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Let x be defined as the number of positive perfect squares less than
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20 Jul 2017, 20:43
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Let @x@ be defined as the number of positive perfect squares less than x Let #x# be defined as the number of primes less than @x@ If #x# = 5, what is the value of #(@x@)#? (A) 7 (B) 5 (C) 4 (D) 2 (E) 1
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Let x be defined as the number of positive perfect squares less than
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02 Apr 2020, 23:35
preetamsaha wrote: pushpitkc VeritasKarishma chetan2u Bunuel If #x# = 5, then @x@ = 12 then @x@ indicates there are 12 no perfect squares less than x , why are we considering the no of perfect squares less than 12 ? I am not getting the problem . can u please explain it ? Hi, VeritasKarishma has explained it very well. Just a point from my side on why you could be going wrong.. Firstly the question is not simple and easy one by any standards, and reading too much into x can get you wrong. Step I : #x#=5 So 5 primes less than @x@. Now five primes are 2,3,5,7 and 11, but the next prime is 13. Thus @x@=12 or 13 but NOT 14, as then primes would become 6. Also solved above...... @x@=12 or 13 Step II : @x@ as 12 makes #(@x@)# = #12#. Now FORGET what you did earlier to get #x#=5. What you know now is that you are working with a new value of x and that is x=12.. Step III : #x# is the number of primes less than @x@ #12# is the number of primes less than @12@, but we do not know what is @12@Step IV : @x@ is the number of positive perfect squares less than x. So @12@ is the number of positive perfect squares less than 12. Thus values are 1, 4, and 9, that is 3 values. Therefore, @12@=3. Step V : Back to step III #x# is the number of primes less than @x@ #12# is the number of primes less than @12@ or 3. Possible value is only 2. Hence #12#=1 E
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Let x be defined as the number of positive perfect squares less than
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20 Jul 2017, 21:18
Given: #x# = 5 #x# is defined as the number of primes less than @x@ where @x@ is the number of positive perfect squares less than x If #x# = 5, then @x@ = 12 which has 5 primes(2,3,5,7,11) less than @x@ and @x@ is number of positive perfect squares less than 12, which is 3(1,4,9) Now, we have been asked to find the value of #(@x@)#  which is the number of primes less than 3 Therefore, #(@x@)# or #3# = 1(Option E) as only 2 is a prime lesser than 3.
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Re: Let x be defined as the number of positive perfect squares less than
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21 Mar 2018, 20:02
pushpitkc wrote: Given : #x# is defined as the number of primes less than @x@ where @x@ is the number of positive perfect squares less than x #x# = 5
If #x# = 5, then @x@ = 12 which has 5 primes(2,3,5,7,11) less than @x@ and @x@ is number of positive perfect squares less than 12, which is 3(1,4,9)
Now, we have been asked to find the value of #(@x@)# or #3# which is the number of primes less than 3 = 1(Option E) as only 2 is a prime lesser than 3. Hi pushpitkcAs per the question statement \(Xmin = 145, @x@ = 12 & #x# = 5.\) Now, As per your explanation  If #x# = 5, then @x@ = 12 which has 5 primes(2,3,5,7,11) less than @x@ and @x@ is number of positive perfect squares less than 12, which is 3(1,4,9)@x@ is number of positive perfect squares less than x but in your above highlighted solution you are considering x itself as 12 for finding the number of @x@. But that is wrong because x will be fixed to minimum of 145. Please correct me if my understanding is wrong.



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Re: Let x be defined as the number of positive perfect squares less than
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01 Apr 2020, 14:03
Bunuel wrote: Let @x@ be defined as the number of positive perfect squares less than x Let #x# be defined as the number of primes less than @x@ If #x# = 5, what is the value of #(@x@)#?
(A) 7 (B) 5 (C) 4 (D) 2 (E) 1 Since #x# = 5, we see that there are 5 primes less than @x@. Since the 5th prime is 11 and the 6th prime is 13, we see that @x@ must be 12 or 13. Therefore, we have: #(@x@)# = #12# or #13# Since #12# is the number of primes less than @12@ and @12@ is the number of positive perfect squares less than 12, @12@ = 3 (since there are 3 perfect squares less than 12: 1, 4 and 9). Notice that since @13@ = 3 also, #13# is equal to the number of primes less than 3 as well. So, #12# = #13# is the number of primes less than 3, and since there is only 1 prime (namely, 2) less than 3, the answer is 1. Answer: E
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Re: Let x be defined as the number of positive perfect squares less than
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02 Apr 2020, 09:49
pushpitkc VeritasKarishma chetan2u Bunuel If #x# = 5, then @x@ = 12 then @x@ indicates there are 12 no perfect squares less than x , why are we considering the no of perfect squares less than 12 ? I am not getting the problem . can u please explain it ?



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Re: Let x be defined as the number of positive perfect squares less than
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02 Apr 2020, 22:48
Bunuel wrote: Let @x@ be defined as the number of positive perfect squares less than x Let #x# be defined as the number of primes less than @x@ If #x# = 5, what is the value of #(@x@)#?
(A) 7 (B) 5 (C) 4 (D) 2 (E) 1 Focus on how the operators are defined: Positive perfect squares = 1, 4, 9, 16, 25 etc So @2@ = 1 (only 1 perfect square less than 2 i.e. 1) @10@ = 3 (3 perfect squares less than 10 i.e. 1, 4 and 9) @30@ = 5 (5 perfect squares less than 30 i.e. 1, 4, 9, 16 and 25) etc Primes = 2, 3, 5, 7, 11 etc #2# = Number of primes less than @2@ = Number primes less than 1 = 0 #10# = Number of primes less than @10@ = Number primes less than 2 = 1 (Only 1 such prime i.e. 2) #30# = Number of primes less than @30@ = Number primes less than 5 = 2 (2 such primes i.e. 2 and 3) etc Now let us look at the question: Given #x# = 5 #x# = Number of primes less than @x@ = 5 First 5 primes are 2, 3, 5, 7, 11 So @x@ = 12 or 13. The moment @x@ becomes 14, we have 6 primes less than it. Find #(@x@)# #(@x@)# = #12# or #13# #12# = Number of primes less than @12@ = Number primes less than 3 = 1 #13# = Number of primes less than @13@ = Number primes less than 3 = 1 Answer (E)
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Re: Let x be defined as the number of positive perfect squares less than
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03 Apr 2020, 12:57
VeritasKarishma chetan2u I'm thankful to your posts, but still the following thinking is spinning into my head . #x#=5 = { 2, 3, 5, 7, 11} = 5 = Number primes less than 12 = Number of primes less than @145@ [ @145@ = 12 (12 perfect squares less than 145 i.e. 1, 4, 9, 16, 25, ..., 144) ] so, #(@x@)# #(@145@)# =5 where am I wrong ?



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Re: Let x be defined as the number of positive perfect squares less than
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03 Apr 2020, 19:21
preetamsaha wrote: VeritasKarishma chetan2u I'm thankful to your posts, but still the following thinking is spinning into my head . #x#=5 = { 2, 3, 5, 7, 11} = 5 = Number primes less than 12 = Number of primes less than @145@ [ @145@ = 12 (12 perfect squares less than 145 i.e. 1, 4, 9, 16, 25, ..., 144) ] so, #(@x@)# #(@145@)# =5 where am I wrong ? Hi, You are not understanding the wordings completely. Would request you to go through above solutions and then ask any query if you have. I am sure if you go over the solutions given slowly, you will understand.
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Re: Let x be defined as the number of positive perfect squares less than
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Re: Let x be defined as the number of positive perfect squares less than
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06 Apr 2020, 01:41
preetamsaha wrote: VeritasKarishma chetan2u I'm thankful to your posts, but still the following thinking is spinning into my head . #x#=5 = { 2, 3, 5, 7, 11} = 5 = Number primes less than 12 = Number of primes less than @145@ [ @145@ = 12 (12 perfect squares less than 145 i.e. 1, 4, 9, 16, 25, ..., 144) ] so, #(@x@)# #(@145@)# =5 where am I wrong ? Not correct. You haven't understood the defn of #x#. #x# is the number of primes less than @x@, not the number of primes less than x. #x# = 5 = Number of primes less than @x@ = Number of primes less than 12 So @x@ = 12 x = 145 or 146 or 147 till 169 Quote: so, #(@x@)# #(@145@)# =5 #(@145@)# = #(12)# = Number of primes less than @12@ (not number of primes less than 12) @12@ = Number of perfect squares less than 12 = 3 #12# = Number of primes less than 3 = 1
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Re: Let x be defined as the number of positive perfect squares less than
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06 Apr 2020, 01:42
preetamsaha wrote: VeritasKarishma chetan2u I'm thankful to your posts, but still the following thinking is spinning into my head . #x#=5 = { 2, 3, 5, 7, 11} = 5 = Number primes less than 12 = Number of primes less than @145@ [ @145@ = 12 (12 perfect squares less than 145 i.e. 1, 4, 9, 16, 25, ..., 144) ] so, #(@x@)# #(@145@)# =5 where am I wrong ? Also note that you don't need to get the values of x here. Getting @x@ is enough to proceed with what is asked. Check my solution.
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Re: Let x be defined as the number of positive perfect squares less than
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06 Apr 2020, 06:17
VeritasKarishma ok . now i have understood it. thanks .




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