hi puys,
can any one explain euler method of finding remainder....as i didnt find anywhere one the net and i asked the same one number system forum....but
answer posted were contrary
so i cant get it so i find difficulty with finding remainder of triple power type question.......9^11^13/7?
thanx
It's like this, buddy.
Consider three numbers: n1,n2,t.
Now suppose n1 when divided by t leaves a remainder of "a", and n2 when divided by t leaves a remainder of "b", then n1 X n2 when divided by t will leave the same remainder as when a X b is divided by t.
Eg. 9 when divided by 7 leaves remainder 2, and 8 when divided by 7 leaves 1, so 9 X 8 (=72) leaves remainder 2, which is also left when 2 X 1 is divided by 7.
Now, 9^11^13.
STEP 1:
Find the remainder when 9^11 is divided by 7.
9 when divided by 7 leaves remainder 2.
9^2...........4 (2 X 2 divided by 7).
9^3........1
9^4.........2(2 X 2 X 2 X 2)
Ah! There is a pattern. The remainders always follow a cycle, and as soon as a remainder is repeated (as in this case, 2), then just follow the cycle.
9^4 leaves a remainder 2.
9^9 will leave remainder 1 (follow the cycle 2-4-1,2-4-1 etc)
9^11 will leave remainder = 4.
STEP 2:
9^11^1 when divided by 7 leaves remainder 4.
9^11^2 will leave remainder = (4^2 divided by 7 leaves remainder) 2.
9^11^3 ......(4^3)...1
9^11^4....(4^4).....4
Here comes our pattern 4-2-1,4-2-1...
So, 9^11^13 will leave remainder
4....
Plz correct if wrong...
