I am a three digit number. \nMy tens digit is five more than my ones digit. \nMy hundreds digit is eight less than my tens digit. \nWhat number am I?
131913
I guess you are 194...not sure
190549
Oh man, that was way harder than I thought. This is what I got:\n\nONE NINETY ONE ZERO NINETEEN EIGHT SIXTY TWO ZERO SIXTY ONE\n\nI have no idea what it means or was it even the one you were looking for, because there might be more "correct" solutions.
246678
Ok, because the numbers are from 00-99, you know there are a total of 100 possible numbers.\n\nA. By chance, the probability of getting each digit correct is 1/100. Because there are three digits, you must perfrom (1/100) * (1/100) * (1/100), which is 1/1000000. You don't have to worry about less possible numbers because the same number can be used more than once.\n\nB. If you already know the first digit, you only need to know the other two digits. Thus, it's (1/100) * (1/100) which is 1/10000.
19666
Since you know that the next to the last digit is 4 or 5 and the last digit is odd (1,3,5,7,9), so there are 10 possibilities for the last two digits: 41, 43, 45, 47, 49, 51, 53, 55, 57, 59.\n\nYou 1 chance in 10 or 10% of getting it right.
218951
The first three digits are actually multiplied by (80 x 250) = 20000. And you are adding 250 which you have subtracted latter.\n\nIn effect two times the first three digits followed by four zeros. You are adding two times the four digits so you get two times the original number. Dividing it by 2 u get the original value.\n\nYou can do this with 40 and 500, 20 and 1000, 10 and 2000 etc.
340337
You're buying digits.\nEach digit cost 5 cents.\n1 is one digit.\n21 consists of 2 digits.\n347 consists of 3 digits.
378090
The answer is 119\n\nTake 2 at a time end up with 1 (59 times=118 one left)\nTake 3 at a time end up with 2 (39 times=117 two left)\nTake 4 at a time end up with 3 (29times =116 three left)\nTake 5 at a time end up with 4 (23times=115 four left) \nTake 6 at a time end up with 5 (19times=114 five left)\nTake 7 at a time end up with 0 (17times=119 o left)\n\nI am dead sure this time = ]\n\nThe trick is look at 5 at a time. there is 4 left this means the last digit must be 9. So look at all possible answer of 7s that ends with 9 and work it out from that!!!!!!!!!!
543372
♥♥♥ you mean like those?  just hit Alt 3, you have to use the number pad though.\nhave fun, ♥
202066
you know... I don't think that's very fair... You asked a question and then answered it. So now I must think of something else that is special about number 9...\n\nYou can multiply by 9 on your fingers and always get the right answer. For example, say you want to multiply 4 and 9. Holding all ten fingers up, put down the fourth finger from the left. Tada! 3 fingers are on the left of the finger you put down, and 6 fingers are on the right. the answer? 36! Just put down the finger corresponding to the number you are multiplying by 9.
365955
The original number is 82\n\nIf the tens digit is 4 times the unit digit then the only numbers that satisfy this condition are 82 and 41.\n\nTurning them round gives 28 and 14.\n\nOnly 28 satisfies the condition that the new number is 54 less than the original number.
151832
8\ntake away o from top and you're left with o on bottom - half - nothing\nor i guess 10 could work, too (or any double digit ending in 0) take away the first number and you're left with 0
498549
If John had a 1 then Sheila's number would need to be 2\n\nIf John had a 5 then Sheila's number would need to be 4\n\nSo when John says "I don't know your number", he has just effectively told Sheila he does not have 1 or 5\n\nThe same logic dictates Sheila cannot have 1 or 5 and also lets you say:-\n\nIf Shiela had a 2, she would know John had a 3 (Since she knows he does not have a 1)\n\nIf Shiela had a 4, she would know John had a 3 (Since she knows he does not have a 5)\n\nSo when Sheila says "I don't know your number" she is effectively saying, "I don't have 2 or 4"\n\nSo Sheila's number must be 3
67815
5000                  AHHHHHHHHHHHHHH!\nIt is 4100!\nGreat Question!
337544
No dash. Only use a hyphen between the tens and units of numbers between 21 and 99. Easy! :)
608574
should be a three digit number at the end of the long one
365944
I feel your pain. This is busy work, to say the least. Renaming in this context refers to whether or not the various place-holders are changed (or "renamed") when you add the numbers together. In this example, when you add the digits in the "ones" column together, you get 13, so you "rename" that as 3 ones and 1 ten and add the 1 ten to the tens column. \n\nTo me, fussing with this concept, rather than just concentrating on the HOW, is busy work and just makes it more confusing. (That's a vent to our public education system -- not to you.)\n\nAnyway, I hope that helped shed some light. Please feel free to email me additional items and I'll help work them out and explain whether or not "renaming" was involved.
473629
Are we counting the number of times the digit zero appears?  For example, would 1,000,000 result in 6 zeroes?  If so, you can count the zeroes as follows:\n\nFirst, count the numbers of the form nnnnnn0... 10, 20, 30, ..., 100, 110, 120, ..., 999990, 1000000.  Every 10th number will have this form, so that is 100,000 zeroes in the units place (don't count zero, but do count 1 million).\n\n1,000,000 * 1/10 = 100,000\n100,000 - 1 + 1 = 100,000\n\nNext, count nnnnn0n... 100, 101, 102, 103...  109, 200, 201, 202, ... 209, .... 999900, 999901, 999902, ..., 999,999, 1,000,000.  This would be 10 out of every hundred numbers, so that is 99,991 zeroes in the tens place.  (You don't count 0 to 9), but you do count 1 million.\n\n1,000,000 * 10/100 = 100,000\n100,000 - 10 + 1 = 99,991\n\nNext, count nnnn0nn.  This would be 100 out of every thousand numbers, but not counting 0 to 99, but counting 1 million.  So that is 99,901.\n\n1,000,000 * 100/1000 = 100,000\n100,000 - 100 + 1 = 99,901\n\nNext, count nnn0nnn.  This would be 1000 out of every ten thousand numbers, but not counting 0 to 999, but counting 1 million.  So that is 99,001.\n\n1,000,000 * 1000/10000 = 100,000\n100,000 - 1,000 + 1 = 99,001\n\nNext, count nn0nnnn.  This would be 10000 out of every hundred thousand numbers, but not counting 0 to 9999, but counting 1 million.  So that is 90,001.\n\n1,000,000 * 10000/100000 = 100,000\n100,000 - 10,000 + 1 = 90,001\n\nFinally, count n0nnnnn.  Without a leading digit, none of the numbers except 1 million would have this pattern:\n\n1,000,000 * 100000/1000000 = 100,000\n100,000 - 100,000 + 1 = 1\n\n100,000 + 99,991 + 99,901 + 99,001 + 90,001 + 1 = 488,895\nSo the answer is that the digit zero appears 488,895 times in the numbers from 1 to 1 million, inclusive.