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# If p=1^3+3^3+5^3 +7^3 +9^3 +11^3 +13^3, what will be the

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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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Updated on: 25 Nov 2016, 06:39
5
00:00

Difficulty:

45% (medium)

Question Stats:

61% (01:31) correct 39% (01:46) wrong based on 106 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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Originally posted by stonecold on 24 Nov 2016, 12:32.
Last edited by stonecold on 25 Nov 2016, 06:39, 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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24 Nov 2016, 18:39
3
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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24 Nov 2016, 19:42
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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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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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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25 Nov 2016, 02:47
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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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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16 Dec 2017, 05:43
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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 &nbs [#permalink] 16 Dec 2017, 05:43
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