Advent calendar 2018

10 December

The equation \(x^2+1512x+414720=0\) has two integer solutions.
Today's number is the number of (positive or negative) integers \(b\) such that \(x^2+bx+414720=0\) has two integer solutions.

Show answer


Show me a random puzzle
 Most recent collections 

Sunday Afternoon Maths LXVII

Coloured weights
Not Roman numerals

Advent calendar 2018

Sunday Afternoon Maths LXVI

Cryptic crossnumber #2

Sunday Afternoon Maths LXV

Cryptic crossnumber #1
Breaking Chocolate
Square and cube endings

List of all puzzles


planes grids money complex numbers square numbers multiples dodecagons shape palindromes perfect numbers books wordplay fractions advent ellipses factorials balancing differentiation circles routes parabolas star numbers number folding tube maps functions bases shapes clocks time indices crosswords dates dice mean quadratics taxicab geometry trigonometry geometry scales partitions 3d shapes prime numbers proportion triangles numbers volume lines pascal's triangle integers probability digits coordinates games calculus sum to infinity hexagons angles logic crossnumbers symmetry cryptic crossnumbers cube numbers cryptic clues algebra graphs ave rugby odd numbers arrows chalkdust crossnumber colouring probabilty sums 2d shapes rectangles chess menace multiplication integration coins christmas addition averages sequences irreducible numbers division speed spheres square roots means remainders doubling triangle numbers area squares unit fractions regular shapes percentages people maths factors cards polygons sport floors perimeter surds chocolate


Show me a random puzzle
▼ show ▼
© Matthew Scroggs 2019