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![The function f(x) is defined as x - 10[x10]. ([x] is the greatest int](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "The function f(x) is defined as x - 10[x10]. ([x] is the greatest int"){kind=icon}
22 Mar 2020, 20:15
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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64% (02:53) correct
36%
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The function \(f(x)\) is defined as \(x - 10[\frac{x}{10}]\). ([\(x\)] is the greatest integer less than or equal to \(x\)). What is \(f(7) + f(7^2) + f(7^3) +...+ f(7^{2020})\)?
A. \(10036\)
B. \(10039\)
C. \(10040\)
D. \(10100 \)
E. \(10107\)
A. \(10036\)
B. \(10039\)
C. \(10040\)
D. \(10100 \)
E. \(10107\)
22 Mar 2020, 20:17
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\(f(x) = x – 10[\frac{x}{10}]\) means the remainder of \(x\) when \(x\) is divided by \(10.\)
For example, \(f(1234) = 1234 – 10*[\frac{1234}{10}] = 1234 – 10*[123.4] = 1234 – 10*123 = 1234 – 1230 = 4.\)
Units digit of Powers of base \(7\) repeats as follows.
\(f(7^1) = 7 ~ 7^5 ~ 7^9 ~ …\)
\(f(7^2) = f(49) = 9 ~ 7^6 ~ 7^{10} ~ …\)
\(f(7^3) = f(343) = 3 ~ 7^7 ~ 7^{11} ~ …\)
\(f(7^4) = f(2401) = 1 ~ 7^8 ~ 7^{12} ~ …\)
\(f(7) + f(7^2) + f(7^3) +…+ f(7^{2020})\)
\(= f(7) + f(7^2) + f(7^3) + f(7^4) + f(7^5) +…+ f(7^{2017}) + f(7^{2018}) + f(7^{2019}) + f(7^{2020})\)
\(= (7 + 9 + 3 + 1) + (7 + 9 + 3 + 1) +…+ (7 + 9 + 3 + 1)\)
\(= 20 * \frac{2020}{4} = 20*505 = 10100\)
Therefore, D is the answer.
Answer: D
\(f(x) = x – 10[\frac{x}{10}]\) means the remainder of \(x\) when \(x\) is divided by \(10.\)
For example, \(f(1234) = 1234 – 10*[\frac{1234}{10}] = 1234 – 10*[123.4] = 1234 – 10*123 = 1234 – 1230 = 4.\)
Units digit of Powers of base \(7\) repeats as follows.
\(f(7^1) = 7 ~ 7^5 ~ 7^9 ~ …\)
\(f(7^2) = f(49) = 9 ~ 7^6 ~ 7^{10} ~ …\)
\(f(7^3) = f(343) = 3 ~ 7^7 ~ 7^{11} ~ …\)
\(f(7^4) = f(2401) = 1 ~ 7^8 ~ 7^{12} ~ …\)
\(f(7) + f(7^2) + f(7^3) +…+ f(7^{2020})\)
\(= f(7) + f(7^2) + f(7^3) + f(7^4) + f(7^5) +…+ f(7^{2017}) + f(7^{2018}) + f(7^{2019}) + f(7^{2020})\)
\(= (7 + 9 + 3 + 1) + (7 + 9 + 3 + 1) +…+ (7 + 9 + 3 + 1)\)
\(= 20 * \frac{2020}{4} = 20*505 = 10100\)
Therefore, D is the answer.
Answer: D
General Discussion
23 Mar 2020, 03:41
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MathRevolution
Given: The function \(f(x)\) is defined as \(x - 10[\frac{x}{10}]\). ([\(x\)] is the greatest integer less than or equal to \(x\)).
Asked: What is \(f(7) + f(7^2) + f(7^3) +…+ f(7^{2020})\)?
f(x) = x - 10[x/10]
f(7) = 7 - 10[7/10] = 7
f(7^2) = 7^2 - 10[7^2/10] = 7^2 - 40 = 9
f(7^3) = 7^3 - 10[7^3/10] = 7^3 - 340 = 3
f(7^4) = 7^4 - 10[7^4/10] = 7^4 - 2400 = 1
f(7^5) = 7^5 - 10[7^5/10] = 7
The function f(x) provides the unit digit of 7^x
\(f(7) + f(7^2) + f(7^3) +…+ f(7^{2020})\) = 7 + 9 + 3 + 1 + 7 + 9 + 3 + 1 + ..... 2020 terms
7 + 9 + 3 + 1 = 20
20 + 20 + .... 505 terms = 20 *505 = 10100
IMO D
