Puzzles
16 December
Arrange the digits 1-9 in a 3×3 square so that:
the median number in the first row is 6;
the median number in the second row is 3;
the mean of the numbers in the third row is 4;
the mean of the numbers in the second column is 7;
the range of the numbers in the third column is 2,
The 3-digit number in the first column is today's number.
median 6 | |||
median 3 | |||
mean 4 | |||
today's number | mean 7 | range 2 |
11 December
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.
+ | ÷ | = 2 | |||
+ | ÷ | ÷ | |||
÷ | ÷ | = 3 | |||
÷ | - | ÷ | |||
÷ | ÷ | = 1 | |||
= 2 | = 1 | = 1 |
9 December
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.
all odd | |||
all even | |||
all multiples of 3 | |||
today's number | all >6 | all non-prime |
3 December
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 largest number you can make with the digits in the red boxes.
+ | + | = 21 | |||
+ | × | × | |||
+ | + | = 10 | |||
+ | ÷ | × | |||
+ | + | = 14 | |||
= 21 | = 10 | = 14 |
21 December
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 smallest number you can make using the digits in the red boxes.
+ | ÷ | = 2 | |||
× | + | - | |||
× | - | = 31 | |||
+ | + | - | |||
- | × | = 42 | |||
= 37 | = 13 | = -2 |
16 December
Arrange the digits 1-9 in a 3×3 square so that the first row makes a triangle number, the second row's digits are all even, the third row's digits are all odd; the first column makes a square number, and the second column makes a cube number.
The number in the third column is today's number.
triangle | |||
all digits even | |||
all digits odd | |||
square | cube | today's number |
14 December
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 numbers in the red boxes.
- | + | = 10 | |||
÷ | + | ÷ | |||
÷ | + | = 3 | |||
+ | - | ÷ | |||
+ | × | = 33 | |||
= 7 | = 3 | = 3 |
8 December
Arrange the digits 1-9 in a 3×3 square so: each digit the first row is the number of letters in the (English) name of the previous digit, each digit in the second row is one less than the previous digit, each digit in the third row is a multiple of the previous digit; the second column is an 3-digit even number, and the third column contains one even digit.
The number in the first column is today's number.
each digit is the number of letters in the previous digit | |||
each digit is one less than previous | |||
each digit is multiple of previous | |||
today's number | even | 1 even digit |
Edit: There was a mistake in this puzzle: the original had two solutions. If you entered the wrong solution, it will automatically change to the correct one.