Free Sample Math Olympiads E-Level

Let’s break it down step by step:

1. First, calculate the sum of the numbers in the first set:

   $$(10+9+8+7+6)=40$$

2. Next, calculate the sum of the numbers in the second set:

   $$(5+4+3+2+1)=15$$

3. Finally, subtract the second sum from the first sum:

   $$(40-15)=25$$

So, the value of ((10 + 9 + 8 + 7 + 6) - (5 + 4 + 3 + 2 + 1)) is 25.

Twenty-eight days (4 wks = 7 days/wk) from Wednesday will be another Wednesday. Since the problem states that the game will be available in 33 days, subtract 33 – 28 = 5, so in five more days, it would be Monday. However, we need 2 days left for the sale, counting back 2 days is a Saturday.

Starting with the ones column, since we know the sum of all the digits in the ones column will result in a 5 in the ones place of the sum, A has to be 5. The sum of the ones column is 25. Carry the 2 and add it to the sum of the numbers in the tens column to get a sum ending with 4, so B is 3 and the sum is 24. Carry the 2 and add it to the sum of the numbers in the hundreds column. Since the sum ends in a 3, C is 2 and the sum is 13. Carry the 1, so D is 0. The four-digit number is 5320.

Call the number A B C D. Since D is equal to the sum of the other three digits, and the sum of all four digits is 16, D must be 16/2 = 8. Now C is half of the remaining value (16 – 8 = 8) or 8/2 = 4. The only pair of increasing numbers that adds to the remaining value (8 – 4 = 4) is 1 and 3. Therefore, John’s number is 1348.

Consider the sum of the digits 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36. The corner digits labeled A, B, C, and D will each be used twice, once in each row and once in each column. The sum of the rows and columns is 9 + 13 + 15 + 21 = 58. Therefore, the four corner numbers must have a sum of 58 – 36 = 22. That means that A + B + C + D = 22.