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What is the largest seven-digit number divisible by 44 that can be formed by the digits 1, 2, 3, 4, 6, 7, and 8 each used exactly once?

Feb 26, 2020

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What is the largest seven-digit number divisible by 44 that can be formed by the digits 1, 2, 3, 4, 6, 7, and 8 each used exactly once?

Well it must be divisible by 4 which means the last two digits must be divisible by 4

So it could end in 12, 16, 24, 28, 32, 36, 48, 64, 68, 72, 76, 84,

And

It must be divisible by 11

https://math.hmc.edu/funfacts/divisibility-by-eleven/

If that is divisible by 11, so is the original number. So, for instance, 2728 has alternating sum of digits 2 – 7 + 2 – 8 = -11. Since -11 is divisible by 11, so is 2728.

For the divisibility test, 4 of the digits must be pos and 3 negative and they must add to a multiple of 11

8-7+6-4+3-1+2 = 7  that is no good

8-6+7-3+4-1+2 = 11    that works

So maybe it is  8673412

I wonder if it could be 87----- ?

8-7=1

6 4 3 2 1     can these be put together to form 10 or -12   (3 plus and 2 minus)      6+4+3+2+1=16

5C2=10 combinations, but most won't work

-10  +6   no

-9    +7    no

-8   +8     no

-7   +9     no

-7   +9     no

-6  +10     no

-5   +11    no

-5  +11    no

-4  +12    no

-3  + 13   maybe

8-7  +6+4+3  -2-1=11

8-7  +6 -2+4 -1+3 =11

that is no good the last digit must be even    maybe 6 or 4   so the last 2 would be  16  or  24

so maybe

8 -7  +6 -1+3 -2+4 =11        8761324         check       8761324/44 = 199121

8-7 +4 -2 +3 -1+6 =11          8742316         check       8742316/44 = 198689

So the biggest one is  8761324

Feb 26, 2020