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25 Jan 2021, 03:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
72% (02:57) correct
28%
(02:54)
wrong
based on 189
sessions
History
Date
Time
Result
Not Attempted Yet
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. \(2^{87}*3^{59}*5^{13}*7^{11}\)
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
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. \(2^{87}*3^{59}*5^{13}*7^{11}\)
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
25 Jan 2021, 03:40
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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
exponent of 7 = (63-49)+1 = 15
C
Bunuel
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 he 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. \(2^{87}*3^{59}*5^{13}*7^{11}\)
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
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. \(2^{87}*3^{59}*5^{13}*7^{11}\)
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
shubhamjha22
Current Student
Joined: 11 Dec 2016
Last visit: 11 May 2025
Posts: 27
Own Kudos:
Given Kudos: 60
Location: India
Schools: ISB '23 (A) IIMA PGPX'23 (I)
GMAT 1: 740 Q50 V41
GPA: 3
25 Jan 2021, 06:04
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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
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
