How can you tell that 7 x 23 will be at least 3 digits?
437453
because 7*2=14 which means the answer is going to be over 100
359942
If you mean, how do you check to see if the number is valid, here's a simple regular expression:\n\n"^[0-9]{3}&#92;-[0-9]{2}&#92;-[0-9]{4}$"\n\nThe above matches:\n\n000-00-0000\n\nit does NOT match:\n\n0000-00-000\n00-000-000\n000000000\n\nJust use a regular expression validator against that and you should get match / no match results.
930433
The answer is 393.\n\n Start with 333.\nIt's an easy 3-digit number that is divisible by 3\n Tens digit id 3 times more than hundreds.\nSo 393. \n Does it add up to 15? \nYes it does!\n\n:) :) :)
675348
I agree with chapin4ever's answer.\nand the three digit number is the ASCII equivalent of the symbol.
306730
well we know the 3 space must be even, in order to avoid the last space resulting in another 5\n\nhmmm... i'l work on it, provided you're working as well!\n\n..and #3 mustn't be 0, to avoid another 0\n..lol i'd have had it if i had been home...i made a program to do it, and then left before i finished/ran it lol\n\nbut that, is of course, the correct asnwer, 396*45=17820\n\nhaha thanx for the challenge!
133768
Assuming each digit is used no more than once in any given integer, this one would be fairly simple (and perhaps time-consuming) to draw out on paper:\n\n1. All integers have to end with the 3. For single-digit integers you of course have:\n\n3\n\n2. For two-digit integers, you have:\n\n43\n63\n83\n\n3. Three-digit integers:\n\n403\n603\n803\n463\n643\n483\n843\n683\n863\n\n4. Four-digit integers:\n\n4063\n4083\n4603\n4803\n6043\n6083\n6403\n6803\n8043\n8063\n8403\n8603\n8463\n8643\n4683\n4863\n6483\n6843\n\nI count 31.
250820
you divide the 3 digits by the 2 digits!
861916
I can't see the rest of your question, but even numbers can be multiples of three. Like 6 (2x3). They are not mutually exclusive.
1169470
792\n\n1 thru 9 = 9 digits\n10 thru 99 = 180 digits (90 numbers with 2 digits each)\n100 thru 300 = 603 digits (201 numbers with 3 digits each)\nTOTAL = 792
94104
u can do it by trial and error. ok the possible digits for ones are just 1,3,5,7,9 since the statement says it is an odd number. ok then try to add 2 and equate it to 12. if it is = that's is the answer. 7+5= 12. so and 7 is greater than 5 by 2. the answer is 75 degrees.
280927
let the first 3 digits of your number be x = abc and the last four digits be y =defg\n\nwhere a-g are the digits\n\nThe sequence of operations can be rewritten as:\n\n((x*80 + 1) * 250 + 2 * y - 250)/2\n\nIf you distribute the first 250 you get\n\n(20000x + 250 + 2y - 250)/2\n\nCancel the 250's\n\n(20000x+2y)/2\n\nDivide both elements of the numerator by two (distribute the 2)\n\n10000x + y\n\nSince x is a three digit number abc, abc * 10000 = abc0000\n\nSince y is a four digit number defg, 10000x + y = abcdefg.  \nThus we see using algebra that you end up with the same 7 digit number you started out with.
974244
Its just the whole numbers!\n0000\n0001\n0002\n0003\netc.\nup to\n9999\nremembering that if the number has less than 3 digits, you need to add the appropriate number of zeroes at the beginning.\n(i.e,. 23 would be 0023 and 456 would be 0456)\n(why anyone would want to 'prove' this or generate them in excel is beyond me. its a fabulous waste of time, unless you needed to print them out.) But if so,maybe you dont have excel:\n0000 0001 0002 0003 0004 0005 0006 0007 0008 0009 0010 0011 0012 0013 0014 0015 0016 0017 0018 0019 0020 0021 0022 0023 0024 0025 0026 0027 0028 0029 0030 0031 0032 0033 0034 0035 0036 0037 0038 0039 0040 0041 0042 0043 0044 0045 0046 0047 0048 0049 0050 0051 0052 0053 0054 0055 0056 0057 0058 0059 0060 0061 0062 0063 0064 0065 0066 0067 0068 0069 0070 0071 0072 0073 0074 0075 0076 0077 0078 0079 0080 0081 0082 0083 0084 0085 0086 0087 0088 0089 0090 0091 0092 0093 0094 0095 0096 0097 0098 0099 0100 0101 0102 0103 0104 0105 0106 0107 0108 0109 0110 0111 0112 0113 0114 0115 0116 0117 0118 0119 0120 0121 0122 0123 0124 0125 0126 0127 0128 0129 0130 0131 0132 0133 0134 0135 0136 0137 0138 0139 0140 0141 0142 0143 0144 0145 0146 0147 0148 0149 0150 0151 0152 0153 0154 0155 0156 0157 0158 0159 0160 0161 0162 0163 0164 0165 0166 0167 0168 0169 0170 0171 0172 0173 0174 0175 0176 0177 0178 0179 0180 0181 0182 0183 0184 0185 0186 0187 0188 0189 0190 0191 0192 0193 0194 0195 0196 0197 0198 0199 0200 0201 0202 0203 0204 0205 0206 0207 0208 0209 0210 0211 0212 0213 0214 0215 0216 0217 0218 0219 0220 0221 0222 0223 0224 0225 0226 0227 0228 0229 0230 0231 0232 0233 0234 0235 0236 0237 0238 0239 0240 0241 0242 0243 0244 0245 0246 0247 0248 0249 0250 0251 0252 0253 0254 0255 0256 0257 0258 0259 0260 0261 0262 0263 0264 0265 0266 0267 0268 0269 0270 0271 0272 0273 0274 0275 0276 0277 0278 0279 0280 0281 0282 0283 0284 0285 0286 0287 0288 0289 0290 0291 0292 0293 0294 0295 0296 0297 0298 0299 0300 0301 0302 0303 0304 0305 0306 0307 0308 0309 0310 0311 0312 0313 0314 0315 0316 0317 0318 0319 0320 0321 0322 0323 0324 0325 0326 0327 0328 0329 0330 0331 0332 0333 0334 0335 0336 0337 0338 0339 0340 0341 0342 0343 0344 0345 0346 0347 0348 0349 0350 0351 0352 0353 0354 0355 0356 0357 0358 0359 0360 0361 0362 0363 0364 0365 0366 0367 0368 0369 0370 0371 0372 0373 0374 0375 0376 0377 0378 0379 0380 0381 0382 0383 0384 0385 0386 0387 0388 0389 0390 0391 0392 0393 0394 0395 0396 0397 0398 0399 0400 0401 0402 0403 0404 0405 0406 0407 0408 0409 0410 0411 0412 0413 0414 0415 0416 0417 0418 0419 0420 0421 0422 0423 0424 0425 0426 0427 0428 0429 0430 0431 0432 0433 0434 0435 0436 0437 0438 0439 0440 0441 0442 0443 0444 0445 0446 0447 0448 0449 0450 0451 0452 0453 0454 0455 0456 0457 0458 0459 0460 0461 0462 0463 0464 0465 0466 0467 0468 0469 0470 0471 0472 0473 0474 0475 0476 0477 0478 0479 0480 0481 0482 0483 0484 0485 0486 0487 0488 0489 0490 0491 0492 0493 0494 0495 0496 0497 0498 0499 0500 0501 0502 0503 0504 0505 0506 0507 0508 0509 0510 0511 0512 0513 0514 0515 0516 0517 0518 0519 0520 0521 0522 0523 0524 0525 0526 0527 0528 0529 0530 0531 0532 0533 0534 0535 0536 0537 0538 0539 0540 0541 0542 0543 0544 0545 0546 0547 0548 0549 0550 0551 0552 0553 0554 0555 0556 0557 0558 0559 0560 0561 0562 0563 0564 0565 0566 0567 0568 0569 0570 0571 0572 0573 0574 0575 0576 0577 0578 0579 0580 0581 0582 0583 0584 0585 0586 0587 0588 0589 0590 0591 0592 0593 0594 0595 0596 0597 0598 0599 0600 0601 0602 0603 0604 0605 0606 0607 0608 0609 0610 0611 0612 0613 0614 0615 0616 0617 0618 0619 0620 0621 0622 0623 0624 0625 0626 0627 0628 0629 0630 0631 0632 0633 0634 0635 0636 0637 0638 0639 0640 0641 0642 0643 0644 0645 0646 0647 0648 0649 0650 0651 0652 0653 0654 0655 0656 0657 0658 0659 0660 0661 0662 0663 0664 0665 0666 0667 0668 0669 0670 0671 0672 0673 0674 0675 0676 0677 0678 0679 0680 0681 0682 0683 0684 0685 0686 0687 0688 0689 0690 0691 0692 0693 0694 0695 0696 0697 0698 0699 0700 0701 0702 0703 0704 0705 0706 0707 0708 0709 0710 0711 0712
678254
187\n3,7,18,42,56 "33"
917451
well Dear Shwanta \n\nif we get  ' x '  as a number we are looking for so we have ;\n\n7x - 4 \n\nGood Luck darling.
1018404
The number is too small for a real "pattern" to emerge.\nThe digits could be the first 10 values of some sequence given by some ( many ) recurrence relation, could be ...\n\nIn order for a proper pattern to emerge, it should be much larger...
544625
3 and 23/32nds\nor simply\n119/32
409160
This is possible only if at least half of those 256 values will not be used at all, and that you know which ones won't be used beforehand. If that's the case, you can map the remaining values that will be used with 7 bits.\n\nSee Shannon's entropy equation...