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Query:Not a Question
[#permalink]
25 Apr 2012, 03:16
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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Re: Query:Not a Question
[#permalink]
25 Apr 2012, 03:42
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.
Re: Query:Not a Question
[#permalink]
25 Apr 2012, 04:21
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.
Re: Query:Not a Question
[#permalink]
27 Apr 2012, 20:02
Expert Reply
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? ______________________________________________________________________________________________
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.
\(643^4 = .......1\) \(873^4 = ........1\)
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?
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.