In the sequence 1,2,2,4,8,32,256...

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 14'th term is close to some estimates of the number of particles in the observable universe.)

What is the last digit of the term of the sequence

Guest Jan 19, 2019

#2**0 **

I assume he/she means the 14th term. If so, then the 14th is:

1

2

2

4

8

32

256

8192

2097152

1 7179869184

3602879 7018963968

6189700 1964269013 7449562112

2230 0745198530 6231415357 1827264836 1505980416

**1 3803492693 5811275748 6951172455 4050904902 2179443407 7311032504 8447598592**

Guest Jan 19, 2019

#3**+1 **

This is simple pattern recognizing

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 14th term. You divide 14 by 6 and find the remainder, which is 2.

So look at the 2nd term in the pattern. which is 2.

But remember the first term is 1, so you have to count back a term

So the answer is 2

CalculatorUser Jan 19, 2019

#7**+7 **

**In the sequence 1,2,2,4,8,32,256... 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 14'th term is close to some estimates of the number of particles in the observable universe.) What is the last digit of the term of the sequence**

\(\begin{array}{|lcll|} \hline a_1 &=& 1 \\ a_2 &=& 2\\ a_3 = a_2\cdot a_1 &=& a_2^1 \\ a_4 = a_3\cdot a_2 = a_2^1\cdot a_2 &=& a_2^2 \\ a_5 = a_4\cdot a_3 = a_2^2\cdot a_2^1 &=& a_2^3 \\ a_6 = a_5\cdot a_4 = a_2^3\cdot a_2^3 &=& a_2^5 \\ a_7 = a_6\cdot a_5 = a_2^5\cdot a_2^3 &=& a_2^8 \\ a_8 = a_7\cdot a_6 = a_2^8\cdot a_2^5 &=& a_2^{13}\\ a_9 = a_8\cdot a_7 = a_2^{13}\cdot a_2^8 &=& a_2^{21} \\ \ldots \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && \text{Let } \mathcal{F} \text{ are the Fibonacci number }\\\\ \mathcal{F}_1 &=& 1\\ \mathcal{F}_2 &=& 1\\ \mathcal{F}_3 &=& 2\\ \mathcal{F}_4 &=& 3\\ \mathcal{F}_5 &=& 4\\ \mathcal{F}_6 &=& 5\\ \mathcal{F}_7 &=& 13\\ \mathcal{F}_8 &=& 21\\ \mathcal{F}_9 &=& 34\\ \mathcal{F}_{10} &=& 55\\ \mathcal{F}_{11} &=& 89\\ \mathcal{F}_{12} &=& 144\\ \mathcal{F}_{13} &=& 233\\ \mathcal{F}_{14} &=& 377\\ \ldots \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a_n &=& a_2^{\mathcal{F}_{n-1}} \quad & | \quad a_2 = 2 \\ a_n &=& 2^{\mathcal{F}_{n-1}} \\ \hline \end{array} \)

**If n = 14:**

\(\begin{array}{|rcll|} \hline a_{14} &=& 2^{\mathcal{F}_{14-1}} \\ a_{14} &=& 2^{\mathcal{F}_{13}} \quad & | \quad \mathcal{F}_{13} = 233 \\ a_{14} &=& 2^{233} \\\\ && \text{ The last digit of the term } a_{14} \\ && 2^{233} \pmod{10} \\ &\equiv & 2^{13\cdot 17+12} \pmod{10}\\ &\equiv & \left(2^{13}\right)^{17} 2^{12} \pmod{10} \quad & | \quad 2^{13} \equiv 2 \pmod{10} \\ &\equiv & 2^{17}\cdot 2^{12} \pmod{10} \\ &\equiv & 2^{29} \pmod{10} \\ &\equiv & 2^{13\cdot 2+3} \pmod{10} \\ &\equiv & \left(2^{13}\right)^{2} 2^{3} \pmod{10} \quad & | \quad 2^{13} \equiv 2 \pmod{10} \\ &\equiv & 2^2\cdot 2^3 \pmod{10} \\ &\equiv & 2^{5} \pmod{10} \\ &\equiv & 32 \pmod{10} \\ &\mathbf{\equiv} & \mathbf{ 2 \pmod{10}} \\ \hline \end{array}\)

\(\text{ The last digit of the term $a_{14}$ is $\mathbf{2}$}\)

heureka Jan 21, 2019