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What is the Last Digit in the Result of the given Expression? [closed]

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2

$begingroup$

$Given$:

$ASC$ is a concatenated number with distinct digits.

$S$ is square of $A$, $C$ is cube of $A$

Deduce the last digit of the following Expression through Deductive Reasoning only:

$$beginalignA^S&times A^ASC\ +space S^C&times S^ASC\ +,C^A&times C^ASCendalign$$

mathematics logical-deduction no-computers

share|improve this question

edited Jul 7 at 10:45

Uvc

asked Jul 7 at 10:36

UvcUvc

2,79955 silver badges3131 bronze badges

$endgroup$

closed as off-topic by greenturtle3141, Glorfindel, Peregrine Rook, w l, Omega Krypton Jul 8 at 11:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – greenturtle3141, Glorfindel, w l, Omega Krypton

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm VTC because determining A, S, C is completely trivial, and this reduces to a very routine number theory problem, so this is not a puzzle.
  $endgroup$
  – greenturtle3141
  Jul 7 at 16:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (I'm also inclined to close as not-a-puzzle on different grounds: that this is something to recognize more than something to actually solve.)
  $endgroup$
  – Rubio♦
  Jul 7 at 22:22
  

  
  
  
  

add a comment | 

2

$begingroup$

$Given$:

$ASC$ is a concatenated number with distinct digits.

$S$ is square of $A$, $C$ is cube of $A$

Deduce the last digit of the following Expression through Deductive Reasoning only:

$$beginalignA^S&times A^ASC\ +space S^C&times S^ASC\ +,C^A&times C^ASCendalign$$

mathematics logical-deduction no-computers

share|improve this question

edited Jul 7 at 10:45

Uvc

asked Jul 7 at 10:36

UvcUvc

2,79955 silver badges3131 bronze badges

$endgroup$

closed as off-topic by greenturtle3141, Glorfindel, Peregrine Rook, w l, Omega Krypton Jul 8 at 11:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – greenturtle3141, Glorfindel, w l, Omega Krypton

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm VTC because determining A, S, C is completely trivial, and this reduces to a very routine number theory problem, so this is not a puzzle.
  $endgroup$
  – greenturtle3141
  Jul 7 at 16:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (I'm also inclined to close as not-a-puzzle on different grounds: that this is something to recognize more than something to actually solve.)
  $endgroup$
  – Rubio♦
  Jul 7 at 22:22
  

  
  
  
  

add a comment | 

$begingroup$

$Given$:

$ASC$ is a concatenated number with distinct digits.

$S$ is square of $A$, $C$ is cube of $A$

Deduce the last digit of the following Expression through Deductive Reasoning only:

$$beginalignA^S&times A^ASC\ +space S^C&times S^ASC\ +,C^A&times C^ASCendalign$$

mathematics logical-deduction no-computers

share|improve this question

edited Jul 7 at 10:45

Uvc

asked Jul 7 at 10:36

UvcUvc

2,79955 silver badges3131 bronze badges

$endgroup$

share|improve this question

edited Jul 7 at 10:45

Uvc

asked Jul 7 at 10:36

UvcUvc

2,79955 silver badges3131 bronze badges

share|improve this question

edited Jul 7 at 10:45

Uvc

asked Jul 7 at 10:36

UvcUvc

2,79955 silver badges3131 bronze badges

asked Jul 7 at 10:36

UvcUvc

2,79955 silver badges3131 bronze badges

asked Jul 7 at 10:36

UvcUvc

2,79955 silver badges3131 bronze badges

1 Answer
1

active

oldest

votes

3

$begingroup$

As A, S, C are single digit numbers, $A=2,S=4,C=8$.

So we have $$2^4 2^248 + 4^8 4^248 + 8^2 8^248 =4^126 + 4^256 + 4^375 equiv 6+6+4 equiv 6$$.

Reason

Odd power of $4 equiv 4$ (last digit) and

Even power of $4equiv6$ (last digit)

share|improve this answer

edited Jul 7 at 11:12

answered Jul 7 at 11:01

Ak19Ak19

2,33533 silver badges2626 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

As A, S, C are single digit numbers, $A=2,S=4,C=8$.

So we have $$2^4 2^248 + 4^8 4^248 + 8^2 8^248 =4^126 + 4^256 + 4^375 equiv 6+6+4 equiv 6$$.

Reason

Odd power of $4 equiv 4$ (last digit) and

Even power of $4equiv6$ (last digit)

share|improve this answer

edited Jul 7 at 11:12

answered Jul 7 at 11:01

Ak19Ak19

2,33533 silver badges2626 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

As A, S, C are single digit numbers, $A=2,S=4,C=8$.

So we have $$2^4 2^248 + 4^8 4^248 + 8^2 8^248 =4^126 + 4^256 + 4^375 equiv 6+6+4 equiv 6$$.

Reason

Odd power of $4 equiv 4$ (last digit) and

Even power of $4equiv6$ (last digit)

share|improve this answer

edited Jul 7 at 11:12

answered Jul 7 at 11:01

Ak19Ak19

2,33533 silver badges2626 bronze badges

$endgroup$

add a comment | 

$begingroup$

As A, S, C are single digit numbers, $A=2,S=4,C=8$.

So we have $$2^4 2^248 + 4^8 4^248 + 8^2 8^248 =4^126 + 4^256 + 4^375 equiv 6+6+4 equiv 6$$.

Reason

Odd power of $4 equiv 4$ (last digit) and

Even power of $4equiv6$ (last digit)

share|improve this answer

edited Jul 7 at 11:12

answered Jul 7 at 11:01

Ak19Ak19

2,33533 silver badges2626 bronze badges

$endgroup$

share|improve this answer

edited Jul 7 at 11:12

answered Jul 7 at 11:01

Ak19Ak19

2,33533 silver badges2626 bronze badges

answered Jul 7 at 11:01

Ak19Ak19

2,33533 silver badges2626 bronze badges

answered Jul 7 at 11:01

Ak19Ak19

2,33533 silver badges2626 bronze badges