   SEARCH HOME Math Central Quandaries & Queries  Question from Nickie, a student: I am a 4-digit number with no repeating digits. I am divisible by 5, my first two digits (left to right) make a number divisible by 3, and my first three digits make a number divisible by 4. Also, my digits have a sum of 19 and I have the digit 7 in the thousands place. Who am I? Hi Nickie,

I can give you some help getting started. This is a 4 digit number and its thousands digit is 7 so I know it looks like

7 _ _ _

It is divisible by 5 so the units digit is 0 or 5. Now I know it looks like

7 _ _ 0
or
7 _ _ 5

The number formed by the first two digits is divisible by 3. There is a really nice test to see if an integer is divisible by 3. An integer is divisible by 3 exactly if the sum of its digits is divisible by 3. Thus the number formed by the first two digits can be 72 (7 + 2 = 9), 75 (7 + 5 = 12 or 78 (7 + 8 = 15). Now you have 6 different forms your number can take.

7 2 _ 0
or
7 2 _ 5
or
7 5 _ 0
or
7 5 _ 5
or
7 8 _ 0
or
7 8 _ 5

For each of the six possibilities above the missing digit can only have one value since the sum of the digits is 19. In each case fill in the blank and then check the last clue, "my first three digits make a number divisible by 4".

Be careful! There may not be a solution, there may be one solution or there may be more than one solution.

Have fun with this,
Penny     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.