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In sequence Q, the first number is 3 and each subsequent number in the

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Bunuel
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In sequence Q, the first number is 3 and each subsequent number in the [#permalink] New post   17 Mar 2020, 04:23
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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. How many times does the digit 8 appear in the units digit of the first 10 terms of the sequence?

A. 9
B. 8
C. 7
D. 6
E. 5
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A
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Re: In sequence Q, the first number is 3 and each subsequent number in the [#permalink] New post   17 Mar 2020, 04:28
Bunuel wrote:
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. How many times does the digit 8 appear in the units digit of the first 10 terms of the sequence?

A. 9
B. 8
C. 7
D. 6
E. 5


The sequence = {3,8,18,38,78,158,318,638,1278,2558,....}

The number of times 8 appear in the unit digits of the first 10 terms of the sequence = 9

IMO A
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Re: In sequence Q, the first number is 3 and each subsequent number in the [#permalink] New post   17 Mar 2020, 09:08
Bunuel wrote:
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. How many times does the digit 8 appear in the units digit of the first 10 terms of the sequence?

A. 9
B. 8
C. 7
D. 6
E. 5


from given info we know that
a1= 3 and
a2= 6+2 ; 8
a3= 16+2 ; 18
so the sequence would be unit digit x8*2+2 ; 2*(9) ; x8
where x can be tens digit from 1 to 9 ...
since the values are from a1 to a10 ; so except for a1 ; we get 8 as digit in all i.e 9 times
IMO A
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Re: In sequence Q, the first number is 3 and each subsequent number in the [#permalink] New post   18 Mar 2020, 16:39
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Bunuel wrote:
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. How many times does the digit 8 appear in the units digit of the first 10 terms of the sequence?

A. 9
B. 8
C. 7
D. 6
E. 5


The first few terms of the sequence are the following:

3

3 x 2 + 2 = 8

8 x 2 + 2 = 18

18 x 2 + 2 = 38

38 x 2 + 2 = 78

As we can see, the units digit of every term in this sequence after the first term is 8. This is because the units digit of each subsequent term is obtained by multiplying the units digit of the previous term (which is 8) by 2 and adding 2. So, of the first 10 terms, there are 9 terms with a units digit of 8.

Answer: A
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Re: In sequence Q, the first number is 3 and each subsequent number in the [#permalink] New post   16 Apr 2020, 10:13
The first term of the sequence is 3
The second term of the sequence is 3 x 2 + 2 = 8
The third term of the sequence is 8 x 2 + 2 = 18
The fourth term of the sequence is 18 x 2 + 2 = 38
ans so on,
As we can see, the units digit of every term in this sequence after the first term is 8.
So, of the first 10 terms, there are 9 terms with a units digit of 8.

correct answer A
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Re: In sequence Q, the first number is 3 and each subsequent number in the [#permalink] New post   02 Oct 2022, 11:19
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Given that 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. And we need to find How many times does the digit 8 appear in the units digit of the first 10 terms of the sequence

\(T_1\) = 3
\(T_2\) = 2*3 + 2 = 8
\(T_3\) = 2*8 + 2 = 18
\(T_4\) = 2*18 + 2 = 38
\(T_5\) = 2*38 + 2 = 78
\(T_6\) = 2*78 + 2 = 158
.
.
.

By now you would have realized that the unit's digit is coming 8 every time as we are multiplying the unit's digit of previous number (which is 8) by 2 to get 6 in the unit's digit and then again 2 to make it 8.
And the combination of hundred's and ten's digit is always an odd number 1, 3, 7, 15,...

So, 8 will occur 9 times in the unit's digit in the first ten terms.

So, Answer will be A
Hope it helps!

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Re: In sequence Q, the first number is 3 and each subsequent number in the [#permalink]
02 Oct 2022, 11:19
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