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?

Guest Feb 26, 2020

#1**+1 **

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**

Melody Feb 26, 2020