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m12, #29 [#permalink]
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16 Nov 2008, 23:22
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What is the unit's digit of \(7^{75}+6\)? (A) 1 (B) 3 (C) 5 (D) 7 (E) 9 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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Re: m12, #29 [#permalink]
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17 Nov 2008, 03:49
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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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Re: m12, #29 [#permalink]
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17 Nov 2008, 04:00
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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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Re: m12, #29 [#permalink]
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21 Nov 2008, 21:26
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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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Re: m12, #29 [#permalink]
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19 Apr 2010, 06:08
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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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Re: m12, #29 [#permalink]
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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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Re: m12, #29 [#permalink]
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21 Apr 2011, 05:07
7^75+6 =>7^(4*18+3) + 6 => unit digit is 3 +6 =>9 E
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Re: m12, #29 [#permalink]
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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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Re: m12, #29 [#permalink]
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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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Re: m12, #29 [#permalink]
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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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Re: m12, #29 [#permalink]
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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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Re: m12, #29 [#permalink]
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24 Apr 2011, 23:10
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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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Re: m12, #29 [#permalink]
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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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26 Apr 2013, 06:28



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Re: m12, #29 [#permalink]
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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











