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In the sequence:

each term (starting from the third term) is the product of the two terms before it. For example, the seventh term is 256, which is the product of the fifth term (8) and the sixth term (32).

This sequence can be continued forever, though the numbers very quickly grow enormous! (For example, the 14th term is close to some estimates of the number of particles in the observable universe.)

What is the last digit of the 35th term of the sequence?

 

$$1,2,2,4,8,32,256,\ldots,$$

 May 20, 2019
 #1
avatar+1046 
+3

A similar question was posted a long time ago

 

the first few terms are 1, 2, 2, 4, 8, 32, 256, 8192, 2097152...

 

Notice how the last digits are forming a pattern, (2, 2, 4, 8, 2, 6, 2, 2, 4, 8, 2, 6, 2...

 

The pattern is 2 -> 2 ->  4 -> 8 ->  2 -> 6

 

So if the pattern repeats every 6 terms. And assuming you are trying to find the 35th term. You divide 35 by 6 and find the remainder, which is 5

 

So count up 5 terms in the pattern.

 

[fixed]

 

we count up 4 terms in the pattern because the first term is 1

 

so 8 is the answer

 

Someone check this.

 

Note: In these questions, always check for a pattern!!!

 May 20, 2019
edited by CalculatorUser  May 20, 2019
edited by CalculatorUser  May 20, 2019
 #2
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0

it's not 2..

Guest May 20, 2019
 #3
avatar+1046 
+2

Ah! I realized that the first terms starts as 1. So using the same strategy.

 

If we count up 5 in the pattern, we will end up with the last digit of term 36.

 

So we count back 1 in the pattern

 

the correct answer is 8!

CalculatorUser  May 20, 2019
edited by CalculatorUser  May 20, 2019
 #5
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0

I don't know

Guest May 20, 2019
 #7
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0

Yup! I tried it a different way and got the same thing

Guest May 20, 2019
 #4
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da answer is 6^7a = 5f%

 May 20, 2019
 #6
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0

answer is 5f%?

Guest May 20, 2019
 #8
avatar+23060 
+2

In the sequence:
each term (starting from the third term) is the product of the two terms before it.

For example, the seventh term is 256, which is the product of the fifth term (8) and the sixth term (32).


This sequence can be continued forever, though the numbers very quickly grow enormous! (For example, the 14th term is close to some estimates of the number of particles in the observable universe.)


What is the last digit of the 35th term of the sequence?


\(1,2,2,4,8,32,256,\ldots,\)

 

\(\begin{array}{|rcl|l|} \hline a_n &&& \text{Fibonacci numbers} \\ \hline a_1 &=& 2^{0} & F_{0} = 0 \\ a_2 &=& 2^{1} & F_{1} = 1 \\ a_3 &=& 2^{1} & F_{2} = 1 \\ a_4 &=& 2^{2} & F_{3} = 2 \\ a_5 &=& 2^{3} & F_{4} = 3 \\ a_6 &=& 2^{5} & F_{5} = 5 \\ a_7 &=& 2^{8} & F_{6} = 8 \\ a_8 &=& 2^{13} & F_{7} = 13 \\ a_9 &=& 2^{21} & F_{8} = 21 \\ \ldots \\ a_n &=& 2^{F_{n-1}} \\ \hline a_{35} &=& 2^{F_{34}} & F_{34} = 5702887 \\ \mathbf{a_{35}} &=& \mathbf{2^{5702887}} \\ \hline \end{array} \)

 

The last digit of the 35th term of the sequence is \( \mathbf{2^{5702887}} \pmod{10}\)

\(\begin{array}{|rcll|} \hline && \mathbf{2^{5702887}} \pmod{10} \quad &| \quad \mathbf{2^4}=16 \equiv \mathbf{6} \pmod{10} \\ &\equiv & 2^{4*1425721+3} \pmod{10} \\ &\equiv & \left( 2^{4} \right)^{1425721}*2^3 \pmod{10} \\ &\equiv & 6^{1425721}*2^3 \pmod{10} \quad & | \quad 6^n \equiv 6 \pmod{10} \\ &\equiv & 6 *2^3 \pmod{10} \\ &\equiv & 6 *8 \pmod{10} \\ &\equiv & 48 \pmod{10} \\ &\equiv & \mathbf{8} \pmod{10} \\ \hline \end{array} \)

 

The last digit of the 35th term of the sequence is 8

 

laugh

 May 21, 2019

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