A probability problem by Digi Verse
Probability
Level
3
How many eight-digit numbers are there composed only of the digits 4 and 3 that are divisible by 24?
The answer is 10.
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1 solution
For a number to be divisible by 24, it needs to be divisible by both 8 and 3. By logic, the unit digit has to be 4, since it has to be even. Then, the tens digit has to be 4 too, since the last two digits need to be divisible by 4 in order to be divisible by 8. The hundreds digit has to be 3, since the last 3 digits have to make up a number divisible by 8, and 444 is not divisible by eight. The number also has to be divisible by 3, and by rule of divisibility, the digits of the number has to add up to a number that is divisible by 3. For the rest of the digits, you will find that the only combinations that allow the digits to add up to a multiple of 3 would by one 4 and four 3's, or four 4's and one 3. Then, you can swap their order, which would be 5 choose 1 for both sets of possibility. 5+5=10 numbers.