Question

somebody check my answer please

Answer 1

Let Sm denote the sum of the cubes of the first 'm' natural numbers .For how many values the sum of the cubes of the first m natural numbers be a multiple of 5.(if m<50)

Answer 2

what is the question

Answer 3

a.20
b.21
c.22
d. none of these<------

Answer 4

@phi

Answer 5

this mean like

2^3 + 3^3 = 8+27= 35 = 7*5 is right
35+4^3 =35+64 =99 not is
99 +5^3 = 99+125=224 not is
224+6^3 = 224+216=440 = 88*5 is right

hope you can continue it same above

Answer 6

i think it means 1^3,2^3,3^3,4^4 as a individual term whether multiple of 5 or not

Answer 7

4 is on 3 not on 4 exponent

so yes i have missed the 1^3 ,sorry

then this mean that what i have wrote above will be changed like in this case then you add to 35+1=36 so this not is a multiplie of 5
but the second and 3rd will be right
and the 4rd not is

so continue it .... hope now is correct

Answer 8

ohhhh ! like individual term so then sorry but there have wrote the sum of ...

what mean that you need adding them .... or not ?

Answer 9

,,the sum of the cubes ..." i think this mean how i have wrote above with 1^3

Answer 10

well the interpretation of question is still confusing
i had gone with identical question ,there is seemed to have individual terms
here is a link

http://openstudy.com/users/mathmath333#/updates/53ff3d6ce4b0de29bf9db967

Answer 11

i think u r correct

Answer 12

so i have checked it there yes im correct ,there is sum them

Answer 13

so i think this way how have wrote there will be usefully in this case too ,there divisidle by 4 and here in this case divisibly by 5

Answer 14

so what will be the ans

Answer 15

you need to calculi it same like there ,here now i dont know it

Answer 16

so then you understand the way how was calculated this sum in case of 4 so then i think will be easy

Answer 17

ok thnks

Answer 18

It helps to know a formula for the sum of the first m cubes: $$1^3+2^3+3^3+\cdots +m^3=\sum_{k=1}^m k^3=\left(\frac{m(m+1)}{2}\right)^2.$$ The expression on the right will be divisible by 5 when \(m(m+1)\) is divisible by 5. So either m or m+1 needs to be divisible by 5 and youre gold.

The numbers when this happens <50 are:

4,5 9, 10,14,15,19,20,24,25,29,30,34,35,39,40,44,45,49.

Answer 19

thanks , this is very helpful

Answer 20

m just testing latex here