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

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To satisfy the cubic equation \(a^3 + 23\) being a multiple of 24, \(a\) must be an odd natural number less than 100. The analysis shows that \(a \equiv 1 \mod 24\), leading to potential solutions of \(a = 1, 25, 49, 73, 97\). The expression \(a^2 + a + 1\) does not yield any additional solutions since it results in an odd number, while 24 is even. Therefore, the only valid values for \(a\) are 1, 25, 49, 73, and 97.
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$a>0,\,\, a\in N,\,\, and \,\, a<100$

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

$find \,\, a$
 
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Albert said:
$a>0,\,\, a\in N,\,\, and \,\, a<100$

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

$\chi$ $\sigma$
 
chisigma said:
[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)
 
Albert said:
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

$\chi$ $\sigma$
 
chisigma said:
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 $,min(a)=1,max(a)=97$
$a=1,25,49,73,97$
 
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