Question

24. A three-digit number 4BC is divisible by 15. What are the values of B and C for the
smallest such number.
(2 marks​

Answer 1

Answer:

Step-by-step explanation:

Prime factorisation of 22=2×11

Prime factorisation of 33=3×11

Prime factorisation of 55=5×11

So, LCM 22,33,55=2×3×5×11=330

As 330 is the smallest 3− digit number divisible by 22,33,55, the number 330+5=335 will give a remainder 5 when divided by these numbers.

The highest 3− digit number is 999.

999 when divived by 330 gives a remainder 9, so 999−9=990 is the largest 3− digit number divisible by 22,33,55, the number 990+5=995 will give a remainder 5 when divided by these numbers.

Explanation:

Answer 2

Answer:

405 is the smallest possible number