1960 AMC 12 Problem 19

Question

Consider equation I:

x+y+z=46,

where

x,

y,

and

z

are positive integers, and equation II:

x+y+z+w=46,

where

x,

y,

z,

and

w

are positive integers. Then:

\mathrm{I}

can be solved in consecutive integers

\mathrm{I}

can be solved in consecutive even integers

\mathrm{II}

can be solved in consecutive integers

\mathrm{II}

can be solved in consecutive even integers

\mathrm{II}

can be solved in consecutive odd integers

Write the sums of three and four consecutive integers

Check which required average is compatible with

46

The sum of three consecutive integers is divisible by

3,

so it cannot be

46.

Four consecutive integers beginning with

n

have sum

4n+6.

Setting

4n+6=46

gives

n=10,

and indeed

10+11+12+13=46.

Therefore, the correct answer is C.

Problem 19 in other years