What is the units digit of (71) 5(46) 3(103) 4 + (57)(1088) 3 ?

Here we go....


Thumb Rule - Always pick the digit at the Unit place for such questions...

let's go step by step

(71)^5 ------> 1^5 ----- always result in a number that has 1 at it's unit place ---------- (1)


(46)^3 -------> 6^3 ------ like 1, 6 always result in a number that has 6 at it's unit place ---------- (2)

Check yourself 6*6 = 36*6 => 216*6 => 1296 and so on

(103)^4 -------> 3^4

Here is the pattern in case of 3 that should be considered

3 = 3
3*3 = 9 (3^2)
3*3*3 = 2(7) (3^3)
3*3*3*3 = 8(1) (3^4)
3*3*3*3*3 = last digit 3

so pattern order in case of number 3 is 4

back to our number in the question ---> (103)^4 => 3^4 => last digit would be 1 ---------------- (3)

From equations 1, 2, and 3

1*6*1 would result in 6

so from (71)^5∗(46)^3∗(103)^4 results in a number that has a last digit of 6 ------------------(a)

Apply the above principle to (57)∗(1088)^3

8 = 8
8*8 = 6(4)
8*8*8 = last digit would be (2)

the last digit of the product of (57)∗(1088)^3 ===> 7*2 => 14 => 4 (we are interested in the unit place) ----------(b)


Combine a and b


6 + 4 would result in 0 at the unit place....


Hence option A is correct


Consider Kudos for the post if it helped.

The question is asking for the unit's digit. So we will just consider the unit's digit of all the numbers because the unit's digit of a operation depends on the last digit only.
For example, 22*24 will end in 8 (2*4).

The other rule is to remember the cyclicity of nos.
Cyclicity of 2= 4 (that means when 2 is squared, cubed, and so on the last digit start repeating after four i.e. 2,4,8.6)
Cyclicity of 3= 4 (3,9,7,1 - only the unit's digit)
Cyclicity of 4= 2 (4,6)
Cyclicity of 5= 1 (5)
Cyclicity of 6= 1 (6)
Cyclicity of 7= 4 (7,9,3,1)
Cyclicity of 8= 4 (8,4,2,6)
Cyclicity of 9= 2 (9,1)


Just write the unit's digit:
1*6*3^4 + 7*8^3
First let's determine 3^4 and 8^3
3^4 will end in 1 and 8^3 will end in 2

=6+4 (4 because 7*2=14)
=10
The unit's digit is 0.

Could somebody explain,Please?

We need to determine the units digit of (71)^5 x (46)^3 x (103)^4 + (57) x (1088)^3. Since we only care about the units digits, we can rewrite the expression as:

What is the units digit of (1)^5 x (6)^3 x (3)^4 + (7) x (8)^3 ?

Since 1 raised to any power always has a units digit of 1, 1^5 will have a units digit of 1.

Since 6 raised to any power always has a units digit of 6, 6^3 will have a units digit of 6.

Since (3)^4 = 81, 3^4 has a units digit of 1.

7, of course, has a units digit of 7.

Since 8^3 = 512, 8^3 has a units digit of 2.

Let’s now plug all of this information into the expression:

1 x 6 x 1 + 7 x 2 = 6 + 14 = 20

Thus, the units digit is 0.

Answer: A
_________________

\((71)^{5}*(46)^{3}*(103)^{4} + (57)*(1088)^{3}\)

1 * 6 *1 + 7 * 2

6 + 4

0

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________