Search results
Results from the WOW.Com Content Network
Cycles of the unit digit of multiples of integers ending in 1, 3, 7 and 9 (upper row), and 2, 4, 6 and 8 (lower row) on a telephone keypad. Figure 1 is used for multiples of 1, 3, 7, and 9. Figure 2 is used for the multiples of 2, 4, 6, and 8. These patterns can be used to memorize the multiples of any number from 0 to 10, except 5.
The second column from the right is added: 1 + 0 + 1 = 10 2 again; the 1 is carried, and 0 is written at the bottom. The third column: 1 + 1 + 1 = 11 2. This time, a 1 is carried, and a 1 is written in the bottom row. Proceeding like this gives the final answer 100100 2 (36 10).
0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24, 8, 25, 43, 62, ... "subtract if possible, otherwise add" : a (0) = 0; for n > 0, a ( n ) = a ( n − 1) − n if that number is positive and not already in the sequence, otherwise a ( n ) = a ( n − 1) + n , whether or not that number is already in the sequence.
SPOILERS BELOW—do not scroll any further if you don't want the answer revealed. The New York Times Today's Wordle Answer for #1269 on Monday, December 9, 2024
Answer: 7 × 1 + 6 × 10 + 5 × 9 + 4 × 12 + 3 × 3 + 2 × 4 + 1 × 1 = 178 mod 13 = 9 Remainder = 9 A recursive method can be derived using the fact that = and that =. This implies that a number is divisible by 13 iff removing the first digit and subtracting 3 times that digit from the new first digit yields a number divisible by 13.
6 is the 2nd superior highly composite number, [5] the 2nd colossally abundant number, [6] the 3rd triangular number, [7] the 4th highly composite number, [8] a pronic number, [9] a congruent number, [10] a harmonic divisor number, [11] and a semiprime.
The digit sum of 2946, for example is 2 + 9 + 4 + 6 = 21. Since 21 = 2946 − 325 × 9, the effect of taking the digit sum of 2946 is to "cast out" 325 lots of 9 from it. If the digit 9 is ignored when summing the digits, the effect is to "cast out" one more 9 to give the result 12.
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