Puzzles

Each clue in this crossnumber (except 5A) gives a property of that answer that is true of no other answer. For example: 7A is a multiple of 13; but 1A, 3A, 5A, 1D, 2D, 4D, and 6D are all not multiples of 13. No number starts with 0.

For a general election, the Advent isles are split into 650 constituencies. In each constituency, exactly 99 people vote: everyone votes for one of the two main parties: the Rum party or the Land party. The party that receives the most votes in each constituency gets an MAP (Member of Advent Parliament) elected to parliament to represent that constituency.

In this year's election, exactly half of the 64350 total voters voted for the Rum party. What is the largest number of MAPs that the Rum party could have?

Put the digits 1 to 9 (using each digit exactly once) in the boxes so that the sums are correct. The sums should be read left to right and top to bottom ignoring the usual order of operations. For example, 4+3×2 is 14, not 10. Today's number is the product of the red digits.

For all values of \(x\), the function \(f(x)=ax+b\) satisfies

What is \(f(65)\)?

Arrange the digits 1-9 in a 3×3 square so that: all the digits in the first row are odd; all the digits in the second row are even; all the digits in the third row are multiples of 3; all the digits in the second column are (strictly) greater than 6; all the digits in the third column are non-prime. The number in the first column is today's number.

Carol uses the digits from 0 to 9 (inclusive) exactly once each to write five 2-digit even numbers, then finds their sum. What is the largest number she could have obtained?

The sum of the coefficients in the expansion of \((x+1)^5\) is 32. Today's number is the sum of the coefficients in the expansion of \((2x+1)^5\).

Noel's grandchildren were in born in November in consecutive years. Each year for Christmas, Noel gives each of his grandchildren their age in pounds.

Last year, Noel gave his grandchildren a total of £208. How much will he give them in total this year?