Problem Solving

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]

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}

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"

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:


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]

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

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.

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

Cheers

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 ?

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.