A computer program when fed with a real number x, generates another nu

Question

A computer program when fed with a real number x, generates another number y such that y is the greatest integer less than or equal to x. Homer inputs the values  \sqrt{1}, \ \sqrt{2}, \ \sqrt{3}, \ \sqrt{4}, \ ..., \ \sqrt{100}  sequentially and finds the product of all the numbers that the computer give as output. The product is:

A.  2^{79}*3^{61}*5^{21}*7^{13}

B.  2^{81}*3^{59}*5^{12}*7^{18}

C.  2^{88}* 3^{58}*5^{12}*7^{15}

D.  2^{87}*3^{59}*5^{13}*7^{12}

E.  2^{87}*3^{59}*5^{13}*7^{11}

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nick1816 · Answer

We get the multiple of 7 only when the input is from  \sqrt{49}  to  \ \sqrt{63} .

exponent of 7 = (63-49)+1 = 15

C

shubhamjha22 · Answer

Input: x= 1^(1/2), 2(1/2), 3^(1/2).....
g(x)= Y = greatest integer less than equal to X

g(1^(1/2))= 1; g(2^(0.5)= 1; g(3^0.5)= 1 
so product of outputs of x from 1^0.5 to 3^0.5= 1 * 1* 1= 1^3

Similarly for next set of numbers
g(2^0.5)= 2; g(5^0.5)= 2; g(6^0.5)=2......g(8^0.5)= 2
there for product = 2*2*2*2*2= 2^5

The output value will change to next integer value every time with reach a perfect square (4,9,16,25....100)
and the powers of the integers increases in odd number 
==> 1 repeats 3 time= 1^3
2 repeats 5 times= 2^5
3 repeats 7 times => 3^7
.
.
.
So on,
Therefore the product of all outputs= 1^3 X 2^5 X 3^7 X 4^9 X 5^11....... 9 ^ 19 X 10
Simplifying with prime factors only: 2^88 X ..... (I didnt calculate all since there is only one option where 2 has power of 88, OPTION C

BrushMyQuant · Answer

↧↧↧ Detailed Video Solution to the Problem ↧↧↧

https://www.youtube.com/watch?v=6Xbjtq4WDQo

f( \sqrt{1} ) = f(1) = Greatest integer less than or equal to 1 = 1 
f( \sqrt{2} ) = f(1.414) = Greatest integer less than or equal to 1.414 = 1
f( \sqrt{3} ) = f(1.731) = 1
f( \sqrt{4} ) = f(2) = 2
 . 
 All values will be equal to 2 till we reach f( \sqrt{9} ) 
f( \sqrt{9} ) = f(3) = 3
.
 All values will be 3 till we reach f( \sqrt{Next Square} ) = f( \sqrt{16} ) 
f( \sqrt{16} ) = f(4) = 4
.
 All values will be 4 till we reach f( \sqrt{25} ) 
f( \sqrt{25} ) = f(5) = 5
.
 All values will be 5 till we reach f( \sqrt{36} ) 
f( \sqrt{36} ) = f(6) = 6
.
 All values will be 6 till we reach f( \sqrt{49} ) 
f( \sqrt{49} ) = f(7) = 7
.
 All values will be 7 till we reach f( \sqrt{64} ) 
f( \sqrt{64} ) = f(8) = 8
.
 All values will be 8 till we reach f( \sqrt{81} ) 
f( \sqrt{81} ) = f(9) = 9
.
 All values will be 9 till we reach f( \sqrt{100} ) 
f( \sqrt{100} ) = f(10) = 10
 .

Now, look at the answer choices. Power of 7 is unique in all the options. 
 Let's find the power of 7

Value of the function is 7 from 49 to 63 (Inclusive) 
=> Number of values = 63 - 49 + 1 = 15 values
=> We will have  7^{15}  in the answer

So,  Answer will be C 
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters

https://www.youtube.com/watch?v=b7JoMX0gk8M

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A computer program when fed with a real number x, generates another nu

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A computer program when fed with a real number x, generates another nu

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