17, 19, 23, 29, and 31
17, 19, 23, 29, and 31
(17+17+17+17+17+17+17+17+17+17+17)-(29+29+29) = 100
18 is an even number between 12 and 29 with digits whose difference is 7. To answer this question, we start with the last part "difference is 7" and come up with a set of values that meet the need as follows: 18 29 [29 is in the set only if the question is inclusive] Now, there is only one even number in the pre-solution set and that number is 18.
It is: 11+13+17+19+23+29 = 112
23 is halfway between 17 and 29
twenty-three 23 - 17 = 6 29 - 23 = 6
17, 19, 23, 29, and 31
17, 19, 23, 29, and 31
The two numbers are 23 and 6. Let: x = first number y = second number x + y = 29 (Equation 1) x - y = 17 x = 17 + y (Equation 2) Substitute Eq. (2) in Eq. (1): x + y = 29 (17 + y) + y = 29 17 + 2y = 29 17 -17 + 2y = 29 - 17 2y = 12 2y/2 = 12/2 y = 6 - second number x = 17 + y x = 17 + 6 x = 23 - first number check. 23 - 6 = 17 17 = 17 Therefore, the two numbers are 23 and 6.
(17+17+17+17+17+17+17+17+17+17+17)-(29+29+29) = 100
29 is an odd number.
7 11 13 17 19 23 29.
20
They are: 11 13 17 19 23 and 29 which makes six of them
18 is an even number between 12 and 29 with digits whose difference is 7. To answer this question, we start with the last part "difference is 7" and come up with a set of values that meet the need as follows: 18 29 [29 is in the set only if the question is inclusive] Now, there is only one even number in the pre-solution set and that number is 18.
29