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Comment on Units Digit of 53 to Power of 35
What do you advise?
Of the (13^5)(15^4)(17^5)
(13)^5 = Odd
(17)^5 = Odd
(15)^4 = Unit digit of 5
5*Odd = Unit of 5
That's a great approach!
Question link: https://gmatclub.com/forum/what-is-the-units-digit-of-the-following-expr...
That's a great approach!
A simple way of thinking
Perfect!
Perfect!
Hi Brent,
Very confused with this example. You state that "when N is divisible by 4, the Units Digit of 53^n is 1 given that 53^8 and 53^4 are both divisible by 4 and have a units digit of 1.
However isnt 53^35 divisible by BOTH 5 and 7? We know that 5th units digit in the sequence is "3", whereas the 7th units digit in the sequence is "7".
How would you know to pick 7 over 5 as a divisor of 53^35?
Thank you.
Let's examine the pattern:
Let's examine the pattern:
53^1 = 53
53^2 = --9
53^3 = --7
53^4 = --1
53^5 = --3
53^6 = --9
53^7 = --7
53^8 = --1
53^9 = --3
53^10 = --9
53^11 = --7
53^12 = --1
53^13 = --3
53^14 = --9
53^15 = --7
53^16 = --1
Notice that the pattern repeats ever 4 powers. So, the pattern has cycle 4.
This is why we focus on the powers that are multiples of 4.
We know that 53 to the powers of ANY multiple of 4 will equal ---1
For example:
53^4 = --1
53^8 = --1
53^12 = --1
53^16 = --1
So, without doing any extra work, I know that 53^400 will have units digit 1.
Conversely, if we focus on powers of 5 (or powers of 7), we see that there is no consistency:
53^5 = --3
53^10 = --9
53^15 = --7
Hmm, so what is the units digit of 53^500? It's hard to tell.
So, once we know the cycle of the pattern, we must focus on multiples of that cycle.
Does that help?
Cheers,
Brent
This helps a lot. Thank you
I get it now.
Rather than just stick to the multiplication rule I should identify patterns or behaviors of the cycle.
Thanks Brent.
Hello Brent,
I just used the quotient laws from previous lessons. If we take 53 as X and 35 as "a" and "b" the question simplifies as X^(a.b) that is 53^5*7.
So if 53^5 ends in 3, then any multiple of 5 must end in 3.
You are correct to say that
You are correct to say that 53^35 = (53^5)^7
However, we cannot conclude that, "since 53^5 ends in 3, then any multiple of 5 must end in 3"
To see why this is not true, let's examine the units digits of several powers of 53:
53^1 = 53
53^2 = --9
53^3 = --7
53^4 = --1
53^5 = --3
53^6 = --9
53^7 = --7
53^8 = --1
53^9 = --3
53^10 = --9
53^11 = --7
53^12 = --1
53^13 = --3
53^14 = --9
53^15 = --7
53^16 = --1
Notice that 53^5 has units digit 3, and 53^10 has units digit 9, and 53^15 has units digit 7.
Does that help?
Cheers,
Brent
Hey Brent,
now one solution seems to be:
Of the (13^5)(15^4)(17^5)
(13)^5 = Odd
(17)^5 = Odd
(15)^4 = Unit digit of 5
5*Odd = Unit of 5
But how is an odd number simply irrelevant?
Cheers.
Question link: https:/
Question link: https://gmatclub.com/forum/what-is-the-units-digit-of-the-following-expr...
We know that (13)^5 and (17)^5 are both odd.
So, (13)^5 x (17)^5 = some odd number.
Also, (15^4) has units digit 5
This means (13^5)(17^5)(15^4) = (some odd number)(some number with units digit 5)
If we multiply any odd integer by 5, the units digit is always 5.
For example: (1327)(225) = ------5
And (613)(65435) = ------5
And (807)(1915) = ------5
etc
Does that help?
Cheers,
Brent
Got it, thanks!
that 500-650 one is not
just has to calculate a bit for the unit digit.
81*5 = 5 81 is 9*9
Question link: https:/
Question link: https://gmatclub.com/forum/what-is-the-units-digit-of-the-following-expr...
I'm having a hard time following your solution, but I trust that your calculations are in order :-)