Question

The last digit of $\left| \!{\underline {\,
  583 \,}} \right. +{{7}^{291}}$ is:
(a)1
(b)2
(c)0
(d)3

Answer 1

Hint: In questions like these, we just have to observe a random pattern of the digits at one's place. Here, as there are two terms besides, so we have to separately solve this question.

Complete step-by-step answer:

As the question specifies, we have to find the last digit of $\left| \!{\underline {\,
  583 \,}} \right. +{{7}^{291}}$
So, we will first find the last digit of the separate terms and then we will build a pattern, to find the last digit of the whole term given.
Hence, our main aim will be to find the last digit or the digit at the one's place for both the terms.
So, let us consider the 1st term first i.e. $\left| \!{\underline {\,
  583 \,}} \right. $
Remember, any factorial greater than $\left| \!{\underline {\,
  4 \,}} \right. $ will end in zero. After all, $\left| \!{\underline {\,
  5 \,}} \right. $ it includes 5 and 2 as factors which means that any factorial larger after will at least be a multiple of 10. Therefore, all we need to do is pull and combine the factors that multiply together to become 10.
Simply, we can write $\left| \!{\underline {\,
  583 \,}} \right. $ as $\left| \!{\underline {\,
  583 \,}} \right. =1\times 2\times 3\times 4\times 5\times ................\times 582\times 583$
So, the multiplication of the first 5 terms i.e. 1, 2, 3, 4, 5 is = 120 or 12 x 10.
Now, let us assume that the product of the remaining 578 terms is K.
Therefore, $\left| \!{\underline {\,
  583 \,}} \right. $ can be written as $12\times 10\times K=12K\times 10$
So, It does not matter what number 12K will be, when it is multiplied with 10, the last digit of the whole number will be 0.
Now, let us consider the second term i.e. ${{7}^{291}}$
Here, we have to observe a general pattern and that will help us to identify the last digit of the digit at the one's place for this term.
So, we will observe the powers of the number 7:
$\begin{align}
  & {{7}^{1}}=7 \\
 & {{7}^{2}}=49 \\
 & {{7}^{3}}=343 \\
 & {{7}^{4}}=2401 \\
 & {{7}^{5}}=16807 \\
 & {{7}^{6}}=117649 \\
\end{align}$ Here, we can observe that the last digit starts repeating itself, after the power 4.
And also with a gap of 4.
Therefore, for the power 291, when 291 is divided by 4, 3 is the remainder.
Hence, the last digit ${{7}^{291}}$ will be 3.
Hence, the last digit $\left| \!{\underline {\,
  583 \,}} \right. +{{7}^{291}}$ will be 0+3=3.

Note: In these types of questions, instead of looking at the calculations, one should observe the general pattern to solve the question.