TIFR CSE 2018 | Part B | Question: 1

Question

What is the remainder when $4444^{4444}$ is divided by $9?$

• A. $1$
• B. $2$
• C. $5$
• D. $7$
• E. $8$

Answer 1

(4444)⁴⁴⁴⁴ % 9 = (4444 % 9)^{4444 % 9} = 7⁷ % 9 = (7².7².7².7) % 9 = (343.343.343.7) % 9 = 1.1.1.7 = 7

OPTION (D)

Answer 2

$4444^{4444}\;mod\;9$

$=(4437+7)^{4444}\;mod\;9 $

(4437 is the nearest number(<4444) which is divisible by 9)

$=7^{4444}\;mod\;9$

$=(7^3)^{1481}\times 7\;mod\;9$

$=7\;mod\;9$

$=7$

Answer 3

Answer : D

apply fermat's little theorem , 4444 , 9 are relatively prime , so you can apply 

make sure power of 4444 should be multiple of 8, floor(4444/8) = 555

so , ((4444)^(555*8)) *(4444)^4 mod9 = 1*4444^4 mod9 

4444 mod 9 = 7.

so, 7^4 mod9 = 7

Answer 4

$4444^{4444}\ mod\ 9$

= $4444^{(3*1481)+1}\ mod\ 9$

= $(4444^{(3*1481)} * 4444 ^1)\ mod\ 9$

=$(4444^{(3*1481)}\ mod\ 9 )* (4444\ mod\ 9) $

=$(4444^{3}\ mod\ 9 )^{1481}* (4444\ mod\ 9) $

=$(87765160384\ mod\ 9 )^{1481}* (4444\ mod\ 9) $

=$1^{1481}*7$

=$7$