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13 Apr 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:
45%
(medium)
Question Stats:
70% (01:56) correct
30%
(01:59)
wrong
based on 276
sessions
History
Date
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Not Attempted Yet
If sequence \(x_1, \ x_2, \ x_3, \ ..., \ x_n\) is such that \(x_{n+1}\) is 5 more than \(x_n\) for \(n > \)0, does \(x_{99}\) have a unit’s digit of 9?
(1) The 100th term plus 1 is a multiple of 5.
(2) The first term is 4.
(1) The 100th term plus 1 is a multiple of 5.
(2) The first term is 4.
13 Apr 2022, 07:03
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Bunuel
Given: \(x_1, \ x_2, \ x_3, \ ..., \ x_n\) is such that \(x_{n+1}\) is 5 more than \(x_n\) for \(n > \)0
Let's list a few terms to get a better idea of the sequence.
First, let k = the first term (\(x_1\)) to get:
\(x_1 = k\)
\(x_2 = k+5\)
\(x_3 = k+5+5\)
\(x_4 = k+5+5+5\)
\(x_5 = k+5+5+5+5\) notice that the number of 5's in the sum is 1 less than the term number
This means \(x_n = k+5(n-1)\)
Target question: Does \(x_{99}\) have a unit’s digit of 9?
Statement 1: The 100th term plus 1 is a multiple of 5.
From our earlier work we know that \(x_{100} = k+5(100-1) = k + 495\)
So, it COULD be the case that \(k = 4\), which means \(x_{100} = 4 + 495 = 499\), and \(499 + 1 = 500\), which is a multiple of \(5\). In this case, \(x_{99} = x_{100} - 5 = 499 - 5 = 494\). So, the answer to the target question is NO, \(x_{99}\) does not have units digit 9
However, it could also be the case that, \(k = 9\), which means \(x_{100} = 9 + 495 = 504\), and \(504 + 1 = 505\), which is a multiple of \(5\). In this case, \(x_{99} = x_{100} - 5 = 504 - 5 = 499\). So, the answer to the target question is YES, \(x_{99}\) does have units digit 9
Since we can’t answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: The first term is 4.
In other words, \(k = 5\), which means \(x_99 = 4+5(99-1) = 494\)[/b].
So, the answer to the target question is NO, \(x_{99}\) does not have units digit 9
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Answer: B
Cheers,
Brent
15 Apr 2026, 08:12
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Deconstructing the Question
The sequence satisfies \(x_{n+1}=x_n+5\), so each term is \(5\) more than the previous one.
We are asked whether \(x_{99}\) has a units digit of \(9\).
The key idea is that adding \(5\) repeatedly makes the units digit alternate. Also,
\(x_n=x_1+5(n-1)\)
Step-by-step
Using the pattern,
\(x_{99}=x_1+5(98)=x_1+490\)
Since \(490\) ends in \(0\), \(x_{99}\) has the same units digit as \(x_1\).
Statement (1): The \(100\)th term plus \(1\) is a multiple of \(5\).
This means
\(x_{100}+1 \equiv 0 \pmod 5\)
so
\(x_{100}\equiv 4 \pmod 5\)
But numbers ending in \(4\) or \(9\) both satisfy this, so we still cannot determine the units digit of \(x_{99}\).
Statement (1): Insufficient
Statement (2): The first term is \(4\).
Then
\(x_{99}=4+490=494\)
Its units digit is \(4\), so the answer to the question is No.
Statement (2): Sufficient
Answer: B
The sequence satisfies \(x_{n+1}=x_n+5\), so each term is \(5\) more than the previous one.
We are asked whether \(x_{99}\) has a units digit of \(9\).
The key idea is that adding \(5\) repeatedly makes the units digit alternate. Also,
\(x_n=x_1+5(n-1)\)
Step-by-step
Using the pattern,
\(x_{99}=x_1+5(98)=x_1+490\)
Since \(490\) ends in \(0\), \(x_{99}\) has the same units digit as \(x_1\).
Statement (1): The \(100\)th term plus \(1\) is a multiple of \(5\).
This means
\(x_{100}+1 \equiv 0 \pmod 5\)
so
\(x_{100}\equiv 4 \pmod 5\)
But numbers ending in \(4\) or \(9\) both satisfy this, so we still cannot determine the units digit of \(x_{99}\).
Statement (1): Insufficient
Statement (2): The first term is \(4\).
Then
\(x_{99}=4+490=494\)
Its units digit is \(4\), so the answer to the question is No.
Statement (2): Sufficient
Answer: B
