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D
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Difficulty:
45%
(medium)
Question Stats:
62% (01:35) correct
38%
(01:35)
wrong
based on 334
sessions
History
Date
Time
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If \(p=1^3+3^3+5^3 +7^3 +9^3 +11^3 +13^3\), what will be the remainder when p is divided by 5?
A) zero
B) one
C) two
D) three
E) four
A) zero
B) one
C) two
D) three
E) four
24 Nov 2016, 20:42
Kudos
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stonecold
Hi
The best way would be UNIT digit..
But another easier way would be to know \(a^3+b^3\) is always div by a+b..
So put the equation in that pair
\(p=1^3+3^3+(5^3) +(9^3 +11^3) +(7^3+13^3)\),
Here only 1^3+3^3 is left..
=1+27=28..
Remainder is 28-25=3
D
General Discussion
24 Nov 2016, 19:39
Kudos
Bookmarks
stonecold
Solution 1:
To find the remainder when divide by 5, first we need to find the units digit.
\(1^3=1\) has units digit 1
\(3^3=27\) has units digit 7
\(5^3=125\) has units digit 5
\(7^3=343\) has units digit 3
\(9^3=729\) has units digit 9
\(11^3\) has units digit 1
\(13^3\) has the same units digit as \(3^3\), which has units digit 7
So the units digit of \(p\) is \(1+7+5+3+9+1+7=43\) or the units digit is 3.
So the remainder is 3 when divide \(p\) by 5.
Solution 2:
\(1^3 \equiv 1 \pmod{5}\)
\(3^3=(5-2)^3 \equiv -2^3 \pmod{5}\)
\(5^3 \equiv 0 \pmod{5}\)
\(7^3 =(5+2)^3 \equiv 2^3 \pmod{5}\)
\(9^3 =(5 \times 2-1)^3 \equiv -1^3 \pmod{5}\)
\(11^3 = (5 \times 2 +1)^3 \equiv 1^3 \pmod{5}\)
\(13^3 =(5 \times 3 -2)^3 \equiv -2^3 \pmod{5}\)
So \(p \equiv 1-2^3+0+2^3-1^3+1^3-2^3 \pmod{5}\)
\(\implies p \equiv 1-8=-7 \pmod{5} \implies p \equiv 3 \pmod{5}\)
The answer is D
