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Difficulty:
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Question Stats:
81% (00:47) correct
19%
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wrong
based on 1938
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What is the unit's digit of \(7^{75} + 6\)?
A. 1
B. 3
C. 5
D. 7
E. 9
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.
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.
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14 Sep 2010, 13:42
Kudos
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sandeep800
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.
14 Sep 2010, 13:20
Kudos
Bookmarks
Official Solution:
What is the unit's digit of \(7^{75} + 6\)?
A. 1
B. 3
C. 5
D. 7
E. 9
The unit's digit of 7 raised to a positive integer power cycles in a pattern of four: 7-9-3-1. Given that \(75 = 4 *18 + 3\), the unit's digit of \(7^{75}\) aligns with the unit's digit of \(7^3\), which is 3.
Consequently, the unit's digit of \(7^{75} + 6\) is: 3 + 6 = 9.
Answer: E
What is the unit's digit of \(7^{75} + 6\)?
A. 1
B. 3
C. 5
D. 7
E. 9
The unit's digit of 7 raised to a positive integer power cycles in a pattern of four: 7-9-3-1. Given that \(75 = 4 *18 + 3\), the unit's digit of \(7^{75}\) aligns with the unit's digit of \(7^3\), which is 3.
Consequently, the unit's digit of \(7^{75} + 6\) is: 3 + 6 = 9.
Answer: E
