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Question: What is the unit digit in the product of \(\left( {{3}^{65}}\times {{6}^{59}}\times {{7}^{71}} \righ...

What is the unit digit in the product of (365×659×771)\left( {{3}^{65}}\times {{6}^{59}}\times {{7}^{71}} \right) ?

Explanation

Solution

To find the unit digit in the product of (365×659×771)\left( {{3}^{65}}\times {{6}^{59}}\times {{7}^{71}} \right) , we must consider the unit digit of each term. Let us first consider 365{{3}^{65}} . Using the formulas (am)n=amn{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}} and am×an=am+n{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} we can write 365=(34)16×3{{3}^{65}}={{\left( {{3}^{4}} \right)}^{16}}\times 3 . Now, we must find the unit digit of this. Next, let us consider 659{{6}^{59}} . We know that 6 to the power of any number has unit digit 6. Hence, we will get its unit digit as 6. Lastly, consider 771{{7}^{71}} . We can write this as 771=(74)17×73{{7}^{71}}={{\left( {{7}^{4}} \right)}^{17}}\times {{7}^{3}} . Now, multiply the unit digits of all the three terms and find the unit digit of the resulting number.

Complete step-by-step answer:
We need to find the unit digit in the product of (365×659×771)\left( {{3}^{65}}\times {{6}^{59}}\times {{7}^{71}} \right) .
Let us first consider 365{{3}^{65}}
We know that (am)n=amn{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}} and am×an=am+n{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}
365=(34)16×3\Rightarrow {{3}^{65}}={{\left( {{3}^{4}} \right)}^{16}}\times 3
Now, let us see the unit digit of the above. That is
Unit digit of [(34)16×3]\left[ {{\left( {{3}^{4}} \right)}^{16}}\times 3 \right]
Let us expand 34{{3}^{4}} . We will get
Unit digit of [(3×3×3×3)16×3]\left[ {{\left( 3\times 3\times 3\times 3 \right)}^{16}}\times 3 \right]
This be written as
Unit digit of [(9×9)16×3]\left[ {{\left( 9\times 9 \right)}^{16}}\times 3 \right]
We know that 9×9=819\times 9=81 . Hence, the above form becoms
Unit digit of [(81)16×3]\left[ {{\left( 81 \right)}^{16}}\times 3 \right]
This is same as
Unit digit of(81)16×Unit digit of 3\text{Unit digit of}{{\left( 81 \right)}^{16}}\times \text{Unit digit of }3
We know that 1 to the power of any number is always 1. Hence,
Unit digit of(81)16×Unit digit of 3 =1×3 =3...(i) \begin{aligned} & \text{Unit digit of}{{\left( 81 \right)}^{16}}\times \text{Unit digit of }3 \\\ & =1\times 3 \\\ & =3...(i) \\\ \end{aligned}
Now, let us consider 659{{6}^{59}} .
We know that 6 to the power of any number has unit digit 6. For example
61=6 62=36 63=216 64=1296 65=7776 ... \begin{aligned} & {{6}^{1}}=6 \\\ & {{6}^{2}}=36 \\\ & {{6}^{3}}=216 \\\ & {{6}^{4}}=1296 \\\ & {{6}^{5}}=7776 \\\ & ... \\\ \end{aligned}
Hence, unit digit of 659=6...(ii){{6}^{59}}=6...(ii)
Let us now consider 771{{7}^{71}} .
We know that (am)n=amn{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}} and am×an=am+n{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}
771=(74)17×73\Rightarrow {{7}^{71}}={{\left( {{7}^{4}} \right)}^{17}}\times {{7}^{3}}
Now, let us observe the unit digit.
Unit digit of 771=Unit digit of [(74)17×73]\text{Unit digit of }{{7}^{71}}=\text{Unit digit of }\left[ {{\left( {{7}^{4}} \right)}^{17}}\times {{7}^{3}} \right]
When we expand 74{{7}^{4}} , we will get
Unit digit of 771=Unit digit of [(7×7×7×7)17×73]\text{Unit digit of }{{7}^{71}}=\text{Unit digit of }\left[ {{\left( 7\times 7\times 7\times 7 \right)}^{17}}\times {{7}^{3}} \right]
We can write this as
Unit digit of 771=Unit digit of [(49×49)17×(7×7×7)]\text{Unit digit of }{{7}^{71}}=\text{Unit digit of }\left[ {{\left( 49\times 49 \right)}^{17}}\times \left( 7\times 7\times 7 \right) \right]
Unit digit of 771=Unit digit of [(2401)17×343]\Rightarrow \text{Unit digit of }{{7}^{71}}=\text{Unit digit of }\left[ {{\left( 2401 \right)}^{17}}\times 343 \right]
We can expand the above form as
Unit digit of 771=Unit digit of (2401)17×Unit digit of 343\text{Unit digit of }{{7}^{71}}=\text{Unit digit of }{{\left( 2401 \right)}^{17}}\times \text{Unit digit of 343}
We know that 1 to the power of any number is always 1. Hence,
Unit digit of 771=1×3=3...(iii)\text{Unit digit of }{{7}^{71}}=1\times 3=3...(iii)
Now, let us multiply (i),(ii) and (iii). We will get

& \text{Unit digit of}\left( {{3}^{65}}\times {{6}^{59}}\times {{7}^{71}} \right)=\text{Unit digit of }\left( 3\times 6\times 3 \right) \\\ & =\text{Unit digit of }\left( 54 \right) \\\ & =4 \\\ \end{aligned}$$ Hence, the unit digit in the product of $\left( {{3}^{65}}\times {{6}^{59}}\times {{7}^{71}} \right)$ is 4. **Note:** One must know what a unit digit is when solving this question. Unit digit is the number in one’s place in the place value, that is, the last digit of a number. You must know the rules of exponents to split the numbers. When large exponents are given, you must split it and solve for each part. Just like the last digit of any power of 6 is 6, the last digit of any power of 5 is also 5 and for 1 also it is 1 itself. All other numbers, i.e 2, 3, 4, 7, 8 and 9 have certain numbers repeating themselves. For example, the last digit of powers 1, 2, 3, 4 of 9 repeat in the cycle as 1, 9, 1, 9. Knowing these is a great help in the exams.