+0

+2
1
542
9
+132

What is the remainder when 1^3 + 2^3 +3^3 + ...+100^3 is divided by 7?

a)1               b)2                c)3                 d)4                e)5

Please help. I need to know how to answer similar questions without the use of a calculator. Stating formulae and working would be really appreciated. Please explain every step ( if you can)

PS: This is an old past paper. I am not cheating by requesting for help.

Mar 14, 2018

#1
+102466
+1

Well the answer is 2 but I do not know how to do it.

http://www.wolframalpha.com/input/?i=(sum+from+x%3D1+to+x%3D100+(x%5E3))+mod+7

I had done some other work (which i deleted) to decide that the last digit was 0.

I wonder if I can use that fact...

7,14,21,28,35,42,49,56,63,70  It seems multiples of 7 can end in any digit.  That doesn't seem very helpful.

I think there is some strange variation on the sum of GP that you can do this with but i have forgotten it...

Mar 14, 2018
edited by Guest  Mar 14, 2018
#2
+22572
+2

What is the remainder when 1^3 + 2^3 +3^3 + ...+100^3 is divided by 7?

a)1               b)2                c)3                 d)4                e)5

Formula:
$$\begin{array}{|rcll|} \hline 1^3 + 2^3 +3^3 + \ldots +n^3 &=& \dfrac{1}{4}n^4 + \dfrac{1}{2}n^3 + \dfrac{1}{4}n^2 \\ \hline \end{array}$$

$$\mathbf{n= 100}:$$

$$\begin{array}{|rcll|} \hline 1^3 + 2^3 +3^3 + \ldots +100^3 &=& \dfrac{1}{4}100^4 + \dfrac{1}{2}100^3 + \dfrac{1}{4}100^2 \\\\ &=& \dfrac{1}{4}100^2\left( 100^2+2\cdot 100 + 1 \right) \\\\ &=& \dfrac{1}{4}100^2\left( 10000 +200 + 1 \right) \\\\ &=& \dfrac{1}{4}100^2\cdot 10201 \\\\ &=& 2500\cdot 10201 \\ \hline \end{array}$$

$$\mathbf{\pmod 7}:$$

$$\begin{array}{|rcll|} \hline && 2500 \pmod 7 &=& 1 \\ && 10201 \pmod 7 &=& 2 \\ && 2500\cdot 10201 \pmod 7 \\\\ &=& 1\cdot 2 \pmod 7 \\ &\mathbf{=}& \mathbf{2\pmod 7} \\ \hline \end{array}$$

The remainder is 2

Mar 14, 2018
#3
+2

Hi

here is different approach for you to consider......using patterns.

now the sum of the cubes of the positve integers up to 100 goes like this

1+8+27+64+.......+10 now look at the running totals ....1,9,36,100,.....ok so if you divide the final sum

by 7 it's like dividing each term in the series by 7 and then finding its sum. Anyway looking at the

remainders when dividing by 7 at each step of the series ie 1/7 then 9/7,36/7 and so on we see that

each odd sum ie first,third,fifth etc give a remainder of 1. Each even sum ie second,fourth etc gives a

remainder of 2. Conclusion.....there is an even number of terms so the remainder must be 2 when

dividing by 7

hope this helps

Mar 14, 2018
#8
+101872
0

That's a nice way to see this, guest.....!!!!

CPhill  Mar 14, 2018
#4
0

Sum of cubes up to n =n^2(n + 1)^2 / 4

n = 100

[100^2 x 101^2] / 4 =25,502,500

25,502,500 mod 7 = 2 - the remainder.

Mar 14, 2018
edited by Guest  Mar 14, 2018
#6
+10413
+1

What is the remainder when 1^3 + 2^3 +3^3 + ...+100^3 is divided by 7?

Mar 14, 2018
edited by Omi67  Mar 14, 2018
#7
+1

Here's yet another approach.

Consider the block of seven numbers (7n - 3), (7n - 2), (7n - 1), 7n, (7n + 1), (7n + 2), (7n + 3).

n = 1 will cover the numbers 4 to 10,

n = 2 will cover 11 through to 17, and so on.

Cubing the first one get you (7n)^3 - 3(3)(7n)^2 + 3(3)^2(7n) - 3^3.

The first three of those are divisible by 7 so any surplus mod 7 comes from the last term -27,

and that will be 1, ( -27 = (-4).7 + 1)

From the remaining six terms, the contributions come from  -8, -1, 0, 1, 8, and 27.

These will be -1, -1, 0, 1, 1, and -1, so the total contribution for all seven numbers will be 1 -1 -1 + 0 +1 + 1 -1 = 0.

That will be the case for all complete blocks of seven.

It follows that the the only contributions to the final remainder will come from (each one cubed), 1, 2, 3 ,95,96,97,98,99, and 100.

Writing the final group of six numbers  as (14.7 - 3), (14.7 - 2), (14.7 - 1), 14.7, (14.7 + 1) and (14.7 +2),

the contributions from these nine will be, 1, 8, 27, -27, -8, -1, 0, 1, 8 leading to remainders (mod 7), 1, 1, -1, 1, -1, -1, 0 1, 1,

for a final total of 2.

Tiggsy

Mar 14, 2018
#9
0