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In number theory, Kaprekar's routine is an iterative algorithm named after its inventor, Indian mathematician D. R. Kaprekar. Each iteration starts with a number, sorts the digits into descending and ascending order, and calculates the difference between the two new numbers. As an example, starting with the number 8991 in base 10: 9981 – 1899 ...
Dividing 272 and 8, starting with the hundreds digit, 2 is not divisible by 8. Add 20 and 7 to get 27. The largest number that the divisor of 8 can be multiplied by without exceeding 27 is 3, so it is written under the tens column. Subtracting 24 (the product of 3 and 8) from 27 gives 3 as the remainder.
In this method, each digit of the subtrahend is subtracted from the digit above it starting from right to left. If the top number is too small to subtract the bottom number from it, we add 10 to it; this 10 is "borrowed" from the top digit to the left, which we subtract 1 from.
6174 is a 7-smooth number, i.e. none of its prime factors are greater than 7. 6174 can be written as the sum of the first three powers of 18: 18 3 + 18 2 + 18 1 = 5832 + 324 + 18 = 6174, and coincidentally, 6 + 1 + 7 + 4 = 18. The sum of squares of the prime factors of 6174 is a square: 2 2 + 3 2 + 3 2 + 7 2 + 7 2 + 7 2 = 4 + 9 + 9 + 49 + 49 ...
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Given numbers and , the naive attempt to compute the mathematical function by the floating-point arithmetic ( ()) is subject to catastrophic cancellation when and are close in magnitude, because the subtraction can expose the rounding errors in the squaring.
Using the example above: 16,499,205,854,376 has four of the digits 1, 4 and 7 and four of the digits 2, 5 and 8; Since 4 − 4 = 0 is a multiple of 3, the number 16,499,205,854,376 is divisible by 3. Subtracting 2 times the last digit from the rest gives a multiple of 3. (Works because 21 is divisible by 3) 405: 40 - 5 x 2 = 40 - 10 = 30 = 3 x ...
The first Project Euler problem is Multiples of 3 and 5. If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000. It is a 5% rated problem, indicating it is one of the easiest on the site.