15
15
15
Five
989. If there is a remainder of 2 when divided by 3, the number is one less than a multiple of 3. If there is a remainder of 4 when divided by 5, the number is one less than a multiple of 5. Thus the number required is one less than a multiple of the lowest common multiple of 3 and 5 (that is 15). So what is needed is an even multiple of 15 less than or equal to 1000: 1000 ÷ 15 = 662/3 Thus the highest even multiple of 15 not greater than 1000 is 66 x 15 = 990, and the required number is 989.
15, 30, 45
12 is 3 less than 15.
Either 30 or 15
3/15 is less than 4/7.
3, 6, 9, 12, 15.As in the three times table.
510. To be divisible by 3, it must be a multiple of 3. Thus the required number is a multiple of both 3 and 5, which will be a multiple of their lcm: lcm(3, 5) = 15. → 495 ÷ 15 = 33 → first multiple of 15 greater than 495 is 15 x 34 = 510 → 525 ÷ 15 = 35 → last multiple of 5 less than 525 is 15 x 34 x 510 → number required is 510.
15 Factors of 15: 1, 3, 5, 15 15 = 1 x 15 15 = 3 x 5