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The original number is divisible by 7 if and only if the number obtained using this procedure is divisible by 7. For example, the number 371: 37 − (2×1) = 37 − 2 = 35; 3 − (2 × 5) = 3 − 10 = −7; thus, since −7 is divisible by 7, 371 is divisible by 7.
For example, 3 is a divisor of 21, since 21/7 = 3 (and therefore 7 is also a divisor of 21). If m is a divisor of n , then so is − m . The tables below only list positive divisors.
7 is a divisor of 42 because =, so we can say It can also be said that 42 is divisible by 7, 42 is a multiple of 7, 7 divides 42, or 7 is a factor of 42. The non-trivial divisors of 6 are 2, −2, 3, −3.
It is divisible by 2 and by 7. 224: it is divisible by 2 and by 7. Add the last two digits to twice the rest. The answer must be divisible by 7. 364: (3 × 2) + 64 = 70." Well, I think there is a mistake. Obviously 371 is not divisible by 14 but (3×2)+71=77. Where it is written "The answer must be divisible by 7" it should be written "The ...
The square-free part is 7, the square-free factor such that the quotient is a square is 3 ⋅ 7 = 21, and the largest square-free factor is 2 ⋅ 3 ⋅ 5 ⋅ 7 = 210. No algorithm is known for computing any of these square-free factors which is faster than computing the complete prime factorization.
35 has two prime factors, (5 and 7) which also form its main factor pair (5 x 7) and comprise the second twin-prime distinct semiprime pair. The aliquot sum of 35 is 13 , within an aliquot sequence of only one composite number (35, 13 , 1 ,0) to the Prime in the 13 -aliquot tree. 35 is the second composite number with the aliquot sum 13 ; the ...
A satirical petition ostensibly aiming to crowdfund a trillion dollars to allow Denmark to buy California has received more than 200,000 signatures.
In number theory, a weird number is a natural number that is abundant but not semiperfect. [1] [2] In other words, the sum of the proper divisors (divisors including 1 but not itself) of the number is greater than the number, but no subset of those divisors sums to the number itself.