Friday puzzle -- an ENIGMA

starting

Dan Donaldson

GMA must be one of:

125

216

343 (cannot be because not all unique)

512

729

NI must be one of:

16

25

36

49

64

81

E cannot be 5 because none of the gma possibilities end in 0 and 125 would be a dup.

Next? someone fill in the next blank

Re: starting

Larry Barrett

Something I don't see yet:

The smallest GMA can be is 123; 123 cubed is 1860867 (a 7 digit number)

How can a 7 digit number divide into a 6 digit number evenly with no remainder?

Re: starting

Dan Donaldson

Not gma cubed, gma IS the cube. The same with the square. The word "enigma" is composed of a number (e) followed by a two digit square (ni) followed by a three digit cube (gma)

Clarification

Alex Y

Correct interpretation, and good start.

Re: starting

Larry Barrett

Thanks for clarification.

Since all digits are unique, there are only a few combinations of NI and GMA:

16729

36125, 36512, 36729

49125, 49216, 49512

64125, 64512, 64729

81729

Now add E to each, keeping in mind that if the last digit (the A) is odd, then E can not be even.

There are 29 possible combinations. I do not see any clever way to narrow it down further. A spreadsheet is your friend.

a bit more

Dan Donaldson

ni must be 36, 49, or 64 (why is left to the student ;-))

Come on, someone dig in also

Re: starting

Dan Donaldson

729 cannot be it either

Re: a bit more

Dan Donaldson

Another couple:

e cannot be 1 or 2 and gma cannot be 729

Re: starting

Dan Donaldson

There are a few ways yet to simplify

and the answer is...

Rob Scrimgeour

34912

typo? 

Re: and the answer is...

Larry Barrett

Of the 20 or so possibilites left after Dan's work, the one that looked like it just had to be the one was 864512, and it almost works.

Rob has the answer but just a typo 

Agree, but did not look like a good prospect   

Good job, guys!

Alex Y

Rob's answer, with the addition of the missing digit, "5", is correct. 349125.

Dan's whittling down of the possibilities was correct. It could be carried just a little further, but then you have to resort to trial and error.

GMA = 125, 216, or 512, and

NI = 36, 49, or 64

Note that GMA divides ENIGMA if and only if it divides ENI000.

The lowest multiple of 512 that is also a multiple of 1000 is 128000. Other possibilities are 256000, 384000, 512000, 640000, 768000, and 896000. In none of these are the second and third digits a perfect square (and most are disqualified for other reasons as well).

Similarly, the lowest multiple of 216 that is a multiple of 1000 is 027000. Checking all the multiples of that, only 864000 has 2nd and third digits equal to one of our target squares, but doesn't work because of the 6.

So GMA must be 125, and since 8*125 = 1,000, ANY choice for ENI will satisfy the condition that ENI125 is divisible by 125.

So now, we have to test E = 3,7,9 and NI = 36, 49, 64. Eliminate the two possibilities with shared digits and you are left with only seven possible values for ENIGMA to test.