IIT JEE , AIEEE Forum - Mathematics , Algebra

Sumitc

Find the remainder when 27¹⁰+7⁵¹ is divided by 10.
the answer given in the book(IIT Mathematics by dasgupta) was 2.
But the ans i got was much different.Please could anyone try and see if they get 2 as answer or not.

shakirshafi12

=27^10+7^51
=3^30+7^51
=9^15+7^51
=(10-1)^15+7(49)^25
=(10-1)^15+7(50-1)^25
=just expand and we will get
=an integral multiple of 10- (15c15)-7(25c25)
=an integral multiple of 10-8
=an integral multiple of 10-10 +2
=an integral multiple of 10+2
hence remainder is 2

edison

To find the remainder when 27¹⁰+7⁵¹ is divided by 10.

Remember whatever is the digit at units palce of 27¹⁰+7⁵¹ is the result when

divided by 10.

So we intend to find the digit at units place of 27¹⁰+7⁵¹

27¹⁰+7⁵¹ = 9¹⁵+ 7 x 49²⁵

= 9 x 81⁷+ 7 x 49²⁵

= 9 x 81⁷+ 343 x (49)²⁴

= 9 x 81⁷+ 343 x (2401)¹²

As 9 x 81⁷ = 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

shakirshafi12

but edison sir
the answer written in dasgupta is 2
and if your solution is right
could you please explain what is wrong in
my solution(i did everything correctly)

edison

Dear Shakirshafi12 your approach and solution is perfect and flawless

I have rectified and edited my calculations as above.

thanx for bringing the descrepancy at the earliest.

shakirshafi12

no problems

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