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Comment on Units Digit of a Large Product
if bases are same, exponenet
(34^5)x(9^5) = (34x9)^5??
That's correct.
That's correct.
To see why this is, consider this example:
(5^3)x(7^3) = (5x5x5)x(7x7x7)
= 5x5x5x7x7x7
= 5x7x5x7x5x7
= (5x7)x(5x7)x(5x7)
= (5x7)^3
49^1 = ---9
49^2 = ---1
49^3 = ---9
49^4 = ---1
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When exponent is Even, Unit digit is 1.
49^18 = ---1.
13^1 = ---3
13^2 = ---9
13^3 = ---7
13^4 = ---1
13^5 = ---3
13^6 = ---9
13^7 = ---7
13^8 = ---1
When Exponent is multiple of 4, unit digits is 1.
so, ---1 * ---1 = 1
Answer is A
I tried solving this using
You're correct to say that 13
You're correct to say that 13 (or 3) has a cycle of 4.
We can see this when we list a few powers:
13^1 = ---3
13^2 = ---9
13^3 = ---7
13^4 = ---1
13^5 = ---3
13^6 = ---9
13^7 = ---7
13^8 = ---1
.
.
.
Once we know that 13 has a cycle of 4, we can see that, when the exponent is a multiple of 4, the units digit of that power must be 1.
That is, 13^4 = ---1 and 13^8 = ---1 and 13^12 = ---1 and 13^16 = ---1 etc.
So, we can conclude that 13^36 = ---1 (since 36 is a multiple of 4)
Now let's deal with 49^18 by listing a few powers:
49^1 = ---9
49^2 = ---1
49^3 = ---9
49^4 = ---1
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We can see that 49 has a cycle of 2
So, when the exponent is a multiple of 2, the units digit of that power must be 1.
That is, 49^2 = ---1 and 49^4 = ---1 and 49^6 = ---1 and 49^8 = ---1 etc.
So, we can conclude that 49^18 = ---1 (since 18 is a multiple of 2)
Does that help?
Cheers,
Brent