Find a three digit number that fits the following criteria:

1)The digits are all different.
2)When each digit is squared and added together the number is the same as the number formed by the 2nd and 3rd digits in the number.
3)When the number formed by the last two digits is squared then added to (the first digit squared and added to itself) it is the same as the original number.

How many different three digit numbers are there that fit the criteria?

I agree with the solution - one three digit number 420. Rather than a programmed method I narrowed down the possibilities & then tested the last half dozen or so.
The first or second digits cannot be zero - if the first is zero, squaring the number formed by the last two is greater than the whole. If the second is zero, squaring the individual digits will give a number larger than the number formed from the last two digits.
Also, the number formed from the last two digits is less than 33 - otherwise squaring it gives a 4 digit number.
This has the additional implication that no digit of the three can be greater than 5.
We have now substantially reduced the number of possible cases (to 11). These can be checked against the condition that the sum of the digits, squared have to equal the number formed by the last two. 32,31,30 don't work. 24 does - it could be 224 - but then two digits are the same, so no. 23 - no; 21 works - 421 would do - but 21^2+14+4 is not 421. 20 works too - 420 satisfies all the conditions. The next possibility is 13 - doesn't work; nor does 12; 10 could give 310 - but also fails at the last 10^2+9+3 is not 310. So the only possibility is 420.

Posted by DrBob on 2003-11-09 02:18:28