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Number Theory Level 2

1234567891011 … 20142015 \large 1234567891011\ldots20142015

The gigantic number above is formed by the concatenation of the first 2015 whole numbers. What is the remainder when this number is divided by 9?


The answer is 0.

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Christian Daang by Christian Daang , 20, Philippines

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2 solutions

Christian Daang
May 7, 2015

Base from the proof, we can say that:

( 12345678910...20142015 ) (12345678910...20142015) mod 9

= ( 1 + 2 + 3 + 4 + 5 + 6 + 6 + 8 + 9 + 10 + . . . + 2014 + 2015 ) (1+2+3+4+5+6+6+8+9+10+...+2014+2015) mod 9

= 2016 ∗ 2015 2 \frac{2016*2015}{2} mod 9

= 0 \boxed{0}

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Moderator note:

Your solution is not entirely phrased well. Hint: A B C D E … Y Z ‾ = A × 1 0 25 + B × 1 0 24 + … + Y × 1 0 1 + Z \overline{ABCDE\ldots YZ} = A\times 10^{25} + B\times 10^{24} + \ldots + Y \times 10^1 + Z .

Actually you added numbers upto 2015 and applied mod . But here only the seperate digits are the matter !

Harish Ganesan - 6 years, 1 month ago

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As you can see in my proof, that's the special in modulo 9.

In (abcdefg...) mod 9 = (a+b+c+d+e+...) mod 9

It can be in this form:

(a+b) mod 9 + (c+d+e) mod 9 + ... ---- (ab)mod 9 + (cde) mod 9 + ... Or ...

Since the divisibility rule for 9 is : to know if tge number is divisible by 9, we need to add all the single digits up to the end of that number and if the result is divisible by 9, then, the whole number is divisible by 9. As you can see, if we apply the modulo properties (i.e. Modular Arithmetic- Addition) and that's what I did.

Christian Daang - 6 years ago

Awesome did the same way upvoted

Ayush Sharma - 5 years, 12 months ago

lets calculate sum of digits in 123456789101112131415......20142015 (1+2+3+4.......+9)=45.

(1+0)+(1+1)+(1+2)+(1+3)....+(1+9)=(45+1*10)

(2+0)+(2+1)+(2+2)+(2+3)....+(2+9)=(45+2*10)

till 2009 total 201 series can be made like above i.e (45+0x10) +(45+1x10)+(45+2x10)+(45+3x10)....+(45+200x10) = sum till 2009 term which can be written as = 45x(integer)+(10(20x21)/2 ) = 45x(integer)+2100

now for sum between 2010 to 2015 we get = (2+0+1)x6+(0+1+2+3+4+5) =18+15=33 therefore total sum is = 45x(integer)+2100 +33= completely divisible by 9

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