S is the sum of the reciprocals of the squares of the prime numbers be

ans is b
as the prime nos are 23,29,31,37 from the conditions

so sum = [(1/23)^2+(1/29)^2+(1/31)^2+(1/37)^2]

is approx to Sum=[(1/400)+(1/900)+(1/900)+(1/1600)]
= 5 *10^(-3) approx

S is the sum of the reciprocals of the squares of the prime numbers between 19 and 41, exclusive. Which of the following is closest to the value of S?

A. \(2∗10^−4\)

B. \(5∗10^−3\)

C. \(2.5∗10^−2\)

D. \(5∗10^−2\)

E. \(2∗10^−1\)

\(S = \frac{1}{23^2} + \frac{1}{29^2} + \frac{1}{31^2} + \frac{1}{37^2}\)
\(\frac{1}{23^2} ≈ \frac{1}{625}\), \(\frac{1}{29^2} ≈ \frac{1}{900}\), \(\frac{1}{31^2} ≈ \frac{1}{900}\), \(\frac{1}{37^2} ≈ \frac{1}{1600}\)

\(S ≈ \frac{1}{625} + \frac{1}{900} + \frac{1}{900} + \frac{1}{1600}\)
\(S ≈ \frac{1}{600} + \frac{1}{900} + \frac{1}{900} + \frac{1}{1600}\)
\(S ≈ \frac{1}{100}(\frac{1}{6} + \frac{1}{9} + \frac{1}{9} + \frac{1}{16})\)
\(S ≈ \frac{1}{100}(\frac{65}{144})\)
\(S ≈ \frac{1}{100}*0.45\)
\(S ≈ 4.5*10^-3\)

Answer B.

DhruvS

Your approach is correct (pretty much similar to my approach).

Since, the numbers in the list are close to the mean and options are almost differed by factor of 10, this approach will definitely work.

I guess B ?

S is the sum of the reciprocals of the squares of the prime numbers between 19 and 41, exclusive. Which of the following is closest to the value of S?
S = 1/ (19*19)+1/ (23*23) + 1/ (29*29) + 1/ (31*31) + 1/ (37*37) + 1/(41*41)
=> 1/(41*41) <= S <= (1/(19*19))
=> 0.005 <= S <=0.002

only B is closer that is 5 * 10 ^-3

\(S=\frac{1}{23^2}\)+\(\frac{1}{29^2}\)+\(\frac{1}{31^2}\)+\(\frac{1}{37^2}\)
If we take a look at options, we can see that except for C and D, rest vary by at least a factor of 10.

The max value of \(\frac{1}{23^2}\) is less than \(\frac{1}{20^2}\).
Estimating the max value of S, we get \(4*\frac{1}{23^2}\) < \(4*\frac{1}{20^2}\) => \(S < 10^{-2}\)

Similarly, the min value of \(\frac{1}{37^2}\) is greater than \(\frac{1}{40^2}\)
Estimating the min value of S, we get \(4*\frac{1}{37^2}\) > \(4*\frac{1}{40^2}\) => \(S > 0.25*10^{-2}\)

=> \(0.25*10^{-2} < S < 10^{-2}\)
If we compare option C and D with the range, we can see that, they vary by at least a factor of 10. Optimizing the range further, we get
\(2.5*10^{-3} < S < 10*10^{-3}\). This range is closer to option B.

Ans : B

IMO B

S = (1/23)^2 + (1/29)^2 + (1/31)^2 + (1/37)^2
~ (1/25)^2 + (1/25)^2 + (1/25)^2 + (1/25)^2
~ 4 * (1/25)^2
~ 64 * 10^ (-4)

B is nearest i.e. 5 * 10^(-3)

Ans B
Prime no btw 19 and 41= 23 29 31 37

Reciprocal and square of smallest will be biggest= 1/23*23 = 1.8* 10(-3), others will be smaller and addition to this.
Only one option choice with 10(-3)

Alternatively we can check 1/20*20= .0025= 2.5* 10(-3), eliminate ACDE,

Answer B

Posted from my mobile device

nick1816 can we not solve it like below (please guide):

List of prime numbers: (23,29,31,37)
Since these are 4 numbers of prime numbers.
Hence Mean of middle 2 terms = (29+31)/2
60/2 = 30.
[\((\frac{1}{30})\)^2]*4
\(~ 4.4 * 10^-3\)
= 5*10^3
Hence B

The answer is A. Since if we first divide 1 by the prime numbers that appear between 19 and 41 exclusive we get 1/20+1/30+1/30+1/40= 0.13
If we square 0.13 we get 0.0169 which is approximately equal to 2*10^-4

Posted from my mobile device

Estimating 1/23^2 + 1/29^2 + 1/31^2 + 1/37^2,
= ~1/20^2 + ~1/30^2 + ~1/30^2 + ~1/40^2
= 1/400 + 2/900 + 1/1600
= 0.0025 + ~0.0020 + ~0.0006
= 0.005
= ~5 * 10^-3

FINAL ANSWER IS (B)

Posted from my mobile device

See the attachment. IMO B

Thank you so much for confirming

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