Certain nos. to be remembered
2^10 = 4^5 = 32^2 = 1024
3^8 =9^4 = 81^2 = 6561
7 * 11 * 13 = 1001
11 * 13 * 17 = 2431
13 * 17 * 19 = 4199
19 * 21 * 23 = 9177
19 * 23 * 29 = 12673
The trick to doing this problem is to see that
1000 = -1 mod 1001
1000^2 = 1 mod 1001 (because 1000^2 = (1001-1)^2 and so on)
1000^3 = -1 mod 1001
and so on
Thus 12345678987654321 = (12*1000^5 + 345*1000^4 + 678*1000^3+ 987*1000^2 + 654*1000 + 321)...making groups of three digits from rightmost digit
Thus using the above rules, this expression is equal to
(321 - 654 + 987 - 678 + 345 - 12) mod 1001
= 309 mod 1001
= 309
2^10 = 4^5 = 32^2 = 1024
3^8 =9^4 = 81^2 = 6561
7 * 11 * 13 = 1001
11 * 13 * 17 = 2431
13 * 17 * 19 = 4199
19 * 21 * 23 = 9177
19 * 23 * 29 = 12673
The remainder obtained when 12345678987654321 is divided by 1001
a)10
b)309
c)692
d)995
Hint : Dont use complex remainder theorems iff u have read them.It does not help every time.
Sol:- a)10
b)309
c)692
d)995
Hint : Dont use complex remainder theorems iff u have read them.It does not help every time.
The trick to doing this problem is to see that
1000 = -1 mod 1001
1000^2 = 1 mod 1001 (because 1000^2 = (1001-1)^2 and so on)
1000^3 = -1 mod 1001
and so on
Thus 12345678987654321 = (12*1000^5 + 345*1000^4 + 678*1000^3+ 987*1000^2 + 654*1000 + 321)...making groups of three digits from rightmost digit
Thus using the above rules, this expression is equal to
(321 - 654 + 987 - 678 + 345 - 12) mod 1001
= 309 mod 1001
= 309
No comments:
Post a Comment