Competition problem — Number theory

Library / /4 of 28

Number theory Difficulty 3.9 AMC 10/12 Find the answer Philippines

Problem:
What is the smallest number that is greater than $2015$ and divisible by both $6$ and $35$ ?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:
The desired number is divisible by $2$ , $3$ , $5$ and $7$ . Hence, it is the least multiple of $2 \times 3 \times 5 \times 7 = 210$ that is greater than $2015$ . The desired number is $2100$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.