Written by Colin+ in ask uncle colin, big in finland.

Dear Uncle Colin,

A seven-digit integer has 870,720 as its last six digits. It is the product of six consecutive even integers. What is the missing first digit?

Please Reveal Our Digit! Underlying Calculation Too

Hi, PRODUCT, and thanks for your message!

There are several approaches to this (as usual) - I can think of a rather brutal one, and a very simple one.

The brutal way is to write a Python script to do it.

prod = lambda x: x*(x+2)*(x+4)*(x+6)*(x+8)*(x+10) # prod(x) multiplies x by the next five numbers of the same parity for i in range(2, 20, 2): # 20^6 is way over our number, whatever it is print (i, prod(i))

This prints out:

2 46080 4 322560 6 1290240 8 3870720 10 9676800 12 21288960 14 42577920 16 79073280 18 138378240

There we go, $8\times 10 \times 12 \times 14 \times 16 \times 18 = 3,870,720$.

But that’s really unsatisfying.

Whatever the first of your six factors is, at least two of the factors must be multiples of 3. That means the overall product is a multiple of 9. And what do we know about multiples of 9? Their digit sums are also multiples of 9. Adding up the six digits you’re given gives 24, and the only digit you can prepend to that to give a multiple of 9 is 3.

But can one work out the original factors from that?

“The median of the factors will be roughly the sixth root of the number. Starting with $3.9 \times 10^6$, it’s obvious that the median factor is between 10 and 20.”

“Yes, obvious, sensei.”

“Now, the cube root of 4 is about 1.6, so the sixth root is about the square root of that - about 1 and a quarter, which makes sixth root of the whole thing about 12.5. *But!* The median factor must be odd, so I’d go for 13; the lowest factor would thus be 8 and the highest 18.”

“Fair.”

Dividing the whole thing by 64 gives 60,480, and we’d want six consecutive integers with that product. That’s simple to factorise: it’s $2^6 \times 3^3 \times 5 \times 7$ - and it’s far less than $10^6$, so the smallest factor must be below 10.

Since there’s no 11 in the prime factorisation, the *largest* factor can’t be bigger than 10. Also, any block of six integers must contain exactly two multiples of three, and at most one multiple of 9. To get *three* threes under these restrictions, the factors must include 6 and 9 (as well as 7 and 8, because they’re consecutive).

So, we’ve accounted for $6 \times 7 \times 8 \times 9$, which is $2^4 \times 3^3 \times 7$, leaving $2^2 \times 5$ - or $4 \times 5$ - over.

That leads to $4 \times 5 \times 6 \times 7 \times 8 \times 9 = 60,480$, and $8\times 10 \times 12 \times 14 \times 16 \times 18 = 3,870,720$.

Hope that helps!

- Uncle Colin

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