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question:Let nge 5 be a positive integer and let {a_1,a_2,...,a_n}={1,2,...,n} .Prove that at least lfloor sqrt{n}rfloor +1 numbers from a_1,a_1+a_2,...,a_1+a_2+...+a_n leave different residues when divided by n .
answer:1. **Define the sequence and residues:** Let ( b_i = a_1 + a_2 + dots + a_i ) for ( i = 1, 2, dots, n ). We need to show that at least ( lfloor sqrt{n} rfloor + 1 ) of the numbers ( b_1, b_2, dots, b_n ) leave different residues when divided by ( n ). 2. **Assume the contrary:** Assume, for the sake of contradiction, that there are at most ( lfloor sqrt{n} rfloor ) different residues among ( b_1, b_2, dots, b_n ) modulo ( n ). 3. **Pigeonhole Principle:** By the Pigeonhole Principle, if there are ( n ) numbers and at most ( lfloor sqrt{n} rfloor ) different residues, then some residues must repeat. Let ( r_1, r_2, dots, r_{lfloor sqrt{n} rfloor} ) be the distinct residues. 4. **Counting the differences:** Consider the differences ( b_{i+1} - b_i ) for ( i = 1, 2, dots, n-1 ). Each ( b_{i+1} - b_i = a_{i+1} ) is one of the numbers from the set ({1, 2, dots, n}). 5. **Bounding the differences:** Since there are at most ( lfloor sqrt{n} rfloor ) different residues, the number of possible differences ( b_{i+1} - b_i ) modulo ( n ) is at most ( lfloor sqrt{n} rfloor (lfloor sqrt{n} rfloor - 1) + 1 ). This is because each pair of residues can differ by at most ( lfloor sqrt{n} rfloor - 1 ) and there are ( lfloor sqrt{n} rfloor ) such pairs. 6. **Contradiction:** The total number of possible values for ( a_i ) is ( n ), but our assumption implies that the number of possible values for ( a_i ) is less than ( n ). This is a contradiction because ( a_i ) must take all values from ( 1 ) to ( n ). 7. **Conclusion:** Therefore, our assumption that there are at most ( lfloor sqrt{n} rfloor ) different residues is false. Hence, there must be at least ( lfloor sqrt{n} rfloor + 1 ) different residues among ( b_1, b_2, dots, b_n ) modulo ( n ). (blacksquare)
question:The tallest building in the world is 100 feet tall. If the second tallest is half that tall, and the third tallest is half as tall as the second, and the fourth is onefifth as tall as the third, how tall are all 4 buildings put together?
answer:The tallest building is 100 feet tall. The second tallest building is half as tall as the tallest, so it is 100 / 2 = 50 feet tall. The third tallest building is half as tall as the second, so it is 50 / 2 = 25 feet tall. The fourth tallest building is one-fifth as tall as the third, so it is 25 / 5 = 5 feet tall. To find the total height of all four buildings put together, we add their heights: 100 + 50 + 25 + 5 = 180 feet. Therefore, all four buildings put together are boxed{180} feet tall.
question:Compute the value of frac{1722^2 1715^2}{1731^2 1706^2}.
answer:First, apply the difference of squares formula to the numerator and the denominator: The numerator becomes: 1722^2 - 1715^2 = (1722 - 1715)(1722 + 1715) = 7 cdot (3437). The denominator becomes: 1731^2 - 1706^2 = (1731 - 1706)(1731 + 1706) = 25 cdot (3437). Now, simplify the fraction: frac{7 cdot 3437}{25 cdot 3437} = frac{7}{25}. Thus, the simplified result is: boxed{frac{7}{25}}.
question:The product of 600×50 has ______ zeroes at the end.
answer:Solution: 600×50=30000; 30000 has 4 zeroes at the end; So, the product of 600×50 has 4 zeroes at the end. Hence, the answer is: boxed{4}. According to the multiplication method of integers, calculate the product of 600×50, and then further explain. To find the number of zeroes at the end of the product of two numbers, you can first calculate their product, and then further explain.