Have given below solutions to the questions given here .
1. A number n! is written in base 6 and base 8 notation. Its base 6 representation ends with 10 zeroes. Its base 8 representation ends with 7 zeroes. Find the smallest n that satisfies these conditions. Also find the number of values of n that will satisfy these conditions.
Base 6 representation ends with 10 zeroes, or the number is a multiple of 6^10. If n! has to be a multiple of 6^10, it has to be a multiple of 3^10. The smallest factorial that is a multiple of 3^10 is 24!. So, when n = 24, 25 or 26, n! will be a multiple of 6^10 (but not 6^11).
Similarly, for the second part, we need to find n! such that it is a multiple of 2 ^ 21, but not 2 ^ 24. When n = 24, n! is a multiple of 2^22. S0, when n = 24, 25, 26, 27, n! will be a multiple of 2 ^ 21 but not 2 ^ 24.
The smallest n that satisfies the above conditions is 24. n = 24, 25 or 26 will satisfy the above conditions.
2. [x] is the greatest integer less than or equal to x. Find the number of positive integers n such that [n/11] = [n/13].
This is a classic case of brute-force counting.
When n = 1, 2, 3....10, both values will be equal to 0. 10 possibilities
When n = 13,14, ....21, both values will be equal to 1. 9 possibilities
When n = 26,27, ....32, both values will be equal to 2. 7 possibilities
When n = 39,27, ....43, both values will be equal to 3. 5 possibilities
When n = 52, 53, 54, both values will be equal to 4. 3 possibilities
when n = 65, both will be equal to 5. 1 possibility
So, totally there are 1 + 3 + 5 + 7 + 9 +10 = 35 possibilities.
3. Positive numbers 1 to 55, inclusive are placed in 5 groups of 11 numbers each. What is the maximum possible average of the medians of the 5 groups?
We need to maximise each median in order to have the overall maximum median possible.
The highest possible median is 50 as they should be 5 numbers higher than the median in the group of 11. So, if we have a set that has a, b, c, d, e, 50, 51, 52, 53, 54, 55, the median will be 50. In this set, it is best not to waste any high values on a, b, c, d or e as these do not affect the median. So, a set that reads as 1, 2, 3, 4, 5, 50, 51, 52, 53, 54, 55 will also have median 50.
The next set can be 6, 7, 8, 9, 10, 44, 45, 46, 47, 48, 49. The median will be 44.
The medians of the 5 groups can be 50, 44, 38, 32, 26. The highest possible average of the medians will be 38.Labels: CAT number theory, CAT Solutions