For how many n, 7 <= n <= 100, ,is the base-n number 235236_n a multiple of 2?
+1161 If I understand your question:
This is the "closed form" of each term:
a_n = 1/2 (6 n + (-1)^(n + 1) + 13).
And the number 235236(n) is a multiple of 7 in the following bases:
n = 10, 12, 16, 18, 22, 24, 28, 30, 34, 36, 40, 42, 46, 48, 52, 54, 58, 60, 64, 66, 70, 72, 76, 78, 82, 84, 88, 90, 94, 96, 100.
guest.
+2666 If the base is odd, the number will never be even, because there are 3 odd numbers. The sum of these numbers will always be odd, because\(\text{odd} \times \text{odd} = \text{odd}\), and \(\text{odd} + \text{odd} + \text{odd} = \text{odd}\)
However, if the base is even, all the individual digits will yield an even number because \(\text{even} \times \text{odd} = \text{even}\), and \(\text{even} + \text{even} = \text{even}\)
So, how many even digits are there between 7 and 100?
