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GRE 1: Q169 V154 If p=1^3+3^3+5^3 +7^3 +9^3 +11^3 +13^3, what will be the  [#permalink]

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Question Stats: 60% (01:30) correct 40% (01:38) wrong based on 130 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

_________________

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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Re: If p=1^3+3^3+5^3 +7^3 +9^3 +11^3 +13^3, what will be the  [#permalink]

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stonecold wrote:
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

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 $$p \equiv 1-2^3+0+2^3-1^3+1^3-2^3 \pmod{5}$$
$$\implies p \equiv 1-8=-7 \pmod{5} \implies p \equiv 3 \pmod{5}$$

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Re: If p=1^3+3^3+5^3 +7^3 +9^3 +11^3 +13^3, what will be the  [#permalink]

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1
2
stonecold wrote:
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
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Re: If p=1^3+3^3+5^3 +7^3 +9^3 +11^3 +13^3, what will be the  [#permalink]

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stonecold wrote:
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

When p/5, remainder = 1 + 2 + 0 + 3 + 4 + 1 + 2 = 13
= 13/5

Remainder = 3

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Joined: 12 Sep 2017
Posts: 263
Re: If p=1^3+3^3+5^3 +7^3 +9^3 +11^3 +13^3, what will be the  [#permalink]

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chetan2u wrote:
stonecold wrote:
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!
Math Expert V
Joined: 02 Aug 2009
Posts: 7743
Re: If p=1^3+3^3+5^3 +7^3 +9^3 +11^3 +13^3, what will be the  [#permalink]

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jfranciscocuencag wrote:
chetan2u wrote:
stonecold wrote:
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!

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
_________________ Re: If p=1^3+3^3+5^3 +7^3 +9^3 +11^3 +13^3, what will be the   [#permalink] 15 Jan 2019, 20:23
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