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 

Advent calendar 2020

Advent calendar 2019

Sunday Afternoon Maths LXVII

Coloured weights
Not Roman numerals

Advent calendar 2018

List of all puzzles


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


Show me a random puzzle
▼ show ▼
© Matthew Scroggs 2012–2021