The integers are 99, 100 and 101. There is also a set of consecutive even integers whose sum is 300. That set is 98, 100 and 102.
Simply consider what ARE the sums of four consecutive non-negative integers: 0+1+2+3 = 61+2+3+4 = 102+3+4+5 = 143+4+5+6 = 18.... You can see that it's any number of the form: 6 plus a (non-negative integer) multiple of 4.A more mathematical proof is to consider an arbitrary starting number n, then: n+(n+1)+(n+2)+(n+3) = 4*n + 6 (which is the same as concluded above).
The numbers are -104 and -102.
102% = 1.02 = 102/100 = 51/50 = 11/50
79500 / 2.5 x 102 = 7.95 x 104 / 2.5 x 102 Now divide 7.95 by 2.5 = 3.18 and also divide 104 by 102 = 10(4 - 2) = 102 Thus, 7.95 x 104 / 2.5 x 102 = 3.18 x 102
The integers are 99, 100 and 101. There is also a set of consecutive even integers whose sum is 300. That set is 98, 100 and 102.
101 and 102
101 and 102
101+102=203
The average of the three integers is going to be -306/3 = -102. Therefore, taking one even integer from either side, -306 = (-100) + (-102) + (-104)
The numbers are 101 and 102.
They are (33, 34, 35).
306/3 = 102 so the three numbers are 101, 102 and 103.
102 = 32 + 34 + 36
The largest integer is 103 because 101+102+103 = 306
101 + 102 = 203. (Hint: solve the equation n + n+1 = 203.)
24, 25, 26, 27 (next to each other and add up to 102)