# If N is the remainder when 5*5*5*.....till 40 is divided by 11 and M is the remainder when 2*2*2*....till 2003 is divide by 17 then value of M-N

Akash Kumar Dutta
98 Points
11 years ago

Dear Hemanthkrishna,
applying the concept of mod we get

5^2 is congruent to 3 mod 11  [ means when 3 is subtracted from 5^2=25 it is divisible from 11 ]
raising to 5th power... 5^10 is congruent to 3^5 mod 11
but 3^5=243 is congruent to 1 mod 11
hence 5^10 is congruent to 1 mod 11
raising again to 4th power... 5^40 is congruent to 1 mod 11
HENCE 5^40 = 11a + 1
So N=1

Similarly for 2^2003
WE have.. 2^4 is congruent to -1 mod 17
raising to 500 power.. 2^2000 congruent to 1 mod 17
Hence 2^2000=17b + 1
Multiplying by 2^3 we have
2^2003 = 17.b.8 + 8
2^2003 = 17c + 8
Hence M=8

SO M-N=8-1=7(ANS)

Regards.

Abhishekh kumar sharma
34 Points
11 years ago

5^2 is congruent to 3 mod 11  [ means when 3 is subtracted from 5^2=25 it is divisible from 11 ]
raising to 5th power... 5^10 is congruent to 3^5 mod 11
but 3^5=243 is congruent to 1 mod 11
hence 5^10 is congruent to 1 mod 11
raising again to 4th power... 5^40 is congruent to 1 mod 11
HENCE 5^40 = 11a + 1
So N=1

Similarly for 2^2003
WE have.. 2^4 is congruent to -1 mod 17
raising to 500 power.. 2^2000 congruent to 1 mod 17
Hence 2^2000=17b + 1
Multiplying by 2^3 we have
2^2003 = 17.b.8 + 8
2^2003 = 17c + 8
Hence M=8

SO M-N=8-1=7(ANS)