11 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.What exactly are they asking
Ok, so, the number of ordered quintuples (,,,,) depends only on a1, a2, and a4, because 5 and 9 are both congruent to 1 mod4. I assume that means we have to count the ways in which 2^a1+3^a2+3^a3 = 2 (mod4) but then I got stuck
So they are asking total number of quintuplets possible?
Well, largest possible number of quintuples. Which is what is throwing me off.
Yeah thats what
Ans is 5 ?
Yeah
ok so i did by binomial. write 3 and 7 as 4k-1 and 5 and 9 as 4k+1 the thing will effectively become 2^a1+2+(-1)^a2+(-1)^a4 after this take cases for a1 when its 0 there will be no solutions and when its 1, (-1)^a2+(-1)^a4 should be zero this is possible when a2 and a4 are not odd or even at the same time so you will get (10)(10)(5)(5)2 and then when a1 >=2 it will be divisible by 4 so now both a2 and a4 should be even or odd at the same time so you will get 8(10)(10)(5)(5)2 cases
sum will become 45000
Ah
+solved @hithav
Post locked and archived successfully!
Archived by
<@763645886500175892> (763645886500175892)
Time
<t:1746618713:R>
Solved by
<@726641475080683522> (726641475080683522)