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Is 46^78+89^67 dividable with 5?

 Nov 25, 2014

Best Answer 

 #2
avatar+99356 
+10

Let me see.

Is 46^78+89^67 dividable with 5?

If a number is divisable  by 5 it must end in a 0 or a 5

6^1=6

6^2=36 which ends in a 6

663=36*6 which ends in a 6

6^(any positive integer) must end in 6

6+4=10 (ends in 0) 

6+9=15 (ends in 5)

so the question becomes  does 89^67 end in a 4 or a 9

what could it end in?  9,81,..9,..1..   Ok, it could end in a 1 or a 9  it cannot end in a 4

89^1 ends in a 9

89^2 ends in a 1

89^3 ends in a 9

89^(an odd number) ends in a 9

SO     89^(67) ends in a 9

9+6=15 

So yes that number will be divisable by 5     

 Nov 25, 2014
 #1
avatar+21860 
+10

Is 46^78+89^67 dividable with 5  ?

yes !

$$\boxed{46^{78} + 89^{67} \stackrel{?}{=} 0 \mod 5} \\\\
\small{
\underbrace{(46^{78}\mod 5)}_{ 1^{78} \mod 5} + \underbrace{(89^{67}\mod 5)}_{-1^{67}\mod 5} \stackrel{?}{=} 0 \quad | \quad (46 \mod 5 = 1) \ and \ (89 \mod 5 = 4 = -1)
} \\\\\\
\small{
1^{78} = 1 \ and \ (-1)^{67} = -1}\\\\
\small{
( 1 \mod 5) + (-1 \mod 5) = [ (1-1) \mod 5 ]=( 0 \mod 5) = 0 \quad q.e.d.
}$$

.
 Nov 25, 2014
 #2
avatar+99356 
+10
Best Answer

Let me see.

Is 46^78+89^67 dividable with 5?

If a number is divisable  by 5 it must end in a 0 or a 5

6^1=6

6^2=36 which ends in a 6

663=36*6 which ends in a 6

6^(any positive integer) must end in 6

6+4=10 (ends in 0) 

6+9=15 (ends in 5)

so the question becomes  does 89^67 end in a 4 or a 9

what could it end in?  9,81,..9,..1..   Ok, it could end in a 1 or a 9  it cannot end in a 4

89^1 ends in a 9

89^2 ends in a 1

89^3 ends in a 9

89^(an odd number) ends in a 9

SO     89^(67) ends in a 9

9+6=15 

So yes that number will be divisable by 5     

Melody Nov 25, 2014

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