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Winner for MMM #11

The prize for MMM #11 goes to Chao Xu. Here is Chao's solution.

A number divisible by 11 if it's alternating sum of it's digits is divisible by 11.
We only have 6 numbers, so we have a+b+c-d-e-f divisible by 11, where a to f are variables for each position. The largest possible number formed by a+b+c-d-e-f is 6+5+4-3-2-1 = 9, the smallest number is -9,
Then we have only one number that is divisible by 11 and between 9 and -9 is 0.
We have
a+b+c = d+e+f
a+b+c = (a+b+c+d+e+f)/2
and a+b+c+d+e+f = 21.
a+b+c = 10.5
It's not a whole number
It shows there is no number can be formed that's divisible by 11.

Jonathan, of the jd2718 blog made these observations (I've not checked his Math):

As a consolation, all 720 of them are multiples of 3.
Half of them (360) are even (multiples of 2)
One in 6 (120) are multiples of 5.
Eight in thirty (192) are multiples of 4.
None are multiples of 9.
Fourteen of 120 (84) are multiples of 8.
Multiples of 7?  New puzzle, good place to stop.

Check out Blinkdagger.com on Monday for another puzzle.

Comments (3) Trackbacks (2)
  1. Yeah! finally won 🙂

  2. I do not even know if my answer arrived 🙁

  3. Not surprising why Jonathon covered all the easy ones and did NOT test seven, but thinking about it, it shouldn’t be too hard… Since every increase of one in the units increases the modulus 1, and every increase in the tens increases the modulus three, I figured out that the rest would be an increase of two in the hundreds, six in the thousands (think minus one) four in the ten-thousands (minus three?) and five in the hundred-thousands (minus two….) so the sequence is 546231… and what a coincidence, if you multiply 5(-2)+4(-3)+6(-1)+2(2) +3(3) + 1(1) …YOU GET -14, which is zero mod 7, the sequence of moduli used in each place makes it a multiple of seven…(546231 = 0 [mod 7]… but wait, 231 by itself is a multiple of seven, since4 23 – 2(1) =21… and 546 is also. [54 – 2(6)=42]..

    It is probably easier on retrospect to just to square each digit in the sequence and discard any sevens (kind of like casting-out sevens) 25+16+36+4 + 9 + 1 is simplified to 4+2+1+4+2+1 and it is obvious the mod is seven…
    To check other numbers, you can just write them out multiplied times the correct modulus for that place value and probabily check it in your head,,but you would have to check each one… so making up a sequence, 234561, we would think 2×5 drops to 3, plus 3(second digit)x4(second modulus) is 15 which drops to modulus of one, now add 4(6) to get 25 and drop to mod 4, then add 2×5 to get 14, and we are at mod 0 [so 234500 is divisible by seven] and we know that 61 is NOT divisible by seven, so we can be assured that 234561 == 61 == 5 mod 7… ok, not EASY, but certainly could be done sans calculator…. fun anyway…thanks

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