Track / Stage 1 / 97 of 240 #97 of 2444

Problem 97

Number theory Difficulty 1.2 Find the answer CEMC Pascal · Canada · 2020

Ewan writes out a sequence where he counts by 11s starting at 3. The resulting sequence is $3, 14, 25, 36, \ldots$ . A number that will appear in Ewan’s sequence is

$113$
$111$
$112$
$110$
$114$

Your answer

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

Official solutions — 2

Solution 1

Ewan’s sequence starts with 3 and each following number is 11 larger than the previous number.
Since every number in the sequence is some number of 11s more than 3, this means that each number in the sequence is 3 more than a multiple of 11. Furthermore, every such positive integer is in Ewan’s sequence.
Since $110 = 11\times 10$ is a multiple of 11, then $113 = 110 + 3$ is 3 more than a multiple of 11, and so is in Ewan’s sequence.
Alternatively, we could write Ewan’s sequence out until we get into the correct range: $3, 14, 25, 36, 47, 58, 69, 80, 91, 102, 113, 124, \ldots$

Solution 2

Ewan’s sequence starts with 3 and each following number is 11 larger than the previous number.

Since every number in the sequence is some number of 11s more than 3, this means that each number in the sequence is 3 more than a multiple of 11. Furthermore, every such positive integer is in Ewan’s sequence.

Since $110 = 11\times 10$ is a multiple of 11, then $113 = 110 + 3$ is 3 more than a multiple of 11, and so is in Ewan’s sequence.

Alternatively, we could write Ewan’s sequence out until we get into the correct range: $3, 14, 25, 36, 47, 58, 69, 80, 91, 102, 113, 124, \ldots$

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.