Q: What is the arithmetic median of all positive two digit multiples of 6?

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The smallest 2 digit multiple of 6 is 12 = 6*2 The largest 2 digit multiple of 6 is 96 = 6*16 So the median is 6*(the median of 2,3,4,...,16) that is, the required median is 6*9= 54

10 of them.

You can use the formula for an arithmetic series for that.

There are 30 two-digit multiples of 3 plus 18 two-digit multiples of 5 minus 6 two-digit multiples of both. Answer: 42

-27

No. It depends on how good you are at arithmetic.

Multiples of 30 from 120 to 990

-4

All integers have an infinite amount of multiples.

All of the odd numbers between 11 and 99 inclusive are two digit numbers that are not multiples of 2. There are 45 of them.

There are 898 three-digit even numbers. Nine of them are multiples of 55. That leaves 889 * * * * * There are 450 three-digit even numbers and 17 of them are multiples of 55. So that leaves 433.

All multiples of 3 have digits that add up to a multiple of 3. There are 333 multiples of 3 between 1 and 1000.

100, 110 and 120 are 3 digit multiplies of 10.

They are 40 and 80.

14,21,28,35,42,49,56,63,70,77,84,91,98

1,3,5,7,9

495

60 numbers

150. 100/ 6 = 16 2/3 so 17 x 6 = 102 is the first three digit multiple of 6 1000 / 6 = 166 2/3 so 166 x 6 = 996 is the last three digit multiple of 6 So there are 166 - 17 + 1 = 150 three digit multiples of 6.

If my arithmetic is right, 966.

The two-digit multiples of nine are 18, 27, 36, 45, 54, 63, 72, 81, 90, 99.

lcm(2, 3, 5) = 30 → 2 digit common multiples are 30, 60, 90.

The first 3 digit integer being a positive multiple of 30 is 120. The final 3 digit integer being a positive multiple of 30 is 990. 990 - 120 = 870. 870 ÷ 30 = 29 But, as 29 is the difference between the two limits and the limits themselves are included then there are 29 + 1 = 30 such numbers.

The smallest positive 4-digit multiple of seven is 1,001.

The largest digit in decimal arithmetic is 9. In hexadecimal, it is F. As far as I am aware, there is no digit that will meet the requirements of this question.

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