Algebra - HOW TO DINF THE REMAINder????

1357

Manish Shankar ·Oct 22 '09 at 2:12

2. (1+x+(1+x)/x³)¹⁰
[(1/x³)(1+x)(1+x³)]¹⁰

see if this leads you somewhere

For 1.

I think you have to find the remainder of both the terms separtely

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21

eragon24 _Retired ·Oct 22 '09 at 2:48

=27^10+7^51
=3^30+7^51
=9^15+7^51
using congruences
$9\equiv -1mod10$

$9^{15}\equiv -1^{15}mod10$

$9^{15}\equiv -1mod10$

$7^{3}\equiv 3mod10$

$7^{2}\equiv -1mod10$

$7 48 = (7 2 ) 24 $

$(7^{2})^{24}\equiv (-1)^{24}mod10$

$7^{48}\equiv 1mod10$

so therefore $7^{48}.7^{3}\equiv (3)(1)mod10$

$7^{51}\equiv 3mod10$

therefore

$(9^{15} + 7^{51})\equiv (3-1)mod10$

$(9^{15} + 7^{51})\equiv (2)mod10$

hence 2 is the remainder.....

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24

eureka123 ·Oct 22 '09 at 6:20

I did this question yesterday only...

write given expression as
$2 7 10 - 7 10 + 7 10 + 7 51 $

$S i n c e 2 7 10 - 7 10 i s a l r e a d y d i v i s i b l e b y 10, s o s o l v i n g f o r 7 10 + 7 51 $

$7 10 + 7 51 = > (50 - 1) 5 + 7 . (50 - 1) 25 = > (10 λ - 1) + 7 . (10 μ - 1) = > 10 (λ + 7 μ) - 8 = > 10 (λ + 7 μ - 1) + 2$

$S o r e m a i n d e r = 2$

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1

Philip Calvert ·Oct 22 '09 at 6:27

arrey use cyclicity na...

it will be the remainder when (9 + 3) is divided by 10 ..

:P

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24

eureka123 ·Oct 22 '09 at 6:29

??????????

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1

Philip Calvert ·Oct 22 '09 at 6:31

4n+1 ?

the last digit of a number k⁴ⁿ⁺¹ = last digit of k

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eragon24 _Retired ·Oct 22 '09 at 6:31

one more way................................................................

To find the remainder when 27^10+7^51 is divided by 10.
Remember whatever is the digit at units palce of 27^10+7^51 is the result when

divided by 10.

So we intend to find the digit at units place of 27^10+7^51

27^10+7^51 = 9^15 + 7 x 49^25

= 9 x 81^7 + 7 x 49^25

= 9 x 81^7 + 343 x (49)^24

= 9 x 81^7 + 343 x (2401)^12

As 9 x 81^7 = number with 9 at units place

and 343 x (2401)12 = number with 3 at units place

so,

9 x 817 + 343 x (2401)^12 = number with digit at units place is carry of (9+3) that is 2

or, digit at units place is 2

So remainder obtained when divided by 10 = 2

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1

Philip Calvert ·Oct 22 '09 at 7:26

why are you all constantly trying to find lengthier methods ...?

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62

Lokesh Verma ·Oct 22 '09 at 7:33

@Philip: It is not about finding lengthier methods all the time...

It is about different tricks and perspectives...

Each method has its use and could help in some other situation....

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1

Philip Calvert ·Oct 22 '09 at 7:36

ok

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21

eragon24 _Retired ·Oct 22 '09 at 8:44

@philip why do u think that i post lengthy methods haan..........[16] if u hav some shorter methods than mine i wud be much glad and wud appreciate u but if u dont hav those jus stop critisizing.......ok Mr Philip

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1

RAY ·Oct 22 '09 at 10:00

thnx people!!

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1

Philip Calvert ·Oct 22 '09 at 10:06

@deepak : Hey dude relax :)
I didn't mean to criticize anyone here.
and as you should've already noticed I have posted my method.
only because you're getting a bit upset i'll post it in full again :

the last digit of 27¹⁰ +7⁵¹ is required.
its clear that the last digit of 27¹⁰ is 9* and the last digit of 7⁵¹ is 3* . so the last digit of their sum is gonna be 2. :)

*This can be seen using what i mentioned in post #7

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23

qwerty ·Oct 22 '09 at 10:20

i m elaborating philip's method ...

see 7¹ will end with 7
7² will end with 9 [ 7 x 7 =49]
7³ will end with 3 [ 49 x 7 = 343 ]
7⁴ will end with 1
7⁵ will end with 7

so the last digit repeats after every 4 powers.
now consider 27¹⁰

as i hav said earlier ....... the last digit 7 will appear after every 4 powers.
now when power is 1 ....last digit = 7
when power is 9 ( = 4{ 2 } +1 ) ...last digit is = 7 [ since it comes after 8 powers ]
and after 7 ...the last digit for the next power i.e 10 has to be 9 ... as it is with the powers 1,2 and 5,6

similarly for 7 ... the power 51= 4(12) + 3

will hav the last digit same as that of power 3 i.e 3

so the last digit of 7⁵¹ + 27¹⁰ = 9 +3 =12

now u can solve

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1

Philip Calvert ·Oct 22 '09 at 10:22

i think you just repeated what i said :D
but nice that you explained it so elaborately

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