# How do i find unit's digit in following:- [(999)^53 + (998)^53 + (997)^53 + .... (1)^53] ?

Relevance

Reason:

You only need to consider the last digit of each term.

last digit last digit after being raised to the power of 53

1 ---------- 1

2 ---------- 8

3 ---------- 7

4 ---------- 4

5 ---------- 5

6 ---------- 6

7 ---------- 3

8 ---------- 2

9 ---------- 9

0 ---------- 0

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sum = 50

The pattern repeats after each 10 numbers. We can add (1000)^53 to the sum because it has the same unit's digit as the original one. Now 1000 numbers repeats 100 times. Therefore, the unit's digit of the sum is 0.

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• I believe the units digit of that is 0. Here is my reasoning. All three digit numbers of the form xxd^53 (x is any digit and d is a fixed digit) have the same units digit, since the units digit depends only on d. Now from the following sum you have 100 numbers of the form xx1, 100 of the form xx2, ..., and 100 of the form xx9 (I ignored the numbers of the form xx0^53, since they do not affect the units digit). So the units digit of that is same as the units digit of:

100*(units_digit(xx1) + ... + units_digit(xx9))

which no matter what is in the parenthesis, is divisible by 10 and thus has a units digit 0.

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