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If 3 < y < 150, for how many values of y is the square root of y twice

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Re: If 3 < y < 150, for how many values of y is the square root of y twice [#permalink]
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Hustler99 wrote:
y must be a perfect square
All the possible values of y are 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144

y must also be even since its square root must be a prime number*2
When y = 4, its sqrt will be 2 and that's not twice a prime number
When y = 16, its sqrt = 4 = 2*2
When y = 36 , its sqrt = 6 = 3*2
When y = 64, its sqrt = 8 and that's not twice a prime
When y = 100, sqrt = 10 = 5*2
When y = 144, sqrt = 12 and that's not twice a prime
So for 3 values of y the sqrt of y is twice a prime (B)

BUT Y>3
SO 02 ONLY
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Re: If 3 < y < 150, for how many values of y is the square root of y twice [#permalink]
Rabab36 wrote:
Hustler99 wrote:
y must be a perfect square
All the possible values of y are 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144

y must also be even since its square root must be a prime number*2
When y = 4, its sqrt will be 2 and that's not twice a prime number
When y = 16, its sqrt = 4 = 2*2
When y = 36 , its sqrt = 6 = 3*2
When y = 64, its sqrt = 8 and that's not twice a prime
When y = 100, sqrt = 10 = 5*2
When y = 144, sqrt = 12 and that's not twice a prime
So for 3 values of y the sqrt of y is twice a prime (B)

BUT Y>3
SO 02 ONLY

Y=36. Its sqr root is equal to 3.
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Re: If 3 < y < 150, for how many values of y is the square root of y twice [#permalink]
We can also think the other way round.
Given that square root of y = twice prime no.
implies y = square of (2 x prime no).
y = 4 x square of prime no.
lets check for 2, 3, 5, 7
y = 16 for prime no 2. Satisfies condition 3<y<150
y = 36 for prime no 3. Satisfies condition 3<y<150
y = 100 for prime no 5. Satisfies condition 3<y<150
y = 196 for prime no 7. Doesn't satisfy 3<y<150

therefore total 3 possible values for y
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Re: If 3 < y < 150, for how many values of y is the square root of y twice [#permalink]
Bunuel wrote:
If 3 < y < 150, for how many values of y is the square root of y twice a prime number?

A. Two
B. Three
C. Four
D. Five
E. Six

Given
$$\sqrt{y}=2*p$$, where p is a prime number.
Thus, $$y=(2p)^2=4*p^2$$
So $$4p^2>3……..p^2>\frac{3}{4}………..0<\frac{3}{4}<1$$…… Clearly $$p\geq 1$$, so smallest value is 2.
$$4p^2<150…..p^2<37.5……….6^2<37.5<7^2……..p\leq 6$$, so largest value is 5.
Possible values: 2, 3 and 5.

B
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If 3 < y < 150, for how many values of y is the square root of y twice [#permalink]
2
Kudos
Bunuel wrote:
If 3 < y < 150, for how many values of y is the square root of y twice a prime number?

A. Two
B. Three
C. Four
D. Five
E. Six

Are You Up For the Challenge: 700 Level Questions

We want y to be perfect square of a number which is 2*Prime. Let's check from the first prime number.

2*2 = 4. Its perfect square = 16 (lies in our range)
2*3 = 6. Its perfect square = 36 (lies in our range)
2*5 = 10. Its perfect square = 100 (lies in our range)
2*7 = 14. Its perfect square = 196 (doesn't lie in our range and neither will greater values)

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Re: If 3 < y < 150, for how many values of y is the square root of y twice [#permalink]
2
Kudos
So the question is how many numbers are there, where sqrt(Y)/2 is a prime number, 3<y<150.

so (2*2)^2=4. so 2 is an option. easy to understand (6*2)^2=144. so numbers from 2-6 fit. there are 3 prime numbers among these. So B is the answer.
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Re: If 3 < y < 150, for how many values of y is the square root of y twice [#permalink]
2
Kudos
Asked: If 3 < y < 150, for how many values of y is the square root of y twice a prime number?

Let the prime number be p

$$2p = \sqrt{y}$$
$$y = 4pˆ2$$

p = {2,3,5,7,9,11,13,,..}
pˆ2 = {4,9,25,49,81,121,..}
y = 4pˆ2 = {16,36,100} : 3 values

IMO B
Re: If 3 < y < 150, for how many values of y is the square root of y twice [#permalink]
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