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All sums of higher powers can be done by the above method. First see the cycle of the last digits, then divide the power by 4 and see the remainder and tick off the answer. There are exceptions to this. For example, (36)n will always end in 6, no matter what the power. A number of the form 74n will only end in 4 or 6. The student should learn these rules, which will help gain speed.

The cyclicity of numbers is given in the following table. The table shows the last digit of the resultant number when the number with the given last digit is raised to a power:

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From the above table we see that cyclicity of most numbers is of 4 digits. For instance, the last digits when 2 is raised to consecutive powers are: 21 =2,22 =4,23 =8,24 =6,thenfor25,26,27,...the same digits will be repeated.

For powers of 3, the last digits will be 3, 9, 7, 1, 3, 9, 7, 1, ...The exceptions are powers of 4, 5, and 9. In these digits, the last digits do not have the cyclicity of 4.

What we have learnt

Since the cyclicity of powers of a digits is known, finding last digits of numbers when they are raised to large powers, becomes easy.

Example: What is the last digit of (1377)273?
Solution: We see that the cyclicity of 7 is given by the digits: 7, 9, 3, 1.
Since the cyclicity is of 4 numbers, we can divide the power, 273, by 4 to get a remainder of 1. As the remainder is 1, the last digit of the number (1377)273 will be 7.
All sums involving higher powers can be done by this method.


Place values: The number 987654321.321 has the following place values:

The number can be written as:

9×109 +8×108 +7×107 +6×106 +5 ×105 +4×104 +3×103 + 2×102 +1×101 +3×10-1 + 2×10-2 +1×10-3.

The place value of 1 in the number is 1 unit = 1.

The place value of 3 in the number is 3 thousand = 3,000.

The place value of 6 in the number is 6 millions = 6,000,000, and so on.

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