# 40

This number is a composite.

The decimal fraction 40999920000041/(999999^3) has the value 0.000 41 000 43 000 47 000 53 000 61 000 71 000 83 000 97 00 113 00 131 00 151 00 173 00 197 00 223 00 251 00 281 00 313 00 347 00 383 00 421 00 461 00 503 00 547 00 593 00 641 00 691 00 743 00 797 00 853 00 911 00 971 0 1033 0 1097 0 1163 0 1231 0 1301 0 1373 0 1447 0 1523 0 1601. Note the 40 prime numbers amongst the zeros! [Keith]

The first 40 digits of Euler's constant form a prime. [Kulsha]

The expression P(n) = n^2 + n + 41 is prime for all natural numbers n < 40. It is composite for n = 40. [Mitchell]

The sequence number for prime numbers is A000040 in The On-Line Encyclopedia of Integer Sequences. [Hartley]

40 = 2^3 * 5 (the first three primes in order). [Beech]

40!+39!+38!+...+3!+2!+1!+0!+1!+2!+3!+...+38!+39!+40! is prime. [Schiffman]

There are 40 possible two-digit endings of primes (with leading zeros). Consider the following "race" after the first 1000 primes:The NASCAR Prime Puzzle

Let a(n) = the number of primes that end in "01" among the first 10^n primes. The sequence begins 0, 2, 25, 254, 2494, 24959, 249814, 2499088, 24998779, ... Let a(n) = the number of primes that end in "03" among the first 10^n primes. The sequence begins 0, 2, 25, 249, 2510, 25056, 250276, 2500054, 24998487, ... Let a(n) = the number of primes that end in "07" among the first 10^n primes. The sequence begins 0, 2, 27, 249, 2459, 24931, 250103, 2500735, 25000294, ... Let a(n) = the number of primes that end in "09" among the first 10^n primes. The sequence begins 0, 3, 24, 245, 2504, 24961, 249670, 2500222, 25001398, ... Let a(n) = the number of primes that end in "11" among the first 10^n primes. The sequence begins 1, 3, 25, 257, 2492, 25048, 249864, 2499701, 25001011, Let a(n) = the number of primes that end in "13" among the first 10^n primes. The sequence begins 1, 3, 23, 256, 2489, 24956, 249883, 2499909, 25002129, ... Let a(n) = the number of primes that end in "17" among the first 10^n primes. The sequence begins 1, 2, 25, 253, 2519, 25001, 250172, 2500991, 24998892, ... Let a(n) = the number of primes that end in "19" among the first 10^n primes. The sequence begins 1, 2, 27, 248, 2514, 25003, 250137, 2499557, 24999197, ... Let a(n) = the number of primes that end in "21" among the first 10^n primes. The sequence begins 0, 2, 27, 250, 2486, 24973, 249850, 2499065, 24999554, ... Let a(n) = the number of primes that end in "23" among the first 10^n primes. The sequence begins 1, 3, 28, 259, 2511, 25012, 249966, 2499856, 24999237, ... Let a(n) = the number of primes that end in "27" among the first 10^n primes. The sequence begins 0, 2, 23, 250, 2504, 24931, 250074, 2499704, 24999642, ... Let a(n) = the number of primes that end in "29" among the first 10^n primes. The sequence begins 1, 2, 26, 255, 2491, 24966, 249873, 2499819, 24999296, ... Let a(n) = the number of primes that end in "31" among the first 10^n primes. The sequence begins 0, 4, 24, 251, 2502, 24973, 249851, 2499874, 25001020, ... Let a(n) = the number of primes that end in "33" among the first 10^n primes. The sequence begins 0, 2, 24, 247, 2509, 24981, 250124, 2499684, 24999545, ... Let a(n) = the number of primes that end in "37" among the first 10^n primes. The sequence begins 0, 3, 25, 254, 2522, 24959, 249813, 2499850, 25001924, ... Let a(n) = the number of primes that end in "39" among the first 10^n primes. The sequence begins 0, 3, 23, 254, 2512, 25041, 249731, 2499788, 25001154, ... Let a(n) = the number of primes that end in "41" among the first 10^n primes. The sequence begins 0, 3, 21, 237, 2505, 24960, 249754, 2500494, 249998836, ... Let a(n) = the number of primes that end in "43" among the first 10^n primes. The sequence begins 0, 2, 24, 246, 2484, 25006, 249884, 2500043, 25002072, ... Let a(n) = the number of primes that end in "47" among the first 10^n primes. The sequence begins 0, 2, 26, 248, 2520, 24992, 249765, 2499987, 25002877, ... Let a(n) = the number of primes that end in "49" among the first 10^n primes. The sequence begins 0, 3, 23, 245, 2516, 24980, 249954, 2499220, 25000053, ... Let a(n) = the number of primes that end in "51" among the first 10^n primes. The sequence begins 0, 2, 28, 253, 2504, 25015, 250178, 2500410, 25001559, ... Let a(n) = the number of primes that end in "53" among the first 10^n primes. The sequence begins 0, 2, 27, 250, 2497, 24974, 250005, 2499972, 24999702, ... Let a(n) = the number of primes that end in "57" among the first 10^n primes. The sequence begins 0, 3, 29, 246, 2528, 25040, 250114, 2499937, 24999120, ... Let a(n) = the number of primes that end in "59" among the first 10^n primes. The sequence begins 0, 2, 26, 254, 2512, 25007, 250264, 2500225, 24999949, ... Let a(n) = the number of primes that end in "61" among the first 10^n primes. The sequence begins 0, 2, 23, 239, 2491, 24951, 250117, 2500005, 24998609, ... Let a(n) = the number of primes that end in "63" among the first 10^n primes. The sequence begins 0, 3, 23, 246, 2516, 25021, 250080, 2500156, 25000087, ... Let a(n) = the number of primes that end in "67" among the first 10^n primes. The sequence begins 0, 4, 25, 250, 2479, 25038, 250019, 2499172, 24999835, ... Let a(n) = the number of primes that end in "69" among the first 10^n primes. The sequence begins 0, 1, 22, 247, 2479, 25021, 249895, 2500196, 24999156, ... Let a(n) = the number of primes that end in "71" among the first 10^n primes. The sequence begins 0, 2, 23, 245, 2500, 24995, 250012, 2500832, 24999429, ... Let a(n) = the number of primes that end in "73" among the first 10^n primes. The sequence begins 0, 3, 27, 244, 2474, 25040, 249980, 2500778, 25001960, ... Let a(n) = the number of primes that end in "77" among the first 10^n primes. The sequence begins 0, 1, 25, 242, 2488, 25099, 250165, 2499766, 25000007, ... Let a(n) = the number of primes that end in "79" among the first 10^n primes. The sequence begins 0, 4, 26, 253, 2503, 25059, 250054, 2500497, 24999733, ... Let a(n) = the number of primes that end in "81" among the first 10^n primes. The sequence begins 0, 2, 24, 251, 2487, 25001, 249942, 2500072, 25000149, ... Let a(n) = the number of primes that end in "83" among the first 10^n primes. The sequence begins 0, 3, 27, 256, 2510, 24978, 249909, 2499642, 25000408, ... Let a(n) = the number of primes that end in "87" among the first 10^n primes. The sequence begins 0, 1, 23, 255, 2516, 25046, 249966, 2500611, 24999550, ... Let a(n) = the number of primes that end in "89" among the first 10^n primes. The sequence begins 0, 2, 25, 247, 2484, 24914, 250134, 2499991, 24998766, ... Let a(n) = the number of primes that end in "91" among the first 10^n primes. The sequence begins 0, 2, 25, 247, 2506, 25059, 250373, 2499894, 24998633, ... Let a(n) = the number of primes that end in "93" among the first 10^n primes. The sequence begins 0, 2, 24, 261, 2506, 25085, 250101, 2500040, 24998592, ... Let a(n) = the number of primes that end in "97" among the first 10^n primes. The sequence begins 0, 3, 25, 260, 2479, 24976, 250091, 2499648, 25000363, ... Let a(n) = the number of primes that end in "99" among the first 10^n primes. The sequence begins 0, 2, 24, 243, 2494, 24988, 250039, 2500511, 24998992, ...... for example "Car Number 57" is leading after the first 10^4 laps (primes). But what about after 10^5 laps? 10^6 laps, etc.? Place your bets now!

**Update**: Thanks to Chuck Gaydos of Arizona, we now know that "Car Number 47" is in the lead with 25002877 primes after 10^9 (one billion) laps.

*Sequences and puzzle proposed by G. L. Honaker, Jr. (June 2016)

The smallest number n such that n!/[R(n)]!-1 is prime, where R(n) is the reversal of n, (i.e., 40!/04!-1). [Loungrides]