1.The sum of n different natural numbers is less than50.the greatest possible value of n is----
2.Let a be the average of all odd prime numbers less than 50.The inreger,most close to a is----
3.If two even natural numbers x,y satisfy equation x的平方加100等于y的平方,x大于y then x乘y等于----
1.Since there are some controversies,I suppose 0 does not belong to natural numbers in this case.The answer is quite straight foward,1+2+3+4+5+6+7+8+9=45.If we add 10,then the sum will exceed 50.So the greatest possible n is 9.
2.First of all,you can list all the odd prime numbers less than 50:3 5 7 11 13 17 19 23 29 31 37 41 43 47.Add them up,and calculate the average.The average a is equal to 23.3.Therefore the integer,most close to a is 23.
3.x^2+100=y^2,then (y-x)(x+y)=100,since x and y are natural numbers,x+y>0,y should be larger than x.I am afraid there is either something wrong with the problem or a typing mistake.
You can show me the original version of the problem 3. 1.The sum of n different natural numbers is less than50.the greatest possible value of n is----
2.Let a be the average of all odd prime numbers less than 50.The inreger,most close to a is----
3.If two even natural numbers x,y satisfy equation x的平方加100等于y的平方,x大于y then x乘y等于----
1.Since there are some controversies,I suppose 0 does not belong to natural numbers in this case.The answer is quite straight foward,1+2+3+4+5+6+7+8+9=45.If we add 10,then the sum will exceed 50.So the greatest possible n is 9.
2.First of all,you can list all the odd prime numbers less than 50:3 5 7 11 13 17 19 23 29 31 37 41 43 47.Add them up,and calculate the average.The average a is equal to 23.3.Therefore the integer,most close to a is 23.
3.x^2+100=y^2,then (y-x)(x+y)=100,since x and y are natural numbers,x+y>0,y should be larger than x.I am afraid there is either something wrong with the problem or a typing mistake.
You can show me the original version of the problem 3.