Sequence and reminder

For students of class 6-8 (age 12 to 14)
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jayon
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Joined: Thu Dec 08, 2016 12:10 am

Sequence and reminder

Unread post by jayon » Thu Dec 08, 2016 2:41 pm

Nafis wrote sequence,terms of which are $1,12,123,1234,12345,\ldots$ and whose $11$th term is $1234567891011$. What will be the remainder if the $2014$th term is divided by $3$?

Please give me the solution..

SYED ASHFAQ TASIN
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Re: Sequence and reminder

Unread post by SYED ASHFAQ TASIN » Fri Dec 30, 2016 9:50 pm

1,get the sum and you will get it!
HELP A MAN , IN DANGER SPECIALLY IN MATH!
! PLEASE!$$…

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jayon
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Re: Sequence and reminder

Unread post by jayon » Wed Jan 11, 2017 9:12 pm

SYED ASHFAQ TASIN wrote:1,get the sum and you will get it!
About which sum are you talking about??
plz make it clear...

dshasan
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Location: Dhaka,Bangladesh

Re: Sequence and reminder

Unread post by dshasan » Thu Jan 12, 2017 6:23 pm

Note that $2014 ^ th $ term consists of $2014$ consecutive integer starting from $1$. And any number $ABC........XYZ\equiv A+B+C+......+Y+Z(Mod 3)$. So $123456789......2014 \equiv 1+ 2 + 3 +.......+2014(mod 3)$.
Now figure out this sum.
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

- Charles Caleb Colton

zarif
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Joined: Sat Jan 14, 2017 4:29 pm

Re: Sequence and reminder

Unread post by zarif » Sat Jan 14, 2017 6:43 pm

প্রতি ক্রমিক ৩ টি সংখ্যার সমষ্টি ৩ দ্বারা বিভাজ্য
এখন প্রতি ৩ টি সংখ্যা করে জোড়া তৈরি করি ১,২,৩;৪,৫,৬;......... ২০১১,২০১২,২০১৩
পূর্বের সংখ্যাগুলোর সমষ্টি ৩ দ্বারা বিভাজ্য
এখন ২০১৪ কে ৩ দ্বারা ভাগ করলে ১ ভাগশেষ সুতরাং ভাগশেস ১

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