advanced algebra

[manip]

Let n be the largest integer such that n <= 2017, and the sum of its digits equals the product of its digits , Find the sum of the digits of n

 

[Deleted member 4993]

Let n be the largest integer such that n <= 2017, and the sum of its digits equals the product of its digits , Find the sum of the digits of n

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[Cubist]

If any digit of n is 0 then all remaining digits of n must be what?

If you have a set of digits (the set is ANY size, not just 4) AND (product of the digits) > (sum of digits) then the set can be adjusted such that (product of the digits) = (sum of digits) by placing a number of extra "1" digits into the set. How many extra "1" are required? Can you see how this can be used to help with the original question?

Spoiler: hint

Produce a table...

Digits in the original set (order of digits doesn't matter here, AND don't include any "0" or "1" digits in this column, why?)	EXTRA number of "1" digits reqd to make (product of the digits) = (sum of digits)	Number of digits in the final, adjusted, set (how many rows can you find with this column equal to 4)
2	0	1+0=1
...	...	...
2,2,2	2	3+2=5