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Exponents again?

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Exponents again?

by AIM TO CRACK GMAT » Mon Oct 21, 2013 12:00 am
What is the last digit of the number 537^1279089028?

(A) 1
(B) 3
(C) 6
(D) 7

Hi guys i tried the method where u follow the 7^2 n get the series of 7,9,3,1,7... N then tried to divide it! somehow i get 7 but that is the wrong answer!!

[spoiler]OA = 1[/spoiler]
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by theCodeToGMAT » Mon Oct 21, 2013 12:06 am
537^1279089028

Consider 1279089028

Perform divisibility test by "4" .. as we know that for "4" we check last 2 digits.. so 28/4 = 0

When the result is "0".. then the power is "4"

So, (7)^4 = xxx1

Answer {A}
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by theCodeToGMAT » Mon Oct 21, 2013 12:09 am
General Steps to such question:

- Consider the power and divide it by "4"

if remainder is "1" -> Take Unit digit of the base number to power "1"
if remainder is "2" -> Take Unit digit of the base number to power "2"
if remainder is "3" -> Take Unit digit of the base number to power "3"
if remainder is "0" -> Take Unit digit of the base number to power "4"
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by mevicks » Mon Oct 21, 2013 12:21 am
AIM TO CRACK GMAT wrote:What is the last digit of the number 537^1279089028?

(A) 1
(B) 3
(C) 6
(D) 7
To simplify and speed up the process use the property of cyclicity on these problems:
Image

Consider the last digit of 537
This number raised to any power would yield a number having the units digit as per the above table.

thus we need to find : 7^1279089028

Now, notice that the last two digits are 28.
there is a neat divisibility property for the numbers divisible by 4.
"If the last two digits of a number are divisible by four then the complete number is divisible by 4"

Using this property we know that our very weird looking number "1279089028" is a multiple of 4.

Thus considering the cyclicity for 7 we know that 7(raised to multiples of 4) would yield a unit digits 1
[spoiler]
Answer : (A) 1[/spoiler]
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by sahilchaudhary » Mon Oct 21, 2013 12:59 am
Take it as 7^8 (We have to take only the last digits, rest doesn't mattter)

Last digit for exponents are:-
7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1...

So, 7^8 = 1

So, the correct answer is A
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by mevicks » Mon Oct 21, 2013 1:28 am
sahilchaudhary wrote:Take it as 7^8 (We have to take only the last digits, rest doesn't mattter)
Hi Sahil,

Unfortunately this approach cannot be generalized.

Consider this case
7^18

Now if you just consider the last digit "8" the units digit of the answer would be 1
but actually 7^18 = 1628413597910449 (used the calculator to explain the point :) )

So basically you have to consider the complete number (in our example its 18, and in the original question it is 1279089028)

Applying our shortcut to 7^18 --> 18 is a multiple of 9 --> check the cyclicity table --> according to the table the units digit should be 9

Hope it helps.
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by ganeshrkamath » Mon Oct 21, 2013 1:28 am
AIM TO CRACK GMAT wrote:What is the last digit of the number 537^1279089028?

(A) 1
(B) 3
(C) 6
(D) 7

Hi guys i tried the method where u follow the 7^2 n get the series of 7,9,3,1,7... N then tried to divide it! somehow i get 7 but that is the wrong answer!!

[spoiler]OA = 1[/spoiler]
7^1 = 7
7^2 = 49
7^3 = 343
7^4 = ...1
7^5 = ...7
and so on.

Clearly, there's a pattern here: 7,9,3,1,7,9,3,1,...
This pattern has a cyclicity of 4.
Now, divisibility rule of 4 is that the number formed by the last 2 digits of the given number should be divisible by 4.
In this case: 1279089028 => 28 should be divisible by 4 (and it is)
So the units digit is 1.

Choose A

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by Brent@GMATPrepNow » Mon Oct 21, 2013 6:07 am
If anyone is interested, we have a free video on this topic: https://www.gmatprepnow.com/module/gmat- ... ts?id=1031

I also wrote an article for BTG called "The Units Digits of Big Powers": https://www.beatthegmat.com/mba/2012/10/ ... big-powers

Cheers,
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by sahilchaudhary » Mon Oct 21, 2013 6:30 am
mevicks wrote:
sahilchaudhary wrote:Take it as 7^8 (We have to take only the last digits, rest doesn't mattter)
Hi Sahil,

Unfortunately this approach cannot be generalized.

Consider this case
7^18

Now if you just consider the last digit "8" the units digit of the answer would be 1
but actually 7^18 = 1628413597910449 (used the calculator to explain the point :) )

So basically you have to consider the complete number (in our example its 18, and in the original question it is 1279089028)

Applying our shortcut to 7^18 --> 18 is a multiple of 9 --> check the cyclicity table --> according to the table the units digit should be 9

Hope it helps.
Thanks mevicks for telling me.
Does using the last 2 digits works all the time ?
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by theCodeToGMAT » Mon Oct 21, 2013 6:40 am
sahilchaudhary wrote: Thanks mevicks for telling me.
Does using the last 2 digits works all the time ?
Yes, since the procedure requires divisibility by "4" and to check any number's divisibility by "4" we consider last two digits.
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