kyro wrote:
I was Going through a question when I came to a conclusion, just wanna verify if its true.
PART I
Suppose n is any digit(0-9), and Given the units digit for n^k is 5,
then for any number(p) that has units digit same as n,
we have,the units digit for p^k as '5'.
Right?
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PART II
Suppose n is any number and Given the units digit for n^k is 5,
then for any number(p) that has its ending digits same as n,
we have,the units digit for p^k as '5'.
I am unsure of this.
Please enlighten me.
PS: Reply seperately for Part I and Part II.
The unit's digit of n^k is decided by only the unit's digit of n.
Part 1: n is any digit and n^k has a unit's digit of say 'a'. Then if p has n as the unit's digit, p^k will also have the unit's digit of 'a'. Only the unit's digit of p decides the unit's digit of p^k
e.g.
3^4 = 81 (unit's digit 1)
643^4 = ........1 (unit's digit will be 1. When you multiply 643 by 643, the unit's digit is obtained by multiplying 3 by 3 only)
Part 2: n is any number and n^k has unit's digit of 'a'. If p is another number with the same unit's digit as n, then p^k will also have 'a' as the unit's digit.
643^4 = .......1873^4 = ........1Only unit's digit of n decides the unit's digit of n^k.
Something to think further:
If unit's digit of n^k is 'a' and unit's digit of p^k is also 'a', does it mean the unit's digits of n and p are the same?
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Karishma
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