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Ekland
Joined: 15 Oct 2015
Last visit: 30 Apr 2023
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Concentration: Finance, Strategy
Schools: Fordham '18 Schulich Jan '18 Desautels '19 Haskayne(Calgary) Alberta"19 Jindal '19 Owen '19 Olin '19 (I) Queens"18
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17 Apr 2021, 02:40
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
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Question Stats:
70% (02:16) correct
30%
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wrong
based on 138
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If \(P = 478+477!+476!\), which of the following must have a remainder when it divides into P?
I. \(239\)
II. \(10+9\)
III. \(5!-90\)
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
I. \(239\)
II. \(10+9\)
III. \(5!-90\)
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
17 Apr 2021, 04:54
Kudos
Bookmarks
Ekland
If \(P = 478+477!+476!\), which of the following must have a remainder when it divides into P?
I. \(239\)
II. \(10+9\)
III. \(5!-90\)
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
I. \(239\)
II. \(10+9\)
III. \(5!-90\)
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
\(478+477!+476!=478+476!(477+1)=478+478*476!=478(1+476!)\)
Now, 476! is product of all numbers till 476, so when we add 1 to 476!, the number is not divisible by all numbers till 476.
So 478(1+476!) will not be divisible by any number till 476 except the factors of 478 that are 2 and 239.
I. \(239\).....478=2*239, so the number is divisible by 239
II. \(10+9\)=19.....19 is not a factor of 478 and is less than 476. Thus it will leave a remainder
III. \(5!-90\)=120-90=30.... .19 is not a factor of 478 and is less than 476. Thus it will leave a remainder
II and III
E
23 Dec 2024, 01:45
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chetan2u
Ekland
If \(P = 478+477!+476!\), which of the following must have a remainder when it divides into P?
I. \(239\)
II. \(10+9\)
III. \(5!-90\)
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
I. \(239\)
II. \(10+9\)
III. \(5!-90\)
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
\(478+477!+476!=478+476!(477+1)=478+478*476!=478(1+476!)\)
Now, 476! is product of all numbers till 476, so when we add 1 to 476!, the number is not divisible by all numbers till 476.
So 478(1+476!) will not be divisible by any number till 476 except the factors of 478 that are 2 and 239.
I. \(239\).....478=2*239, so the number is divisible by 239
II. \(10+9\)=19.....19 is not a factor of 478 and is less than 476. Thus it will leave a remainder
III. \(5!-90\)=120-90=30.... .19 is not a factor of 478 and is less than 476. Thus it will leave a remainder
II and III
E
I can see how 476! will have a few tailing 0's. And adding a 1 to this makes it an odd number, and can be a prime too. But not sure how to prove the above statement.
