 elements belong to a subset B of U

Number of elements in U, n(U) = 2009

Number of elements in A, n(A) = 1681

Number of elements in B, n(B) = 1075

• (A - B) implies the element that are in A but not in B.

n(A - B) = n(A) - n(B) = 1681 - 1075

n(A - B) = 606

• Given n(A - B) = $m^{2}+P_{1} P_{2} P_{3}$ where $P_{1}, P_{2}, P_{3}$ are distinct prime numbers and m is a positive integer (m > 0).

$m^{2}+P_{1} P_{2} P_{3}$ = 606

By trial and error method,

Since m is a positive integer

• let's substitute m = 1 in the above equation

$P_{1}P_{2}P_{3}$ = 605

Substituting different prime numbers for $P_{1}$, 605 is divisible by 11.

But that leaves $P_{1} = 11, P_{2} = 11, P_{3} = 5$

As $P_{1}, P_{2}, P_{3}$ must be distinct, this m ≠ 1.

• let's substitute m = 2 in the above equation

$P_{1}P_{2}P_{3}$ = 602

• Substituting different prime numbers for $P_{1}$, 602 is divisible by 2 and leaves 301.

• 301 is divisible by 7 and leaves 43.

That leaves $P_{1} = 2, P_{2} = 7, P_{3} = 43$

All are prime numbers.

Therefore $P_{1} +P_{2} +P_{3}$ = 2 + 7 + 43 = 52

Know more about sets:

Examples of sets

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Subsets

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