Olympiad Maths Prep

Track / Stage 4 / 322 of 340 #582 of 2000

Problem 582

AMC 12 late, AIME early
Number theory Difficulty 5.0 Find the answer

Subject I
Determine the remainder of the division of the number 1⋅2⋅3⋅4⋅…⋅69⋅70+12341 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 69 \cdot 70 + 1234 by 2013.

Official solution

Solution:
2013=3⋅11⋅612013=3 \cdot 11 \cdot 61 2 points
1⋅2⋅3⋅4⋅…⋅69⋅70=1⋅2⋅3⋅4⋅…⋅11⋅12⋅…⋅61⋅62⋅…⋅69⋅70=1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 69 \cdot 70=1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 11 \cdot 12 \cdot \ldots \cdot 61 \cdot 62 \cdot \ldots \cdot 69 \cdot 70=
=2013⋅1⋅2⋅4⋅…⋅10⋅12⋅…⋅60⋅62⋅…⋅69⋅70=2013 \cdot 1 \cdot 2 \cdot 4 \cdot \ldots \cdot 10 \cdot 12 \cdot \ldots \cdot 60 \cdot 62 \cdot \ldots \cdot 69 \cdot 70
2 points
1234<20131234<2013 1 point
1⋅2⋅3⋅41 \cdot 2 \cdot 3 \cdot 4.
69⋅70+1234=2013⋅(1⋅2⋅4⋅69 \cdot 70+1234=2013 \cdot(1 \cdot 2 \cdot 4 \cdot
10⋅12⋅…⋅60⋅6210 \cdot 12 \cdot \ldots \cdot 60 \cdot 62 ..... 1 point
according to the Division Theorem with Remainder, the required remainder is 1234 1 point

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.