IMC 2020 Q8
8. One of these options gives the value of \(17^2 + 19^2 + 23^2 + 29^2\). Which is it?
- A 2004
- B 2008
- C 2012
- D 2016
- E 2020
Intermediate Mathematical Challenge 2020, Question 8. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: E 2020
Since the units digits of all the options are different, it is sufficient to add the units digits of the four squares in the sum. As \(7^2 = 49\), \(17^2\) has a units digit of 9. In a similar way, we calculate that \(19^2\) has a units digit of 1, \(23^2\) has a units digit of 9 and \(29^2\) has a units digit of 1.
Therefore the units digit of the given sum is the units digit of 9 + 1 + 9 + 1, that is 0.
As we are told that the correct answer is one of the options, we may deduc that it is 2020.
Solution © UK Mathematics Trust, from the IMC 2020 solutions. Longer versions with investigations: Solutions and investigations (PDF).