Enigma 1304 - Some obvious facts

Enigma 1304 - Some obvious facts New Scientist magazine, 28 August 2004, by Susan Denham.

You know that:

* NINETY is divisible by 9,

* TEN is 1 more than a perfect square divisible by 9,

* there are SIX perfect squares between TEN and NINETY.

But in those displayed words each capital letter consistently represents a digit, with different letters used for different digits.

Which number should be SENT?

Ciao,

S p o i l e r

S p a c e

Start by looking at 3 digit numbers that are 1 more than a complete square divisible by 9.

Squares that are divisible by 9 and have 3 digits are 144, 225, 324, 441, 576, 729 and 900. Since TEN must contain 3 different digits, we can reduce TEN to 4 possibilities: 145, 325, 730 and 901. We can then eliminate 730 since NINETY cannot start with 0.

If TEN = 145, then NINETY (must be divisible by 9) can be either: 505413, 515412, 525411, 535410, 535419, 545418, 555417, 565416, 575415, 585414 or 595413 For each value of NINETY, the following values of SIX are given 505413 698 515412 705 525411 712 535410 719 535419 719 545418 726 555417 733 565416 739 575415 746 585414 753 595413 759

Checking if the I in NINETY ever equals I in SIX eliminates all possiblities, thus TEN cannot equal 145

If TEN = 345.... Values for SIX could be.... 505233 692 515232 699 525231 706 535230 713 535239 713 545238 720 555237 727 565236 733 575235 740 585234 747 595233 753

Again, the I in NINETY isn't matched in SIX, thus TEN can't be 345

If TEN = 901 - values for SIX are...

191097 407 181098 395 171099 383 101097 287 111096 303 121095 317 131094 332 - * 141093 345 - * 151092 358 - * 161091 371 171090 383

The * items are where the I in NINETY = I in SIX, Thus T = 9, E = 0 and N = 1. Since all possible values for SIX give S = 3,

S+E+N+T = 3+0+1+9 = 13

S p o i l e r

S p a c e

Start by listing all square numbers that are divisible by 9 and have 3 digits. Adding 1 to each of these gives us the following possible values of TEN: 145, 226, 325, 442, 577, 730 and 901.

From these, we can eliminate all that have a duplicate digit (since all numbers refer to different digits) leaving us with 145, 325, 730 and 901.

Since NINETY cannot start with 0, TEN cannot be 730.

If TEN = 145, then the following values for NINETY are possible (since NINETY must be divisible by 9): 505413, 515412, 525411, 535410, 535419, 545418, 555417, 565416, 575415, 585414, 595413. For each value of NINETY, SIX would be….

505413 698 515412 705 525411 712 535410 719 535419 719 545418 726 555417 733 565416 739 575415 746 585414 753 595413 759

Since the I in each SIX doesn't match the I in NINETY, we deduce that TEN cannot be 145.

If TEN = 325, then NINETY and SIX could be….

505233 692 515232 699 525231 706 535230 713 535239 713 545238 720 555237 727 565236 733 575235 740 585234 747 595233 753

Again, I in NINETY doesn't math I in SIX anywhere, so TEN can't be 325.

If TEN = 901, NINETY and SIX could be…

191097 407 181098 395 171099 383 101097 287 111096 303 121095 317 131094 332 * 141093 345 * 151092 358 * 161091 371 171090 383

The * items show where the I in NINETY matches the I in SIX, thus TEN = 901. In all three possibilities, S = 3,

thus S+E+N+T = 3 + 0 +1 + 9 = 13.

post as spoiler space

9 NINETY so as every accountant knows, 9 (2N+I+E+T+Y)

Three digit perfect squares divisible by 9 = { 144, 225, 324, 441, 576, 729, 900 }. 225, 441, 576 don't work because TEN has all different digits. 729 doesn't work because N <> 0 (usual convention in such problems).

So TEN is one of { 145, 325, 901 }.

And 2N+E+T is respectively { 15, 15, 11 }

And NINETY looks like respectively { 5I541Y, 5I523Y, 1I109Y }.

case TEN = 145: 2N+E+T = 15 so I+Y = 3 or 12. if I+Y = 3 then {I,Y}={0,3} since 1 is taken, then NINETY = 505413 or 535410, then SIX is 698 or 719, don't work if I+Y = 12 then {I,Y} is {3,9} since 4 and 5 are taken, then NINETY = 535419 or 595413, then SIX is 719 or 759, don't work

case TEN = 325: 2N+E+T = 15 so I+Y = 3 or 12. I+Y = 3 would require the use of 2 or 3, doesn't work So I+Y = 12, {I,Y} is {4,8} since 2 3 5 are taken, then NINETY is 545238 or 585234, then SIX is 720 or 747, don't work

case TEN = 901: 2N+E+T = 11 so I+Y = 7 or 16 if I+Y = 7 then {I,Y} = {2,5} or {3,4} if {I,Y} is {2,5} then NINETY = 121095 or 151092, SIX is 317 or 358, second looks good (*) if {I,Y} is {3,4} then NINETY = 131094 or 141093, SIX is 332 or 345, don't work if I+Y = 16 then nothing for I,Y to be

So we just have the one solution (*) :

0123456789 ENYS I XT

check:

NINETY = 151092 = 9 * 16788 TEN = 901 = 1 + 30^30 = 1 + 9*100 There are SIX = 358 = (388-31+1) perfect squares between 901 (between the squares of 30 and 31) and 151092 (between the squares of 388 and 389)

And the required answer: SENT = 3019

Oops - misread problem, SENT = 3019!! PS. Sorry for double post