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In the sequence {A_n}, A1=1 and An = 3An-1 + 1 for all posit

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by Max@Math Revolution » Fri Jun 01, 2018 1:24 am

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[GMAT math practice question]

In the sequence {A_n}, A1=1 and An = 3An-1 + 1 for all positive integers n. What is the units digit of A100?

A. 0
B. 1
C. 2
D. 3
E. 4
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by GMATGuruNY » Fri Jun 01, 2018 4:04 am
Max@Math Revolution wrote:[GMAT math practice question]

In the sequence {A_n}, A1=1 and An = 3An-1 + 1 for all positive integers n. What is the units digit of A100?

A. 0
B. 1
C. 2
D. 3
E. 4
A� = 1.
A₂ = 3A� + 1 = (3*1) + 1 = 4.
A₃ = 3A₂ + 1 = (3*4) + 1 = 13.
A₄ = 3A₃ + 1 = (3*13) + 1 = 40.
Aâ‚… = 3Aâ‚„ + 1 = (3*40) + 1 = 121.
A₆ = 3A₅ + 1 = (3*121) + 1 = 364.

The results above indicate that the units digits repeat in a CYCLE OF 4:
1, 4, 3, 0...1, 4, 3, 0...
Implication:
Every 4th term -- A₄, A₈, A�₂...A₉₂, A₉₆, A�₀₀ -- will have a units digit of 0.
Thus, the units digit for A�₀₀ is 0.

The correct answer is A.
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by Jeff@TargetTestPrep » Sun Jun 03, 2018 5:31 pm
Max@Math Revolution wrote:[GMAT math practice question]

In the sequence {A_n}, A1=1 and An = 3An-1 + 1 for all positive integers n. What is the units digit of A100?

A. 0
B. 1
C. 2
D. 3
E. 4
A(1) = 1

A(2) = 3 + 1 = 4

A(3) = 12 + 1 = 13

A(4) = 39 + 1 = 40

A(5) = 120 + 1 = 121

The pattern of the units digits is 1-4-3-0.

So if k is a multiple of 4, then A(k) has a units digit of zero.

Since 100 is a multiple of 4, the units digit of A(100) = 0.

Answer: A

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by Max@Math Revolution » Mon Jun 04, 2018 11:23 am
=>
The units digit of A_{100} will be the remainder when A_{100} is divided by 10.
Using the formula A_n = 3A_{n - 1} + 1 yields
A2 = 3A1 + 1 = 3*1 + 1 = 3 + 1 = 4 ~ 4
A3 = 3A2 + 1 = 3*4 + 1 = 13 ~ 3
A4 = 3A3 + 1 ~ 3*3 + 1 = 10 ~ 0
A5 = 3A4 + 1 ~ 3*0 + 1 ~ 1
In general, we have:
A1 ~ A5 ~ ... ~ A4k+1 ~ 1
A2 ~ A6 ~ ... ~ A4k+2 ~ 4
A3 ~ A7 ~ ... ~ A4k+3 ~ 3
A4 ~ A8 ~ ... ~ A4k ~ 0
Since the index (100) of A100 is a multiple of 4, the units digit of A100 is 0.
Therefore, the answer is A.
Answer: A
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