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14,179: the number of digits is odd (5) → 417 − 1 − 9 = 407: 0 − 4 − 7 = −11 = −1 × 11. 12: It is divisible by 3 and by 4. [6] 324: it is divisible by 3 and by 4. Subtract the last digit from twice the rest. The result must be divisible by 12. 324: 32 × 2 − 4 = 60 = 5 × 12. 13: Form the alternating sum of blocks of three from ...
d() is the number of positive divisors of n, including 1 and n itself; σ() is the sum of the positive divisors of n, including 1 and n itselfs() is the sum of the proper divisors of n, including 1 but not n itself; that is, s(n) = σ(n) − n
2008 – number of 4 × 4 matrices with nonnegative integer entries and row and column sums equal to 3 [8] 2009 = 7 4 − 7 3 − 7 2; 2010 – number of compositions of 12 into relatively prime parts [9] 2011 – sexy prime with 2017, sum of eleven consecutive primes: 2011 = 157 + 163 + 167 + 173 + 179 + 181 + 191 + 193 + 197 + 199 + 211; 2012 ...
In the past 500 years, there was no leap day in 1700, 1800 and 1900, but 2000 had one. In the next 500 years, if the practice is followed, there will be no leap day in 2100, 2200, 2300 and 2500 ...
For example, there are six divisors of 4; they are 1, 2, 4, −1, −2, and −4, but only the positive ones (1, 2, and 4) would usually be mentioned. 1 and −1 divide (are divisors of) every integer. Every integer (and its negation) is a divisor of itself. Integers divisible by 2 are called even, and integers not divisible by 2 are called odd.
For example, 6 is highly composite because d(6)=4 and d(n)=1,2,2,3,2 for n=1,2,3,4,5 respectively. A related concept is that of a largely composite number , a positive integer that has at least as many divisors as all smaller positive integers.
12 (twelve) is the natural number following 11 and preceding 13. Twelve is the 3rd superior highly composite number , [ 1 ] the 3rd colossally abundant number , [ 2 ] the 5th highly composite number , and is divisible by the numbers from 1 to 4 , and 6 , a large number of divisors comparatively.
Two properties of 1001 are the basis of a divisibility test for 7, 11 and 13. The method is along the same lines as the divisibility rule for 11 using the property 10 ≡ -1 (mod 11). The two properties of 1001 are 1001 = 7 × 11 × 13 in prime factors 10 3 ≡ -1 (mod 1001) The method simultaneously tests for divisibility by any of the factors ...