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stonecold
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If p=1^3+3^3+5^3 +7^3 +9^3 +11^3 +13^3, what will be the Updated on: 25 Nov 2016, 07:39
Originally posted by stonecold on 24 Nov 2016, 13:32.
Last edited by stonecold on 25 Nov 2016, 07:39, edited 2 times in total.
Renamed the topic and edited the question.
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62% (01:35) correct 38% (01:35) wrong based on 334 sessions

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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
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D
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chetan2u
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If p=1^3+3^3+5^3 +7^3 +9^3 +11^3 +13^3, what will be the 24 Nov 2016, 20:42
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stonecold
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
Show more

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
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If p=1^3+3^3+5^3 +7^3 +9^3 +11^3 +13^3, what will be the 24 Nov 2016, 19:39
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stonecold
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
Show more

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
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If p=1^3+3^3+5^3 +7^3 +9^3 +11^3 +13^3, what will be the 25 Nov 2016, 03:47
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stonecold
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
Show more
When p/5, remainder = 1 + 2 + 0 + 3 + 4 + 1 + 2 = 13
= 13/5

Remainder = 3

Sent from my GT-N7100 using GMAT Club Forum mobile app
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If p=1^3+3^3+5^3 +7^3 +9^3 +11^3 +13^3, what will be the 15 Jan 2019, 18:04
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chetan2u
stonecold
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

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
Show more

Hello chetan2u !

I found very interesting your other approach, why can't 1^3+3^3 be another pair?

I mean, why should we have to leave those outside the \(a^3+b^3\) form?

Kind regards!
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If p=1^3+3^3+5^3 +7^3 +9^3 +11^3 +13^3, what will be the 15 Jan 2019, 20:23
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chetan2u
stonecold
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

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

Hello chetan2u !

I found very interesting your other approach, why can't 1^3+3^3 be another pair?

I mean, why should we have to leave those outside the \(a^3+b^3\) form?

Kind regards!
Show more


Hi..

The reason is we are grouping the terms such that the sum is a multiple of 5..
5^3 will be divisible by 5..
9^3+11^3 will be divisible by (9+11) or 20 and hence by 5
(7^3+11^3) will be divisible by 5 similarly

So we are left with 1^3+3^3, this will be divisible by (1+3) or 4, but we are looking for divisibility by 5..
So we solve it 1^3+3^3=1+27=28..
28 divided by 5 gives a remainder of 3
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If p=1^3+3^3+5^3 +7^3 +9^3 +11^3 +13^3, what will be the 06 Jan 2025, 02:40
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