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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? ______________________________________________________________________________________________
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.
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Ummm.......I am quite confused by the round-about logic u have used....but i'll tell u what I know:
for ANY number n ending in 5, n^k will ALWAYS end in 5; provided k is an integer greater than or equal to 1.
Analysis of PART I Suppose n is any digit(0-9), and Given the units digit for n^k is 5, -->unit digit of n^k will be 5 only for the digit 5 and NO OTHER digit from 0-9. then for any number(p) that has units digit same as n, we have,the units digit for p^k as '5'. --> this statement doesnt make any sense since the only digit n for which n^k will end in 5 is 5 itself.
same logic applies to Part 2.....only a number n ending in 5 (say 15) will have a n^k form which ends in 5.
Hi Kyro, The answer to both your parts is the same. All powers of a no. ending in 5 also end in 5. eg. 5^2= 25, 5^4=625 etc. Also no other no has 5 as a units digit in its power cycle. Thus if n^k = xxx5 it means, n will have 5 in its units place. Hope it clarifies.
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? ______________________________________________________________________________________________
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.
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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.
\(643^4 = .......1\) \(873^4 = ........1\)
Only 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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Hi there,
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