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There are no such integers. In fact, there are no real numbers that satisfy the requirements.

Q: What two integers have a sum of ten and a product of seventyfive?

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+5 and -15 do.

10+(X)+(X+1)=196 10+2X+1=196 11+2X=196 2X=185 X=92.5 Therefore, ten plus the sum of two consecutive integers can not be even.

48

public class sum { public static void main(String args[]){ int sum=0; for(int i=1;i<=10;i++){ sum=sum+i; } System.out.println(sum); } }

5,000,000*2=10,000,000 5,000,000 and 2

Related questions

The sum of the first ten positive integers is: 55

+5 and -15 do.

negative 15 and positive 5

The sum of the first ten positive integers, i.e. 1,2,3,4,5,6,7,8,9, and 10, is 55. The sum of the first ten negative integers, i.e. -1,-2,-3,-4,-5,-6,-7,-8,-9, and -10 is -55. The sum of the first ten positive integers plus the sum of the first ten negative integers is 0.

the product of 8 and the sum of 10 and -7

sum = 10 1/2 product = 5 difference = 51/2

The integers are -4, -3, -2, -1, 0, 1, 2, 3, 4 and 5. The largest is five.

10+(X)+(X+1)=196 10+2X+1=196 11+2X=196 2X=185 X=92.5 Therefore, ten plus the sum of two consecutive integers can not be even.

The product is the result of a multiplication sum. Since this sum has just one number in the question part, it is impossible to find the product of this single number.

48

5+5 = 10 (Sum is ten)5*5 = 25 (Product is 25)*this product is maximum for all any 2 real numbers that == 10

public class sum { public static void main(String args[]){ int sum=0; for(int i=1;i<=10;i++){ sum=sum+i; } System.out.println(sum); } }