Which is the smallest multiple of 7, which leaves 5 as remainder in each case, when divided by 8, 9, 12 and 15?

2525 721 1085 736

1085

8, 9, 12, 15 → LCM = 360 Required number = \(\frac{(360\;×\;R) + 5}{7}\) Now put value of x as 3 = \(\frac{1080 + 5}{7}\) .... (divided completely) Therefore, the number is = 1085