IIT JEE , AIEEE Forum - Mathematics , Algebra

johri_anshuman

Prove that _32³²³²when divided by 7 leaves the remainder 4.

vineetvsb

firstly find the remainder when u devide 3232by 7
then 3232 =(-2)(mod7)
-->(3232)^6 = (-2)^6(mod7)
--> (3232)^6= 64(mod7)
but 64 = 1(mod7)
-->(3232)^6= 1(mod7)
-->[(3232)^6]^5=1(mod7)
-->(3232)^30=1(mod7)----------------------1
now u now that
3232 =(-2)(mod7)
(3232)^2 = 4(mod 7)------------------------------2
now multiply both 1&2
u get
(3232)^32=4(mod7)
hence this is the right method
plz.......... rate me

johri_anshuman

vineetvsb, i think that u made a mistake. the remeainder of 3232^32 is not to be found.

The remainder of ((32^32)^32)/7 is to be found.

DON007

hey ..............sorry................but...........as it is a binomial theorem question.................. but i m doing it by different method...........

thanku...............

look.........

2=2 (mod 7) (means when 2 / 7 leaves remainder 2)

2⁵= 32 (mod 7)

2⁵ = 4 ( mod 7) ( 32 divided by 7 gives 4)

32³²= 4 ³² (mod 7) ( raising power of 32 on both sides)

32³² = 2⁶⁴ (mod 7) .................... ( 1)

now lets find remainder obtained by 7 of 2⁶⁴

2 = 2 ( mod 7)

2²=4 (mod 7)

2⁴= 2 ( mod 7) (as 16 divided by 7 gives remainder 2)

2⁸= 2²=4 (mod 7)

2¹⁶= 2 ( mod 7) ( as 16 divided by 7 gives 2 rem)

2³²= 4 ( mod 7)................................. ( 2 )

2 ⁶⁴= 2 ( mod 7) ( as 16 gives 2 rem)..............( 3)

putting (3) in (1)............... we get ............

32³²= 2 ( mod 7 )

₃₂32³² = 2³²

now putting 2 in 1...............

we get

₃₂32³²= 4 ( mod 7)..........................

ya.. ya i know it its sometime confusing...................but............i have tried my best to explain it hope u understands...........

thanku.............

anuj_jamwal

tis is on using binomial theorem

nunoxic

Nice solution anuj.......and good handwriting too :D

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