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 mar​

Answer 1

Answer:

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.

Answer 2

Answer:

given 4BC

Then it should be divisible by 15

15×1=15

15×2=30

15×3=45

15×4=60

15×5=75 so B and C are 0,5

405÷14=25