find 4 consecutive integers such that 3 times the sum of the last two greater integers is bigger than 5 times the first integer by 100

Answer 1

x+ x+1 +x+2 +x +3 = 4 consecutive integers

3(x+2 +x+3) = 5x +100

3(2x+5) = 5x +100

6x +15 = 5x + 100

x = 85

the integers are

85, 86, 87 and 88

check the answer

3(87+88) = 5(85) +100

3(175) = 425 +100

525 = 525