The digit ‘5’ is in the Hundreds place, ‘6’ is in the Tens place, and ‘2’ is in the Ones place. The number 562 can be expanded as 5 × 100 + 6 × 10 + 2 × 1.
Similarly, numbers such as 2345 can be expanded as 2 × 1000 + 3 × 100 + 4 × 10 + 5 × 1. In this case, the digit ‘2’ is said to be in the Thousands place, ‘3’ to be in the Hundreds place, ‘4’ to be in the Tens place, and ‘5’ to be in the Ones place.
(Practice: What is the Ones and Hundreds digit of the number 3990?)
I. When the number ‘a’ ends in 1: e.g. ‘a’ = 81
II. When the number ‘a’ ends in 3, 7 and 9: e.g. ‘a’ = 653 or ‘a’ = 817 or ‘a’ = 399
III. When the number ‘a’ ends in 2, 4, 6, 8: e.g. or ‘a’ = 72
IV. When the number ‘a’ ends in 5: e.g. ‘a’ = 55
It’s best to present the method with examples. We encourage you to follow the procedure on your own simultaneously on paper.
Examples:
1) Find the last two digits of \(41^{8899}\).
Solution: Here a = 41 and b = 8899. Since a is ending in 1, we have Ones digit of our answer as ‘1’. Next using Step 2, Tens digit is calculated as 4 (Tens digit of a) × 9 (Ones digit of b) = 36 (product). Pick the Ones digit of this Step. So, the solution is 61.
2) Find the last two digits of \(121^{45867}\).
Solution: Here a = 121 and b = 45867. Since a is ending in 1, we have Ones digit of our answer as ‘1’. Next using Step 2, Tens digit is calculated as 2 × 7 = 14. Pick the Ones digit of this product, which is ‘4’. So, the solution is 41.
3) Find the last two digits of \(5145651^{456}\) × \(61^{567}\).
Solution: The last two digits of \(5145651^{456}\) is 01 and the last two digits of \(61^{567}\) is 21. Therefore last two digits of \(5145651^{456}\) × \(61^{567}\) is 01 × 21 that is 21.
Before we learn the next set of cases, let us quickly look at the table of cyclicity of unit digits.
\(Digit\) | \(Digit^{1}\) | \(Digit^{2}\) | \(Digit^{3}\) | \(Digit^{4}\) |
---|---|---|---|---|
2 | 2 | 2 | 8 | 6 |
3 | 3 | 9 | 7 | 1 |
4 | 4 | 6 | ||
5 | 5 | |||
6 | 6 | |||
7 | 7 | 9 | 3 | 1 |
8 | 8 | 4 | 2 | 6 |
9 | 9 | 1 |
This table represents the cyclicity of the unit digits till power of 4. Observe that cyclicity of 2, 3, 7, and 8 is 3. Cyclicity of 4, and 9 is 1. Cyclicity of 0, 1, 5, and 6 is 0.
Let us now remember the following two rules to solve for this set.
Rule 1: \(2^{10}\) raised to the even power always give last two digits as 76.
e.g. \(2^{260}\) = \((2^{10})^{26}\). Therefore, last two digits of \(2^{260}\) will be 76. The same holds true for \(2^{320}\), that is, last two digits of \(2^{320}\) will be 76.
Rule 2: \(2^{10}\) raised to the odd power always give last two digits as 24. e.g. \(2^{70}\) = \((2^{10})^{7}\). Therefore, last two digits of \(2^{70}\) will be 24. The same holds true for \(2^{130}\), that is, last two digits of \(2^{130}\) will be 24.
Let us look at an example where we can utilize these rules. Suppose we would like to know the last two digits of \(36^{199}\). So, \(36^{199}\) = \((2^{2} \times 3^{2})^{199}\) = \(2^{2 × 199} × 3^{2 × 199}\) = \(2^{398} × 3^{398}\) = \(2^{(390+8)} × 3^{398}\). This is a good example to apply the above things we have learned.
A. \(2^{(390 + 8)}\) = \(2^{390} × 2^{8}\) = \((2^{10})^{39} × 256\). Keeping last two digits of \((2^{10})^{39} × 256\), that is, 24 × 56. So, we get last two digits of \(2^{398}\)equal to 44.
B. \(3^{398}\) = \((3^{4})^{99} × 3^{2}\) = \((81)^{99} × 9\). Keeping last two digits of \((81)^{99} × 9\), that is, 21 × 09. So, we get last two digits of \(3^{398}\) equal to 89.
Therefore, last two digits of \(36^{199}\) = 44 × 89. Keeping last two digits of this, we get the answer as 16.
III. When the number ‘a’ ends in 5
The last two digits for numbers \(a^{b}\), with a ending with ‘5’ are always 25 or 75.
Rule 1: If the product of Ones digit of the power b and digit left to the ‘5’ in a (Tens digit of a), is EVEN, then last two digits of the expression is 25.
Rule 2: If the product of ones digit of the power b and digit left to the ‘5’ in a (Tens digit of a), is ODD, then last two digits of the expression is 75.
Here’s an example for this set. We will calculate the last two digits of \(785^{587}\). Here a = 785 and b = 587. The Ones digit of b is 7 and Tens digit of a is 8. This means 7 × 8 is 56 which is EVEN. Therefor the last two digits of \(785^{587} \) is 25. Similarly, the last two digits of \(1475^{589}\) is 75.