# Math

Determine the largest possible integer n such that 942! is divisible by 15n.

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asked by Judy
1. 942!/(15n)
= 942x941x940x939x938!/(15n)
= 942(941)(188)(313)(938!)/n

I divided by 5 and by 3 leaving the division by n

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posted by Reiny
2. The answer to the question as posted is n=942!/(15), which is a large number approximately equal to: 1.45*10^2393.

I suspect the factor 15n should read 15n.
If that's the case, I offer the following.

942! is the product 942*941*940*....3*2*1

If n=1, then 942!/(15) should give a whole number, i.e. 942! contains at least the factor 5 and 3.

If n=2, then 942!/(15²) should give a whole number, i.e. 942! contains at least the factor 5² and 3².

If n=3, then 942!/(15³) should give a whole number, i.e. 942! contains at least the factor 5³ and 3³.

... and so on.

So the problem reduces to counting the number of factors of 5 in 942! and number of factors of 3 in 942!.

The number of factors of 3 is likely more numerous in 942! since 3 is a smaller number, so it would appear more frequently. We therefore concentrate on the power of 5 which divides 942!.

942/5= 188.4,
so there are 188 factors of 5.
942/25=37.68, so there are 37 more factors of 5.
942/125=7.5, there are 7 more factors of 5.
942/625=1.5, there is 1 more factor of 5.
In total, there are (188+37+7+1)=233 factors of 5, so
maximum n=233.

As a check, dividing 942!/15^233 gives ...0944 as the last digits, i.e. it does not divide by 5 (nor 15) any more.

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