(9+5+5+1+1) x 1
Working on the .21 only. There are 2 digits in 21. The 1 is in the 100th decimal place.So .21 is also 21/100 which gives a final answer of 6 21/100
3.14159265358979323846
One possible solution is: 4.4 + 4 / .4 = 21
You can make: 1 combination containing 0 digits, 7 combinations containing 1 digits, 21 combinations containing 2 digits, 35 combinations containing 3 digits, 35 combinations containing 4 digits, 21 combinations containing 5 digits, 7 combinations containing 6 digits, and 1 combinations containing 7 digits. That makes 2^7 = 128 in all.
Yes, but only if there are no digits after the decimal point. For example, 18, 19, 20, 21 are consecutive numbers in the decimal system.
Try the different numbers - there are only 11 after all - and multiply their digits. Or analyze the factors of the number 21.
42/40 = 21/20 or 11/20
Working on the .21 only. There are 2 digits in 21. The 1 is in the 100th decimal place.So .21 is also 21/100 which gives a final answer of 6 21/100
987-65-43+21 = 900
3.14159265358979323846
21
It is impossible. The digits are all odd. Adding six odd digits gives an even digit so their sum cannot be 21.
2+3=5... 5x4=20... 20+1=21 !
One possible solution is: 4.4 + 4 / .4 = 21
There are 9 pages with a single digit (pages 1-9) = 9 digits There are 30 - 9 = 21 pages with two digits = 21 × 2 = 42 digits → There are 9 + 41 = 51 digits in total.
You can make: 1 combination containing 0 digits, 7 combinations containing 1 digits, 21 combinations containing 2 digits, 35 combinations containing 3 digits, 35 combinations containing 4 digits, 21 combinations containing 5 digits, 7 combinations containing 6 digits, and 1 combinations containing 7 digits. That makes 2^7 = 128 in all.
21