Can you find the smallest number?
 

This one is really simple. It should take no more than 10 minutes to solve. Your I.Q. can be indirectly extapolated by how long it takes you to figure out this simple mathematical puzzle. ( yeah, if you really believe this, let me sell you a lot I own with ocean frontage :) )

3o minutes or more = I.Q. 100
20 to 30 minutes = I.Q. 120
10 to 20 minutes = I.Q. 140
< 10 minutes = I.Q. 150+

Find the two smallest whole numbers that:
divided by seven has a remainder of 4
divided by eight has a remainder of 5
divided by nine has a remainder of 6

Use any tools, programs, etc of your choice to solve.

Fusion :cool:

Re: Can you find the smallest number?
 

[quote="Fusion_power"] let me sell you a lot I own with ocean frontage :) [/quote]

Hey, there is some very nice coastal property in Alabama.

And they even wrote a song about such property in Arizona....

But South Dakota ????? :(


As for the puzzle, do you really mean smallest non-negative integers?
If so, it took much longer to type this than it took to find such numbers.

501 and 1005?

Why did I feel compelled to use a spreadsheet to do this one? Must be all those years as an accountant :redface:

Re: Can you find the smallest number?
 

Those IQ's are unrealistic....try these

30 minutes or more = I.Q. 70
20 to 30 minutes = I.Q. 80
15 to 20 minutes = I.Q. 90
10 to 15 minutes = I.Q. 100
7 to 10 minutes = I.Q. 110
4 to 7 minutes = I.Q. 120
1 to 4 minutes = I.Q. 130
< 1 minute = I.Q. 140+

Find the two smallest whole numbers that:
divided by seven has a remainder of 4
divided by eight has a remainder of 5
divided by nine has a remainder of 6

Use any tools, programs, etc of your choice to solve.

EASY!
 

That was so easy - 501 and 1005
Found using:
<script>
var x=0
var y=0
var curnum=1
while(!y && curnum<2000) {
if (curnum%7==4 && curnum%8==5 && curnum%9==6) {
if (!x) x=curnum
else y=curnum
}
curnum++
}
alert("FOUND: "+x+","+y)
</SCRIPT>

Written and exectued in under a minute! :)

easy:
just do (7*8*9)-9+6 to get the first number.
7*8*9*2 -9+6 for the second....
just used windows calculator, less then 30 secs....(with check)

[EDIT you can use the 7*8*9 trick, because the greatest common divider is 1...]

isn't this just a chinese remainder theorem problem?

It would be harder if 7, 8 and 9 weren't all relatively prime.