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Problem 88


Problem 88


Product-sum Numbers

A natural number, N, that can be written as the sum and product of a given set of at least two natural numbers, {a1,a2,,ak} is called a product-sum number: N=a1+a2++ak=a1×a2××ak.

For example, 6=1+2+3=1×2×3.

For a given set of size, k, we shall call the smallest N with this property a minimal product-sum number. The minimal product-sum numbers for sets of size, k=2, 3, 4, 5, and 6 are as follows.

k=2:4=2×2=2+2k=3:6=1×2×3=1+2+3k=4:8=1×1×2×4=1+1+2+4k=5:8=1×1×2×2×2=1+1+2+2+2k=6:12=1×1×1×1×2×6=1+1+1+1+2+6

Hence for 2k6, the sum of all the minimal product-sum numbers is 4+6+8+12=30; note that 8 is only counted once in the sum.

In fact, as the complete set of minimal product-sum numbers for 2k12 is {4,6,8,12,15,16}, the sum is 61.

What is the sum of all the minimal product-sum numbers for 2k12000?


积和数

若自然数N能够同时表示成一组至少两个自然数{a1,a2,,ak}的积与和,也即N=a1+a2++ak=a1×a2××ak,则称之为积和数。

例如,6是积和数,因为6=1+2+3=1×2×3

给定这一组自然数的数目k,满足上述性质的最小N值被称为最小积和数。当k=23456时,最小积和数如下所示:

k=2:4=2×2=2+2k=3:6=1×2×3=1+2+3k=4:8=1×1×2×4=1+1+2+4k=5:8=1×1×2×2×2=1+1+2+2+2k=6:12=1×1×1×1×2×6=1+1+1+1+2+6

因此,对于2k6,所有的最小积和数之和为4+6+8+12=30;注意8只被计算了一次。

已知对于2k12,所有最小积和数构成的集合是{4,6,8,12,15,16},这些数之和是61

对于2k12000,所有最小积和数之和是多少?


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