Nicholas challenges Rajiv with these instructions for finding his house after basketball practice.

Get on the subway at 14th Street. The subway stops at 23rd, 28th, 33rd, 42nd, 51st, 59th, 86th, and 96th Streets. Get off at the first street you come to that is divisible by both 3 and 7.

How can Rajiv solve this problem? At which stop should he get off the subway?

Question

Corinne

Mathematics

19 May, 08:22

Nicholas challenges Rajiv with these instructions for finding his house after basketball practice.

Get on the subway at 14th Street. The subway stops at 23rd, 28th, 33rd, 42nd, 51st, 59th, 86th, and 96th Streets. Get off at the first street you come to that is divisible by both 3 and 7.

How can Rajiv solve this problem? At which stop should he get off the subway?

+5

Answers (1)

Know the Answer?

Caden Garrison · Answer 1

Make use of the LCM of 3 and 7 42nd street

Step-by-step explanation:

A number divisible by both 3 and 7 will be divisible by their least common multiple, which happens to be 3·7 = 21. Rajiv can examine the street numbers to see which are divisible by 21. (Those would be in the list 21, 42, 63, 84.}

Of the street numbers listed, only 42nd street has a number divisible by 21.

Rajiv should get of at 42nd street.