Only people with the highest IQ can crack infamous maths puzzle that leaves many stumped

How do you really measure intelligence? While modern IQ tests were introduced in 1905 and supposedly give us an official marker of how 'smart' someone is, that hasn't come without its controversy.

As well as a debate about emotional intelligence versus cognitive intelligence, the fact that there are so many IQ tests out there means it's almost impossible to measure.

The likes of the Wechsler Scales and the Stanford-Binet are held as the gold standards of IQ testing, and in terms of who the world's smartest person is, that honor technically goes to Marilyn vos Savant.

While vos Savant held the Guinness World Record from 1985 to 1989, thanks to her impressive 228, the organization retired the record in 1990 due to controversy.

Plenty have supposedly overtaken vos Savant since then, with YoungHoon Kim regularly boasting about his 276.

Kim has his skeptics, but we wonder whether he could crack this notorious maths problem that's apparently only for the brightest of the bright.

Math problems that only geniuses can supposedly solve are nothing new, but this one seems to have more people than usual stumped.

Known as the Circle of Death or the Josephus Problem, this theoretical problem is a counting-out game that promises to push your brain to the limit.

Named after Flavius Josephus, the Jewish historian and leader who lived in the 1st century, it uses the problem to explain his first-hand account of the Siege of Yodfat.

Josephus says he and 40 of his men were trapped in a cave by Roman soldiers, but chose to take their own lives instead of being captured. They drew lots on who would go first, although according to Josephus, either sheer luck or intervention from God left him and one other man alive to eventually surrender.

This story has evolved into the Circle of Death, as Mr Barton Maths explains how the Josephus Problem works. We hope you're paying attention...

The site writes: "There are one hundred people standing in a circle. They count off beginning at one and ending at one hundred. Since they are in a circle, One is next to Two and One Hundred.

“One has a sword and kills Two. He passes the sword to Three who kills Four. And so forth."

While this was hard enough to keep up with and has some of us already tapping out, the problem continues: "Ninety-Nine kills One Hundred and passes the sword to One. Then One kills Three and passes the sword to Five. This goes on until only one person is left standing.

"Which number is he?”

Whether you want to test your friends and family on Christmas Day, or have simply had enough of scratching your head, Geeks For Geeks has the answer: "If the number of people (n) is a power of 2, the first person will survive.

"After each round, half the people are eliminated, and the person who started the game survives. When n is not a power of 2: Let 2m be the largest power of 2 less than n (100). The formula for the survivor is : Survivor = 2 x (n - 2m) + 1."

Crunching the numbers, things go as follows

• Round 1: Remove all even numbers → remaining:
• Round 1: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99
• Round 2: Remove every second → remaining:
• Round 2: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 53, 57, 61, 65, 69, 73, 77, 81, 85, 89, 93, 97
• Round 3: Remove every second → remaining:
• Round 3: 1, 9, 17, 25, 33, 41, 49, 57, 65, 73, 81, 89, 97
• Round 4: Since they are in a circle, 97 killed 1, and in the last move, 89 killed 97, leaving us with these survivors.
• Round 4: 9, 25, 41, 57, 73, 89
• Round 5: Now, 9 eliminates 25, 41 eliminates 57, and 73 eliminates 89, leaving us with these survivors.
• Round 5: 9, 41, 73
• Round 6: Now, 9 eliminates 41, leaving 9 and 73. The next turn will be 73’s.
• Round 6: 9, 73
• Round 7: Now 73 eliminated 9 , And 73 is the Last Survivor.
• Round 7: 73

See, you don't need the smarts of YoungHoon Kim to solve this one.