- 31 Jul '09 14:00While driving to work this morning I was listening to an AM radio station. The station number is a four digit integer. As I preset the station on one of seven buttons on my radio. Then I started thinking about the number ... if was divisible by seven if the second digit is removed. What is the number of the radio station?
- 31 Jul '09 16:04

Not enough information so will have to do some deduction.*Originally posted by petrovitch***While driving to work this morning I was listening to an AM radio station. The station number is a four digit integer. As I preset the station on one of seven buttons on my radio. Then I started thinking about the number ... if was divisible by seven if the second digit is removed. What is the number of the radio station?**

1. I see you are in US so I will assume you mean frequency when you refer to station number? (We use wavelength for AM in Europe)

2. I will assume frequency range 535-1705 KHz (US )

3. I will assume 10KHz between stations (in Europe it is 9KHz)

Our number is therefore abcd

where

535 < abcd < 1705

acd = 7n

a=1

b=0,1,2,3,4,5,6,7

c=0,7

d=5

That still leaves 15 solutions, what have I missed? - 31 Jul '09 17:48AM stations generally (always?) end in 0 -- and being that it's a 4-digit number, the first number is a 1

if we eliminate the second digit, you get a three digit number 1e0 -- the only number between 100-199 that ends in 0 and is divisible by 7 is 140, so e = 4

if AM stations max out at 1705, that leaves us with the solutions of 1040, 1140, 1240, 1340, 1440, 1540, 1640 - 31 Jul '09 18:17

Given the range I assumed they ended in 5.*Originally posted by Melanerpes***AM stations generally (always?) end in 0 -- and being that it's a 4-digit number, the first number is a 1**

if we eliminate the second digit, you get a three digit number 1e0 -- the only number between 100-199 that ends in 0 and is divisible by 7 is 140, so e = 4

if AM stations max out at 1705, that leaves us with the solutions of 1040, 1140, 1240, 1340, 1440, 1540, 1640

I stand corrected.

The problem with this problem is that you need 'local' knowledge ... and it aslo appears to be cooked!