$9$ Implies $8$ ? No Way!

Number Theory Level 2

We can express 9 positive single-digit numbers of the product $9! = 362880$ and the sum $45$ in at least two distinct ways. Following the factorial, $\begin{array}{rl} 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 &= 45\\ 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 &= 9! \end{array}$ Another way to express the sum and the product with some distinct numbers is $\begin{array}{rl} 1 + 2 + 4 + 4 + 4 + 5 + 7 + 9 + 9 &= 45\\ 1 \times 2 \times 4 \times 4 \times 4 \times 5 \times 7 \times 9 \times 9 &= 9! \end{array}$ Can we also express 8 positive single-digit numbers of the product $8!$ and the sum $36$ in the similar manner?

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1 solution

Vimay MarCisse
Jul 20, 2018

By prime factorisation we find :

8! = 1 × 2^7 × 3^2 × 5 × 7

So let's start our calcul with :

1 + 5 + 7

Because if any of these integers is multiplied by any other, the result is at least a 2 digit number (except for "1" but if i'm correct, it has to stay as "1" and can't be removed)

Then, 36 is even, thus we know that 3^2 has to be as 9 in our calcul (not 6 and 6), else the sum can't be even. So we get :

1 + 5 + 7 + 9

Then there only stay 2^7 which has to be decompsed into four 1 digit numbers. So we can first try :

1 + 5 + 7 + 9 + 2 + 2 + 4 + 8 = 38

Then we just have to adjust a little bit to finally get :

1 + 5 + 7 + 9 + 2 + 4 + 4 + 4 = 36

1 × 5 × 7 × 9 × 2 × 4 × 4 × 4 = 8!

Without loss of generality, we can keep $5$ and $7$ since they are both guaranteed to be single-digit primes that can't be multiplied to be distinct single-digit integers.

Michael Huang - 2 years, 10 months ago

9 9 9 Implies 8 8 8 ? No Way!

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$9$ Implies $8$ ? No Way!