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

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

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New post Updated on: 24 Jan 2017, 21:37
1
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A
B
C
D
E

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Question Stats:

71% (01:07) correct 29% (01:23) wrong based on 210 sessions

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What is the units digit of \((71)^{5}*(46)^{3}*(103)^{4} + (57)*(1088)^{3}\) ?

A. 0
B. 1
C. 2
D. 3
E. 4

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Originally posted by hazelnut on 24 Jan 2017, 19:27.
Last edited by Bunuel on 24 Jan 2017, 21:37, edited 1 time in total.
Added the OA.
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Re: What is the units digit of (71) 5(46) 3(103) 4 + (57)(1088) 3 ?  [#permalink]

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New post 25 Jan 2017, 02:31
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1
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


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Re: What is the units digit of (71) 5(46) 3(103) 4 + (57)(1088) 3 ?  [#permalink]

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New post 25 Jan 2017, 02:08
Could somebody explain,Please?
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Re: What is the units digit of (71) 5(46) 3(103) 4 + (57)(1088) 3 ?  [#permalink]

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New post 03 Apr 2017, 09:41
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.
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Re: What is the units digit of (71) 5(46) 3(103) 4 + (57)(1088) 3 ?  [#permalink]

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New post 06 Apr 2017, 09:05
ziyuen wrote:
What is the units digit of \((71)^{5}*(46)^{3}*(103)^{4} + (57)*(1088)^{3}\) ?

A. 0
B. 1
C. 2
D. 3
E. 4


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
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Re: What is the units digit of (71) 5(46) 3(103) 4 + (57)(1088) 3 ?  [#permalink]

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Re: What is the units digit of (71) 5(46) 3(103) 4 + (57)(1088) 3 ?   [#permalink] 02 Apr 2020, 05:34

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

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