-5 and +5.
The two numbers are five and eleven. 5*5=25 and 11*11=121 and 25+121=146 which is four less than 100+50 which is 150.
-25
call the numbers a & b (a+b)2 = a2+2ab+b2 which is greater than a2 + b2 by twice the product of the numbers. Check: say 3 and 5 32 + 52 = 9 + 25 = 34 (3 +5)2 = 64, greater by twice a x b. QED -------------------- If a and b are the numbers, then (a+b)2 = a2 + 2ab + b2, which is different from a2 + b2 (not necessarily larger). The two quantities are equal only when one (or both) of a,b is zero.
The first digit must be 7 or greater, because the digits of 699 only sum to 24. Of the numbers beginning with 7, only 799 sums to 25, because if either of the other two digits are less than 9, the sum will be less than 25. That's one so far. For number beginning with 8, the other two digits must sum to 17. That means they must be 9 and 8. So there are two suitable numbers beginning with 8, namely 889 and 898. That's a total of three so far. Finally, we have numbers beginning with 9. The other two digits must sum to 16, so the candidates are: 9 and 7, or 8 and 8. That gives us three numbers that fit the bill: 997, 979 and 988. That's six altogether. So the answer is six.
20x5 =100 20+5=25 100+25= 125
14 and 11
25 and 6.
No two odd numbers can sum to 7.
10 and 15
If the sum of 2 numbers is 25 and their product is 156, the numbers are 12 and 13.
The two numbers with a sum of 33 and a product of 200 are 25 and 8. 25 x 8 = 200, 25 + 8 = 33.
If the sum of 2 numbers is 25 and their product is 156, the 2 numbers would be 12 and 13.
25 + 3 = 28 25 x 3 = 75 The two numbers are therefore 25 and 3.
5+5 = 10 (Sum is ten)5*5 = 25 (Product is 25)*this product is maximum for all any 2 real numbers that == 10
It's unclear whether you want two numbers that add to a sum or multiply to a product. 25 x 4 = 100 25 + 75 = 100
25 and 3 25*3 = 75 25+3 = 28
Suppose the two numbers are x and y. Then, the sum of THEIR reciprocals is 1/x + 1/y = y/xy + x/xy = (y + x)/xy = 7/25