Albert1
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$a>0,\,\, a\in N,\,\, and \,\, a<100$
if $a^3+23 $ is a multiple of 24
$find \,\, a$
if $a^3+23 $ is a multiple of 24
$find \,\, a$
Albert said:$a>0,\,\, a\in N,\,\, and \,\, a<100$
if $a^3+23 $ is a multiple of 24
$find \,\, a$
why a<5 ? (a<100 is given)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
$\chi$ $\sigma$
Albert said:why a<5 ? (a<100 is given)
yes ,you got it !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
$\chi$ $\sigma$