#Help me with this grade 9 question pls
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<@&286206848099549185>
<@&286206848099549185>
3^1 = ..3
3^2 = ..9
3^3 = ..7
3^4 = ..1
3^5 = ..3
and then work your way to prove that the numerator is divisible by both 5 and 4
bro
it is 3^2n+1 -7
<@&286206848099549185>
so how tf that thing can help me
this is literally all the hints lol
can you solve it?
that was i need if you can solve it i can solve it also
without -7 it is unsolveable
so which one is it
do you mean 3^(2n) - 1?
I mean (3^(2n+1))-7
don't ping it again and again, doing it once's sufficient
okay
^ not divisible by 20
it gonna be 3^(2(2k+1)+1)-7= 3^(4k+3)-7 = 81^k*27-7
ah
still divisable
81^k * 27 - 7
now consider only the remainders when you divide by 20
?
you mean mod(81,20)?
yep
thanks
np
lol
do you know how to find m to find the prime number?