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m12, #29

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m12, #29  [#permalink]

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New post 16 Nov 2008, 23:22
3
What is the unit's digit of \(7^{75}+6\)?

(A) 1
(B) 3
(C) 5
(D) 7
(E) 9

Spoiler: :: OA
E

Source: GMAT Club Tests - hardest GMAT questions

The solution says that 7^76 ends with 1. how do we come to know this?
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New post 17 Nov 2008, 03:49
2
7^4 will end with 1 and hence any power to 7 that is multiple of 4 will end with 1.
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New post 17 Nov 2008, 04:00
3
ritula wrote:
What is the unit's digit of 7^75+6 ?

1
3
5
7
9

The solution says that 7^76 ends with 1. how do we cum 2 know this?


only units place

7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
7^5 = 7..etc

7,9,3,1 is the pattern.

7^75 will end with 3, i.e 3+6 = 9

E
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New post 21 Nov 2008, 21:26
1
7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
7^5 = 7..etc

7,9,3,1 is the pattern.

7^75 will end with 3(since 75=4(18)+3)

agreed...its 9
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New post 19 Apr 2010, 06:08
2
7^1, 7^2, 7^3, 7^4, ....7^73
....7,...9,...3,...1, ... ....7

75/4 = 18 rem 3 ...the 3rd power in the series takes a unit digit of 3

Therefore, the answer will be ..3 + 6 = ..9

E
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New post 19 Apr 2010, 22:23
7^1 = 7 => unit digit 7
7^2 = 49 => unit digit 9
7^3 = 343 => unit digit 3
7^4 = 2401 => unit digit 1
7^5 = 16807=> unit digit 7
so cyclicity is 4

75/4 has remainder as 3.
so unit digit of 7^75 will be 3
so answer is 9 hence E.
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New post 21 Apr 2011, 05:07
7^75+6
=>7^(4*18+3) + 6
=> unit digit is 3 +6
=>9
E
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New post 21 Apr 2011, 05:12
7 has a cycle of 4
75/4 = 3 Rem

So 7^75 has same last digit as 7^3 = 3

3 + 6 = 9

Answer - E
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New post 21 Apr 2011, 11:33
the exponents of 7 have units digits ending in 7,9,3 and 1 and then series repeat .
So if 7 exponent 75 ( ie 75/4 = 3) which means we need to 3 units digits and add 6 to it
Correct answer E
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New post 23 Apr 2011, 14:14
7 has cyclicity of 4.

=> unit digit of 7^75 is same as unit digit of 7^ Remainder (75/4) = unit digit of 7^3 =3
=> unit digiit of 7^75 +6 = 3 +6 = 9

Answer is E.
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New post 24 Apr 2011, 11:01
7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401
7^5 = 16807 (as we see the cycle starts repeating as far as unit's digit is concerned).

We need to get to the power 75, and 75/4 yeilds a remainder of 3 (4*18 + 3), hence the unit's digit for 7*75 is 3.

The question asks for unit digit when 6 is added, hence 6+3 = 9.
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New post 24 Apr 2011, 23:10
1
Here the first part can be written as:

7^1 = 7 (ending with 7)
7^2 = 49 (ending with 9)
7^3 = 343 (ending with 3)
7^4 = 2401 (ending with 1)
7^5 = some number (ending with 7)

so it will create numbers ending with 7,9,3,1,7,9,3,1,.....and so on

The power 75 can be broken into 18 * 4 (means 19 such groups of four) + 3

so when all those 18 groups are over we will have a number ending with 1. Then follow the pattern to reach:

1 => 7, 9, 3 (stop)

So we have a number ending with 3 and if we add 6 to this we will get a number ending with 9.

So answer is e :).
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New post 25 Apr 2012, 07:44
the unit digits in power repeat after an interval of 4.
hence, 3^1 has same nits digit as 3^5, 3^9, 3^13 ......
this is true for all integers.
hence. 7^75 willhave sam eunits digit as 7^3i.e 3
hence units digit will be 3+6 = 9 option E.
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New post 26 Apr 2013, 06:28
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New post 29 Apr 2013, 11:17
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I think there is an easier/faster method for this one...... we have series of 7,9,3,1,7,9,3,1...... sets of 4 numbers (7,9,3,1).....
closest number to 75 which is divisible by 4 is 76.... therefore 7^76 will have 1 at unit place.........

.that means it will be 3 at the 75th power....so 3+6 will give '9' at unit place for the final answer



for example if we want to find out unit place for 8^9+6

8^1 = 8....8^2=64....8^3=(unitplace)2........8^4=(unitplace)6.....8^5=(unitplace)8....so on......again we have (8,4,2,6,8,4,2,6....) sets of 4.......closest number to 9 which is divisible by 4 is 8.....therefore 8^8 will have 6 at unit place....so 8^9 should have 8 ....and 8+6 gives '4' at unit place for final answer.

Checked on calculator: (8^9)+6 =134217734
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Re: m12, #29   [#permalink] 29 Apr 2013, 11:17
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