1968 AMC 12 Problem 33

Attempt Problem 33 of the 1968 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1968 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

33.

A number

N

has three digits when expressed in base

7.

When

N

is expressed in base

9

the digits are reversed. Then the middle digit is:

0

1

3

4

5

Answer: A

Concepts:number base place value Diophantine Equation

Difficulty rating: 2060

Small hint:

Call the base-

7

digits

a,

b,

c

and equate

49a+7b+c

to

81c+9b+a

Big hint:

Rearrange to

24a-b=40c

and use the digit bounds

Solution:

Let

N=(abc)_7=(cba)_9.

Then

49a+7b+c=81c+9b+a,

so

24a-b=40c.

With

1\leq a\leq6,

0\leq b\leq6,

and

1\leq c\leq6,

the only possibility is

a=5,

c=3,

b=0.

Indeed,

(503)_7=(305)_9=248.

Therefore, the correct answer is A.