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f(2^2006)+f(3^2006)+f(5^2006)+f(7^2006)+f(9^2006)

We have to find the unit digit of the following,
2^2006,3^2006,5^2006,7^2006,9^2006

For, 2^2006 = (2^2004)(2^2) Taking unit digits 6*4=24, unit digit = 4
3^2006 = (3^2004)(3^2) Taking unit digits 1*9=9, unit digit = 9
5^2006 , unit digit = 5
7^2006 = (7^2004)(7^2) Taking unit digits 1*9=9, unit digit = 9
9^2006 = (9^2004)(9^2) Taking unit digits 1*1=1, unit digit = 1

Adding the unit digits from above = 4+9+5+9+1 = 28
Hence D
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Bunuel
\(f(x)\) denotes the remainder when \(x\) is divided by \(10\). For example,\( f(5) = 5, f(92) = 2\) and \(f(271) = 1\). What is the value \(f(2^{2006}) + f(3^{2006}) + f(5^{2006}) + f(7^{2006}) + f(9^{2006})\)?

A. \(2\)4

B. \(2\)6

C. \(2\)8

D. \(3\)0

E. \(3\)2

Note that the unit digit of each option is different. Hence, if calculate the unit digit of the sum, we can narrow down the answer

2006 = 2004 + 2

  • \(f(2^{2006})\) = Unit digit = 4
  • \(f(3^{2006})\) = Unit digit = 9
  • \(f(5^{2006})\) = Unit digit = 5
  • \(f(7^{2006})\) = Unit digit = 9
  • \(f(9^{2006})\) = Unit digit = 1

Sum = 4 + 9 + 5 + 9 + 1 = Unit digit = 8

Option C
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Bunuel
\(f(x)\) denotes the remainder when \(x\) is divided by \(10\). For example,\( f(5) = 5, f(92) = 2\) and \(f(271) = 1\). What is the value \(f(2^{2006}) + f(3^{2006}) + f(5^{2006}) + f(7^{2006}) + f(9^{2006})\)?

A. \(24\)

B. \(26\)

C. \(28\)

D. \(30\)

E. \(32\)


Remainder when divided by 10 means nothing but the unit’s digit of the number.

So we can rephrase the question as
What is the sum of units digit of \(f(2^{2006}) , f(3^{2006}) , f(5^{2006}) , f(7^{2006}) , f(9^{2006})\)?
Now 2006=4*501+2 and the units digit repeats after every 4th successive power. ( It could repeat every time, for example 1, 5, 6 and 0, or repeat after two successive powers, for example 4 and 9.)

\(f(2^{2006}) + f(3^{2006}) + f(5^{2006}) + f(7^{2006}) + f(9^{2006})\) would mean
\(f(2^{2}) + f(3^{2}) + f(5^{2}) + f(7^{2}) + f(9^{2})\)
4+9+25+49+81 = 4+9+5+9+1 = 28

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