Solutions - Mu Alpha Theta
... by 8. Since we are not allowed to use 4, 0, or double digits, we have only five choices for the last two positions: 16, 32, 56, 72, and 96. That leaves six numbers to choose from for the previous six positions. The total therefore is 5 6! 3600 . 7. A If the sum is prime, then the sum must be 2, ...
... by 8. Since we are not allowed to use 4, 0, or double digits, we have only five choices for the last two positions: 16, 32, 56, 72, and 96. That leaves six numbers to choose from for the previous six positions. The total therefore is 5 6! 3600 . 7. A If the sum is prime, then the sum must be 2, ...
Real Number - Study Point
... each of 8, 15 and 21. 12. Consider the numbers 4 n , where n is a natural number. Check whether there is any value of n for which 4 n ends with the digit zero. 13. Check whether 5n can end with the digit 0 for any n N. 14. Check whether 8n can end with the digit 0 for any n N. 15. Check whether ...
... each of 8, 15 and 21. 12. Consider the numbers 4 n , where n is a natural number. Check whether there is any value of n for which 4 n ends with the digit zero. 13. Check whether 5n can end with the digit 0 for any n N. 14. Check whether 8n can end with the digit 0 for any n N. 15. Check whether ...
THE CARMICHAEL NUMBERS UP TO 1015 0. Introduction and
... we proceed by using Proposition 2, looping first over all D in the range 2 to P - 1, and then over all C with CD satisfying the inequalities of Proposition 2(3). For each such pair (C, D), we test whether the values of pd_x and pd obtained from 2(1) and 2(2) are integral and, if so, prime. Finally, ...
... we proceed by using Proposition 2, looping first over all D in the range 2 to P - 1, and then over all C with CD satisfying the inequalities of Proposition 2(3). For each such pair (C, D), we test whether the values of pd_x and pd obtained from 2(1) and 2(2) are integral and, if so, prime. Finally, ...
Factor by Grouping Short-cut
... Even + even or odd + odd. Since the 3 and 63 are both odd, you must find two odd factors. 3 * 63 = 3 * 9 * 7 = 3 * 3 * 3 * 7. List all possible odd factors. Remember odd * odd = odd. 3 * ____, 7 * _____, 9 * _____, 21 * _____, 27 * ____, and 63 * ____. It appears that we may have many possible group ...
... Even + even or odd + odd. Since the 3 and 63 are both odd, you must find two odd factors. 3 * 63 = 3 * 9 * 7 = 3 * 3 * 3 * 7. List all possible odd factors. Remember odd * odd = odd. 3 * ____, 7 * _____, 9 * _____, 21 * _____, 27 * ____, and 63 * ____. It appears that we may have many possible group ...
