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23 Aug 2022, 01:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
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Question Stats:
66% (00:56) correct
34%
(01:04)
wrong
based on 100
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\(20! + 21\) is NOT a multiple of which of the following?
I. 21
II. 7
III. 2
A. I only
B. III only
C. I and II
D. I and III
E. I, II, and III
Are You Up For the Challenge: 700 Level Questions
I. 21
II. 7
III. 2
A. I only
B. III only
C. I and II
D. I and III
E. I, II, and III
Are You Up For the Challenge: 700 Level Questions
23 Aug 2022, 03:22
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To see if a number is a multiple of \(20!+21\), one needs to see if it can divide \(20!+21\). The obvious issue here is addition sign, which prevents straightforward division. In order to divide the equation, one needs to find a common factor on both sides of the addition sign.
\(20!\) is the product of all numbers from 1 to 20, and the factors of \(21\) are \(1, 3, 7, 21\). This means that the only possible multiples of \(20!+21\) are the factors of 21.
Which means that from:
I. 21
II. 7
III. 2
Only 2 will not be a multiple of \(20!+21\).
Answer B
In more detail:
20!+21 can be written as
1*2*3*4*5*6*7*.....*20 + 21.
I. 21
Rewrite 1*2*3*4*5*6*7*.....*20 + 21 as: 1*2*3*4*5*6*7*.....*20 +(3*7)
Take out a common factor of (3*7) to get: (3*7)[1*2*1*4*5*6*1*.....*20 +1]
Now as the addition sign is isolated inside of the bracket, one can divide by 21.
II. 7
Similarly as above, rewrite the equation as: 1*2*3*4*5*6*7*.....*20 +(3*7)
Take out a common factor of 7: 7[1*2*3*4*5*6*1*.....*20 + 3*1]
Which means that \(20!+21\) is divisible by 7.
III. 2
1*2*3*4*5*6*7*.....*20 + 21
The issue here is that 21 is not an even number, and thus there is no way to take 2 out in order for \(20!+21\) to be divisible by 2. (On a side note from 5! onwards all factorials end in 0, which means that adding 21 to 20! will immediately make the number odd and therefore not have 2 as a factor)
\(20!\) is the product of all numbers from 1 to 20, and the factors of \(21\) are \(1, 3, 7, 21\). This means that the only possible multiples of \(20!+21\) are the factors of 21.
Which means that from:
I. 21
II. 7
III. 2
Only 2 will not be a multiple of \(20!+21\).
Answer B
In more detail:
20!+21 can be written as
1*2*3*4*5*6*7*.....*20 + 21.
I. 21
Rewrite 1*2*3*4*5*6*7*.....*20 + 21 as: 1*2*3*4*5*6*7*.....*20 +(3*7)
Take out a common factor of (3*7) to get: (3*7)[1*2*1*4*5*6*1*.....*20 +1]
Now as the addition sign is isolated inside of the bracket, one can divide by 21.
II. 7
Similarly as above, rewrite the equation as: 1*2*3*4*5*6*7*.....*20 +(3*7)
Take out a common factor of 7: 7[1*2*3*4*5*6*1*.....*20 + 3*1]
Which means that \(20!+21\) is divisible by 7.
III. 2
1*2*3*4*5*6*7*.....*20 + 21
The issue here is that 21 is not an even number, and thus there is no way to take 2 out in order for \(20!+21\) to be divisible by 2. (On a side note from 5! onwards all factorials end in 0, which means that adding 21 to 20! will immediately make the number odd and therefore not have 2 as a factor)
