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What is the unit digit of 7^6 -6^7
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nick1816
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What is the unit digit of 7^6 -6^7 [#permalink] New post  22 Apr 2019, 19:28
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What is the unit digit of \(7^6 - 6^7\)?

A. 3
B. 7
C. 5
D. 1
E. 9
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chetan2u
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Re: What is the unit digit of 7^6 -6^7 [#permalink] New post  22 Apr 2019, 20:02
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nick1816 wrote:
What is the unit digit of \(7^6 - 6^7\)?

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


The term \(7^6=7^(4+2)\), so same units digit as 7^2 or units digit is 9.
The term \(6^7\) will end in 6 ..

So, \(7^6 - 6^7\) is same as( ..abc9 - ..xyz6)
So two cases..
(1) \(7^6 >6^7\), then units digit is 9-6 or 3
(2) \(7^6 <6^7\), then units digit is 16-9 or 7

Now \(a^b>b^a\), when a and b are consecutive and b>a, that is b=a+1

So here case II

B
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manu11
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Re: What is the unit digit of 7^6 -6^7 [#permalink] New post  22 Apr 2019, 20:32
chetan2u wrote:
nick1816 wrote:
What is the unit digit of \(7^6 - 6^7\)?

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


The term \(7^6=7^(4+2)\), so same units digit as 7^2 or units digit is 9.
The term \(6^7\) will end in 6 ..

So, \(7^6 - 6^7\) is same as( ..abc9 - ..xyz6)
So two cases..
(1) \(7^6 >6^7\), then units digit is 9-6 or 3
(2) \(7^6 <6^7\), then units digit is 16-9 or 7

Now \(a^b>b^a\), when a and b are consecutive and b>a, that is b=a+1

So here case II

B



Are there any exceptions to \(a^{b}>b^a\), when a and b are consecutive and b>a, that is b=a+1?

I tried on numbers 2 & 3, 3= 2+1 , but \(2^3(8) < 3^2(9)\)
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chetan2u
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Re: What is the unit digit of 7^6 -6^7 [#permalink] New post  22 Apr 2019, 20:40
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manu11 wrote:
chetan2u wrote:
nick1816 wrote:
What is the unit digit of \(7^6 - 6^7\)?

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


The term \(7^6=7^(4+2)\), so same units digit as 7^2 or units digit is 9.
The term \(6^7\) will end in 6 ..

So, \(7^6 - 6^7\) is same as( ..abc9 - ..xyz6)
So two cases..
(1) \(7^6 >6^7\), then units digit is 9-6 or 3
(2) \(7^6 <6^7\), then units digit is 16-9 or 7

Now \(a^b>b^a\), when a and b are consecutive and b>a, that is b=a+1

So here case II

B



Are there any exceptions to \(a^{b}>b^a\), when a and b are consecutive and b>a, that is b=a+1?

I tried on numbers 2 & 3, 3= 2+1 , but \(2^3(8) < 3^2(9)\)


Hi..
Yes the initial two positive integers are exceptions..
1^2<2^1
2^3<3^2
But thereafter the gaps in two keeps increasing the other way..

3^4>4^3...81>64
4^5>5^4..1024>625
And so on..
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nick1816
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Joined: 19 Oct 2018
Posts: 2242
Location: India
Re: What is the unit digit of 7^6 -6^7 [#permalink] New post  22 Apr 2019, 21:52
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Excellent observation buddy.
There is one more way to check which term is greater
assume \(7^6\)>\(6^7\)
or \((7/6)^6\)>6
or \((1.1667)^6\)>6
Now we know \((1.1667)^6\)<\((1.2)^6\)
\((1.1667)\)^6<\((1.44)^3\)<\((1.5)^3\)
\((1.1667)^6\)<3.375
So our assumption was wrong
\(7^6\) must be smaller than \(6^7\)


chetan2u wrote:
nick1816 wrote:
What is the unit digit of \(7^6 - 6^7\)?

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


The term \(7^6=7^(4+2)\), so same units digit as 7^2 or units digit is 9.
The term \(6^7\) will end in 6 ..

So, \(7^6 - 6^7\) is same as( ..abc9 - ..xyz6)
So two cases..
(1) \(7^6 >6^7\), then units digit is 9-6 or 3
(2) \(7^6 <6^7\), then units digit is 16-9 or 7

Now \(a^b>b^a\), when a and b are consecutive and b>a, that is b=a+1

So here case II

B
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