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What is the unit's digit of 7^75 + 6 ?

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What is the unit's digit of 7^75 + 6 ?  [#permalink]

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New post Updated on: 03 Jul 2013, 00:52
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What is the unit's digit of \(7^{75} + 6\) ?

A. 1
B. 3
C. 5
D. 7
E. 9

(C) 2008 GMAT Club - m12#29

I put the official explanation and the part I do not understand (blue text) in a spoiler

\(7^1\) ends with 7

\(7^2\) ends with 9

\(7^3\) ends with 3

\(7^4\) ends with 1

\(7^5\) ends with 7

...

\(7^{76}\) ends with 1. --> ???

So, \(7^{75}\) ends with 3. --> ???
If 7^5 ends with 7, shouldnt 7^75 also end with 7? Hence 7+6=13 - answer b?? Please help!!


\(7^{75} + 6\) ends with 9.
The correct answer is E.

Originally posted by AndreG on 14 Sep 2010, 12:58.
Last edited by Bunuel on 03 Jul 2013, 00:52, edited 1 time in total.
Renamed the topic and edited the question.
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Re: GMAT CLUB TEST m12#29 - last digit  [#permalink]

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New post 14 Sep 2010, 13:42
4
7
sandeep800 wrote:
AndreG wrote:
What is the unit's digit of \(7^{75} + 6\) ?

(C) 2008 GMAT Club - m12#29

* 1
* 3
* 5
* 7
* 9

I put the official explanation and the part I do not understand (blue text) in a spoiler

\(7^1\) ends with 7

\(7^2\) ends with 9

\(7^3\) ends with 3

\(7^4\) ends with 1

\(7^5\) ends with 7

...

\(7^{76}\) ends with 1. --> ???

So, \(7^{75}\) ends with 3. --> ???
If 7^5 ends with 7, shouldnt 7^75 also end with 7? Hence 7+6=13 - answer b?? Please help!!


\(7^{75} + 6\) ends with 9.
The correct answer is E.



well i will say that whatever may be the number if we have to find the last digit of some number whose power isgiven..then the best method is to divide the power by 4 since all the digits from 1...9 will surely repeat after every 4th digit...
then raise the digit to the power of remainder...
here 75/4 remainder=3
7^3=last digit comes out to be 3
now 3+6=9

thanx


The above is correct with a little correction: when remainder is zero, then we should rise to the power not of remainder 0 but to the power of the cyclicity number.

For example las digit of 7^24 is the same as the last digit of 7^4 as the cyclicity of 7 in power is 4 and 24 divided by 4 gives remainder of zero.

From Number Theory chapter of Math Book:

LAST DIGIT OF A POWER

Determining the last digit of \((xyz)^n\):

1. Last digit of \((xyz)^n\) is the same as that of \(z^n\);
2. Determine the cyclicity number \(c\) of \(z\);
3. Find the remainder \(r\) when \(n\) divided by the cyclisity;
4. When \(r>0\), then last digit of \((xyz)^n\) is the same as that of \(z^r\) and when \(r=0\), then last digit of \((xyz)^n\) is the same as that of \(z^c\), where \(c\) is the cyclisity number.

• Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base.
• Integers ending with 2, 3, 7 and 8 have a cyclicity of 4.
• Integers ending with 4 (eg. \((xy4)^n\)) have a cyclisity of 2. When n is odd \((xy4)^n\) will end with 4 and when n is even \((xy4)^n\) will end with 6.
• Integers ending with 9 (eg. \((xy9)^n\)) have a cyclisity of 2. When n is odd \((xy9)^n\) will end with 9 and when n is even \((xy9)^n\) will end with 1.

Example: What is the last digit of \(127^{39}\)?
Solution: Last digit of \(127^{39}\) is the same as that of \(7^{39}\). Now we should determine the cyclisity of \(7\):

1. 7^1=7 (last digit is 7)
2. 7^2=9 (last digit is 9)
3. 7^3=3 (last digit is 3)
4. 7^4=1 (last digit is 1)
5. 7^5=7 (last digit is 7 again!)
...

So, the cyclisity of 7 is 4.

Now divide 39 (power) by 4 (cyclisity), remainder is 3.So, the last digit of \(127^{39}\) is the same as that of the last digit of \(7^{39}\), is the same as that of the last digit of \(7^3\), which is \(3\).

Hope it helps.
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Re: GMAT CLUB TEST m12#29 - last digit  [#permalink]

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New post 14 Sep 2010, 13:20
3
AndreG wrote:
What is the unit's digit of \(7^{75} + 6\) ?

(C) 2008 GMAT Club - m12#29

* 1
* 3
* 5
* 7
* 9

I put the official explanation and the part I do not understand (blue text) in a spoiler

\(7^1\) ends with 7

\(7^2\) ends with 9

\(7^3\) ends with 3

\(7^4\) ends with 1

\(7^5\) ends with 7

...

\(7^{76}\) ends with 1. --> ???

So, \(7^{75}\) ends with 3. --> ???
If 7^5 ends with 7, shouldnt 7^75 also end with 7? Hence 7+6=13 - answer b?? Please help!!


\(7^{75} + 6\) ends with 9.
The correct answer is E.


7 in power repeats pattern of 4: 7-9-3-1. As 75=4*18+3 then the last digit of \(7^{75}\) is the same as the last digit of \(7^3\), which is 3. Units digit of \(7^{75} + 6\) will be: 3 plus 6 = 9.

Answer: E.

For more on this issue check Number Theory chapter of Math Book (link in my signature).

Hope it helps.
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Re: GMAT CLUB TEST m12#29 - last digit  [#permalink]

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New post 14 Sep 2010, 13:26
AndreG wrote:
What is the unit's digit of \(7^{75} + 6\) ?

(C) 2008 GMAT Club - m12#29

* 1
* 3
* 5
* 7
* 9

I put the official explanation and the part I do not understand (blue text) in a spoiler

\(7^1\) ends with 7

\(7^2\) ends with 9

\(7^3\) ends with 3

\(7^4\) ends with 1

\(7^5\) ends with 7

...

\(7^{76}\) ends with 1. --> ???

So, \(7^{75}\) ends with 3. --> ???
If 7^5 ends with 7, shouldnt 7^75 also end with 7? Hence 7+6=13 - answer b?? Please help!!


\(7^{75} + 6\) ends with 9.
The correct answer is E.



well i will say that whatever may be the number if we have to find the last digit of some number whose power isgiven..then the best method is to divide the power by 4 since all the digits from 1...9 will surely repeat after every 4th digit...
then raise the digit to the power of remainder...
here 75/4 remainder=3
7^3=last digit comes out to be 3
now 3+6=9

thanx
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Re: GMAT CLUB TEST m12#29 - last digit  [#permalink]

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New post 14 Sep 2010, 13:55
1
Bunuel wrote:
sandeep800 wrote:
AndreG wrote:
What is the unit's digit of \(7^{75} + 6\) ?

(C) 2008 GMAT Club - m12#29

* 1
* 3
* 5
* 7
* 9

I put the official explanation and the part I do not understand (blue text) in a spoiler

\(7^1\) ends with 7

\(7^2\) ends with 9

\(7^3\) ends with 3

\(7^4\) ends with 1

\(7^5\) ends with 7

...

\(7^{76}\) ends with 1. --> ???

So, \(7^{75}\) ends with 3. --> ???
If 7^5 ends with 7, shouldnt 7^75 also end with 7? Hence 7+6=13 - answer b?? Please help!!


\(7^{75} + 6\) ends with 9.
The correct answer is E.



well i will say that whatever may be the number if we have to find the last digit of some number whose power isgiven..then the best method is to divide the power by 4 since all the digits from 1...9 will surely repeat after every 4th digit...
then raise the digit to the power of remainder...
here 75/4 remainder=3
7^3=last digit comes out to be 3
now 3+6=9

thanx


The above is correct with a little correction: when remainder is zero, then we should rise to the power not of remainder 0 but to the power of the cyclicity number.

For example las digit of 7^24 is the same as the last digit of 7^4 as the cyclicity of 7 in power is 4 and 24 divided by 4 gives remainder of zero.

From Number Theory chapter of Math Book:

LAST DIGIT OF A POWER

Determining the last digit of \((xyz)^n\):

1. Last digit of \((xyz)^n\) is the same as that of \(z^n\);
2. Determine the cyclicity number \(c\) of \(z\);
3. Find the remainder \(r\) when \(n\) divided by the cyclisity;
4. When \(r>0\), then last digit of \((xyz)^n\) is the same as that of \(z^r\) and when \(r=0\), then last digit of \((xyz)^n\) is the same as that of \(z^c\), where \(c\) is the cyclisity number.

• Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base.
• Integers ending with 2, 3, 7 and 8 have a cyclicity of 4.
• Integers ending with 4 (eg. \((xy4)^n\)) have a cyclisity of 2. When n is odd \((xy4)^n\) will end with 4 and when n is even \((xy4)^n\) will end with 6.
• Integers ending with 9 (eg. \((xy9)^n\)) have a cyclisity of 2. When n is odd \((xy9)^n\) will end with 9 and when n is even \((xy9)^n\) will end with 1.

Example: What is the last digit of \(127^{39}\)?
Solution: Last digit of \(127^{39}\) is the same as that of \(7^{39}\). Now we should determine the cyclisity of \(7\):

1. 7^1=7 (last digit is 7)
2. 7^2=9 (last digit is 9)
3. 7^3=3 (last digit is 3)
4. 7^4=1 (last digit is 1)
5. 7^5=7 (last digit is 7 again!)
...

So, the cyclisity of 7 is 4.

Now divide 39 (power) by 4 (cyclisity), remainder is 3.So, the last digit of \(127^{39}\) is the same as that of the last digit of \(7^{39}\), is the same as that of the last digit of \(7^3\), which is \(3\).

Hope it helps.


Thanx a lot bunuel for correcting me..i wud have applied my method in GMAT if u had not corrected me....:)
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Re: GMAT CLUB TEST m12#29 - last digit  [#permalink]

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New post 14 Sep 2010, 14:06
Wow, you guys helped me a lot! THANKS!!
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Re: GMAT CLUB TEST m12#29 - last digit  [#permalink]

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New post 14 Sep 2010, 15:23
Thanks for summarising the concept. I used to calculate what you call cyclicity in every problem and reach my conclusions
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What is the unit's digit of 7^75 + 6 ?  [#permalink]

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New post 09 Mar 2014, 13:09
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Re: What is the unit's digit of 7^75 + 6 ?  [#permalink]

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New post 23 Nov 2016, 11:45
AndreG wrote:
What is the unit's digit of \(7^{75} + 6\) ?

A. 1
B. 3
C. 5
D. 7
E. 9


Since , the cyclicity of 7 is 4

The units digit of \(7^{75} = 3\)

So, Units digit will be 3+ 6 = 9

Hence, answer will be (E) 9...


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Re: What is the unit's digit of 7^75 + 6 ?  [#permalink]

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New post 23 Jan 2017, 18:22
Nice Question.
Here is what i did in this one ->
Cyclicity of 7 is 4 =>
7
9
3
1
Hence the units digit of 7^75 => 7^4m+3 will be 3.
So 7^75+6 will have 3+6=9 as its units digit.

Hence E.

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Re: What is the unit's digit of 7^75 + 6 ?  [#permalink]

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New post 17 Mar 2018, 11:38
Hi All,

The GMAT doesn't expect you to calculate the value of 7^75; instead, let's try to figure out the pattern behind this math…. The question asks for the UNITS DIGIT, so we'll focus on that…

7^1 = 7
7^2 = 49
7^3 = 343
7^4 = ends in a 1

7^5 = ends in a 7
7^6 = ends in a 9
etc.

The pattern is 7931 7931 7931 etc.

So the units digit will follow this repeating pattern of 4 units digits…

75/4 = 18r3

Thus, the 75th number will be the "third number" in the pattern…which is a 3

The question asks us to ADD 6 to this number though….

3+6 = 9

Final Answer:

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What is the unit's digit of 7^75 + 6 ?  [#permalink]

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New post 17 Mar 2018, 11:50
Hi
To solve these types of questions, it is not necessary to remember cyclicity order of different digits,
In fact, all digits follow a common cyclicity order, they repeat itself after 4k+1 power.
Steps to solve such questions:
1) divide the power by 4 and find remainder.
here the remainder is 3
2) Now find the unit digit by raising it to exponent of remainder (if remainder is 0, raise it to exponent 4)
here, it is 7^3 = 3

Now answer = 3+6 =9.

this method works for every digit :cool:

(PS: for some of digits , we have simpler pattern method.
1) 0 - always 0
2) 4 - odd power = 4, even power = 6
3) 5 always 5
4) 6 always 6
5) 9 - odd power 9, even power = 1)
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Re: What is the unit's digit of 7^75 + 6 ?  [#permalink]

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Re: What is the unit's digit of 7^75 + 6 ?   [#permalink] 08 May 2019, 05:15
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