What values of a satisfy the cubic equation $a^3+23$ being a multiple of 24?

[Albert1]

$a>0,\,\, a\in N,\,\, and \,\, a<100$

if $a^3+23 $ is a multiple of 24

$find \,\, a$

 

[chisigma]

$a>0,\,\, a\in N,\,\, and \,\, a<100$

if $a^3+23 $ is a multiple of 24

$find \,\, a$

[sp]Because 24 is even and 23 is odd, a must be odd and is a<5... then 1 is solution and 3 isn't solution...[/sp]

Kind regards

$\chi$ $\sigma$

 

[Albert1]

[sp]Because 24 is even and 23 is odd, a must be odd and is a<5... then 1 is solution and 3 isn't solution...[/sp]

Kind regards

$\chi$ $\sigma$

why a<5 ? (a<100 is given)

 

[chisigma]

why a<5 ? (a<100 is given)

All right!...

[sp]You can find a solving the cubic equation...

$\displaystyle a^{3} - 1 = (a - 1)\ (a^{2} + a + 1) \equiv 0\ \text{mod}\ 24\ (1)$

Now $\displaystyle a - 1 \equiv 0\ \text{mod}\ 24$ has the only solution $\displaystyle a\ \equiv 1\ \text{mod}\ 24$ and that means a=1, 25, 49, 73, 97...

Otherwise $\displaystyle a^{2} + a + 1 \equiv 0\ \text{mod}\ 24$ has no solutions because in both cases 'a even' and 'a odd' the expression $\displaystyle a^{2} + a + 1$ gives an odd number and 24 in even, so that the solution found in the previous step are the only solutions...[/sp]

Kind regards

$\chi$ $\sigma$

 

[Albert1]

All right!...

[sp]You can find a solving the cubic equation...

$\displaystyle a^{3} - 1 = (a - 1)\ (a^{2} + a + 1) \equiv 0\ \text{mod}\ 24\ (1)$

Now $\displaystyle a - 1 \equiv 0\ \text{mod}\ 24$ has the only solution $\displaystyle a\ \equiv 1\ \text{mod}\ 24$ and that means a=1, 25, 49, 73, 97...

Otherwise $\displaystyle a^{2} + a + 1 \equiv 0\ \text{mod}\ 24$ has no solutions because in both cases 'a even' and 'a odd' the expression $\displaystyle a^{2} + a + 1$ gives an odd number and 24 in even, so that the solution found in the previous step are the only solutions...[/sp]

Kind regards

$\chi$ $\sigma$

yes ,you got it !
for $a\in N , 0<a<100$
and $\displaystyle a\ \equiv 1\ \text{mod}\ 24$
so $,min(a)=1,max(a)=97$
$a=1,25,49,73,97$