Some problems

[kate]

Problem 1: Put the mathematical operators + and - so that the equation to be through:
9 8 7 6 5 4 3 2 1 = 32
Is there any way the sum to be 31. How many are the possible combinations where the sum is 31.

Problem 2: Find the sum of all four-digital numbers which are made from the digits 2, 3, 4 and 6. (the digits could be put more than ones, for instance 6666, 6263, i.e. all possible combinations).

Problem 3: Three girls together have 12000 calendars. The first girl divided her into two equal parts, and left one half for herself and the other half divided equally to her friends. This did the other two girls too. At the end all three girls have equal number of calendars. What was the initial quantity of calendars of everyone of the girls?

[shyamjayakannan]

2) number of four digit numbers with 2 in the ones place = [tex]4^3[/tex]. This is the same for 2 in the 10s, 100s and 1000s place. This is further true for the remaining numbers as well. So total sum = [tex]4^3(1+10+100+1000)(2+3+4+6)=\boxed{1066560}[/tex]