The new year has come upon us, and with it a new number to work with. Because of this, I want to compile a list of properties and facts about the number 2018 for use in competitions and problem writing. This can include relationships with other numbers, ways to formulate it as an expression, and expressions involving it; however, any fact about 2018 that could be used in a competition setting can belong here.
I'll start this list off with some basic facts:
Feel free to contribute anything about the number 2018 in the comments!
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Comments
2018 is the sum of 4 distinct nonzero fourth powers:
2^4 + 3^4 + 5^4 + 6^4
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vbnm
cvbnm,.
2018 can be written as a sum of 4 consecutive positive integers, but not as a sum of any other number of consecutive positive integers. This sum is 503+504+505+506=2018.
2018 is the number of ways we can partition 60 into prime parts
here are the first 10 partitions of 60 into prime parts:
{53,7},{53,5,2},{53,3,2,2},{47,13},{47,11,2},{47,7,3,3},{47,7,2,2,2},{47,5,5,3},{47,5,3,3,2},{47,5,2,2,2,2}...
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good solution
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2018 has 4 positive divisors.
Sum of the 4 positive divisor of 2018 is 3030.
2018 is one larger than a prime number.
2018 can be expressed as the sum of two perfect squares. 132+432=2018
Etc...
● 2018 in binary=11111100010 base of two
● Divisors = 1, 2, 1009, 2018
● Count from 1 upto 2018 take 33 minutes.
• 2018 is a deficient number, because the sum of its proper divisors (1012) is less than itself. Its deficiency is 1006.
• 2018 is a UnLucky number.
• 2018 is a UnHappy number.
There are two ways to formulate it as the sum of the difference of two sets of consecutive cubes.
(9^3-8^3) + (25^3 - 24^3)
8&9 are consecutive, 24&25 are consecutive.
and
(15^3-14^3) + (22^3 - 21^3)
2018 is a Squarefree composite number such that the sum of its divisors is also Squarefree
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Indeed. I wrote a little problem about this property: it is quite "rare" among all year numbers.
2018 can be written as a sum 2 perfect squares of which the first digit is perfect square (12,22) repectively and the last digit is both 3 : 2018=132+432.
2018 is a semiprime that is a prime+1
Can 2018 be expressed as the difference of two powers ? 2018 = m^n - p^q ? where m, n, p and q are all prime ?
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I had got this straight away, and posted a true way, but I suddenly realised that it had a very strong correlation to one of the problems in the Popular section, and have removed it.
2018 in different number bases:
base 2: 1111110010
base 3: 2202202
base 4: 133202
base 5: 31033
base 6: 13202
base 7: 5612
base 8: 3742
base 9: 2682
base 10: 2018
base 11: 1575
base 12: 1202
base 13: BC3
base 14: A42
base 15: 8E8
base 16: 7E2