1.Find the remainder when 2710 +751 is divided by 10.
PLEASE TELL ME THE GENERAL METHOD TO THIS SUM
2.determine constant term in expansion ok (1+ x + (1/x)2 + (1/x)3)10
13572. (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
21=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.....
24I did this question yesterday only...
write given expression as
27^{10}-7^{10}+7^{10}+7^{51}
Since \ 27^{10}-7^{10}\ is \ already \ divisible \ by \ 10,so\ solving \ for \ 7^{10}+7^{51}
7^{10}+7^{51} =>(50-1)^5+7.(50-1)^{25}=>(10\lambda -1)+7.(10\mu -1)=>10(\lambda +7 \mu)-8=>10(\lambda + 7\mu -1) +2
So \ remainder \ =2
1arrey use cyclicity na...
it will be the remainder when (9 + 3) is divided by 10 ..
:P
21one 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
1why are you all constantly trying to find lengthier methods ...?
62@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....
21@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
1@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
23i 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
1i think you just repeated what i said :D
but nice that you explained it so elaborately
