2017 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 23 Hard

Let N=123456789101112\ldots 4344 be the 79-digit number that is formed by writing the integers from 1 to 44 in order, one after the other. What is the remainder when N is divided by 45? (2017 AMC 10B Problem, Question#23)

  • A.

    1

  • B.

    4

  • C.

    9

  • D.

    18

  • E.

    44

Answer:C

We only need to find the remainders of N when divided by 5and 9 to determine the answer. By inspection，N=4 (mod5)，The remainder when Nis divided

by 9is1+2+3+4+\ldots +1+0+1+1+1+2+\ldots +4+3+4+4，but

sin c{{e}^{10=1(mod9)}}，we can also write this

a{{s}^{1+2+3+\cdots +10+11+12+\cdots 43+44=\frac{44\cdot 45}{2}=22\cdot 45}}，which has a remairder of 0 rod9. Therefoer，by inspection，the answer is (\text{C})9

Note：the sum of the digits or N is270.

Noting the solution above，we try to find the sum of the digits to figure out its remainder when divided by9. From1thru9，the sum is45，10 thru19，the sum is55，20 thru29is65，and30thru39is75．Thus the sum of the digits is45+55+65+75+4+5+6+7+8=240+30=270，and thus N is divisible by9. Now.refer to the above solution ^{N=4\left( mod5 \right)}an{{d}^{ N=0\left( mod9 \right)}}From this information，we can conclude this information, we can conclude that ^{N\equiv 54 \left( mod 5 \right)}and{{ }^{N\equiv 54\left( mod 9 \right)}}.

Therefore ^{N=54\left( md 45 \right)}an{{d}^{N=9\left( md 45 \right)}} so the remainder is (\text{C})9.

Because a number is equivalent to the sum of its digits modulo9，we have that ^{N\equiv 1+2+3+4+5+\cdots +44\equiv \frac{44\times 45}{2}\equiv 0(mod9)}Furthemore，we see that N-9ends in the digt5and thus is divisible by5，so N-9is divisible by45，meaning the remainder is(\text{C})9.

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