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In sequence Q, the first number is 3, and each subsequent nu
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30 Jul 2018, 10:16
30 Jul 2018, 10:16
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Expert Reply
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Question Stats:
84% (01:11) correct
15% (01:30) wrong based on 46 sessions
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In sequence Q, the first number is 3, and each subsequent number in the sequence is determined by doubling the previous number and then adding 2. In the first 10 terms of the sequence, how many times does the digit 8 appear in the units digit?
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Re: In sequence Q, the first number is 3, and each subsequent nu
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31 Jul 2018, 06:59
31 Jul 2018, 06:59
1
From the question stem:
The first term is 3
The second term is 3*2 + 2 = 8
The third term is 8*2 + 2 = 18
We can see that when the unit digit 8 is multiplied by 2 it ends with 6 and when 2 is added it becomes 8 hence all the terms except the first have their units digit as 8. In total there are 9 such terms
The first term is 3
The second term is 3*2 + 2 = 8
The third term is 8*2 + 2 = 18
We can see that when the unit digit 8 is multiplied by 2 it ends with 6 and when 2 is added it becomes 8 hence all the terms except the first have their units digit as 8. In total there are 9 such terms
Re: In sequence Q, the first number is 3, and each subsequent nu
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12 Aug 2018, 07:04
12 Aug 2018, 07:04
Expert Reply
Explanation
After the first term in the sequence, every term has a units digit of 8:
\(Q_1 = 3\)
\(Q_2 = 2(3) + 2 = 8\)
\(Q_3 = 2(8) + 2 = 18\)
\(Q_4 = 2(18) + 2 = 38\)
\(Q_5 = 2(38) + 2 = 78\)
…
So 8 is the units digit nine out of the first ten times.
After the first term in the sequence, every term has a units digit of 8:
\(Q_1 = 3\)
\(Q_2 = 2(3) + 2 = 8\)
\(Q_3 = 2(8) + 2 = 18\)
\(Q_4 = 2(18) + 2 = 38\)
\(Q_5 = 2(38) + 2 = 78\)
…
So 8 is the units digit nine out of the first ten times.
