Posted 19 Sep 2007 · Report post I just found a number with an interesting property: When I divide it by 2, the remainder is 1. When I divide it by 3, the remainder is 2. When I divide it by 4, the remainder is 3. When I divide it by 5, the remainder is 4. When I divide it by 6, the remainder is 5. When I divide it by 7, the remainder is 6. When I divide it by 8, the remainder is 7. When I divide it by 9, the remainder is 8. When I divide it by 10, the remainder is 9. It's not a small number, but it's not really big, either. When I looked for a smaller number with this property I couldn't find one. Can you find it? 0 Share this post Link to post Share on other sites

0 Posted 29 Jul 2010 · Report post Great riddle, very close to the Chinese remainder theorem, Here's my brute force code for original riddle: int remainder = 1; int answer = 0; string result = string.Empty; for (int number = 1; number <= 10000; number++) { for (int divisor = 2; divisor <= 10; divisor++) { if (number % divisor == remainder) { if (remainder == 9) { answer = number; result += string.Format("{0} ", answer); break; } remainder++; } } remainder = 1; } Code for the second riddle: string result2 = string.Empty; for (int number = 11; number <= 100000; number++) { for (int divisor = 2; divisor <= 10; divisor++) { if (number % divisor == 1) { if (divisor == 10) { ++divisor; if (number % divisor == 0) { result2 += string.Format("{0} ", number); break; } } } else break; } } 0 Share this post Link to post Share on other sites

0 Posted 7 Jul 2011 (edited) · Report post There is a perfect solution. Study the question inteligently. Assume that answer is X. If you add 1 to the answer X = Say X+1 this figur will be the exact divisible by each number from 1 to 10. Therefore you need to find a LCM Lowest common Multiple of numbers 1 to 10. That is 2520. Therefore Answer = 2519 Pramod Kasarekar Edited 7 Jul 2011 by kasarekar 0 Share this post Link to post Share on other sites

0 Posted 8 Oct 2011 · Report post Ohw, I thought it was much easier. Is this possible? x² - x Because: 2² - 2 = 4 - 2 = 2 --> 2/2 = 1 3² - 3 = 9 - 3 = 6 --> 6/3 = 2 4² - 4 = 16 - 4 = 12 --> 12/4 = 3 5² - 5 = 25 - 5 = 20 --> 20/5 = 4 6² - 6 = 36 - 6 = 30 --> 30/6 = 5 7² - 7 = 49 - 7 = 42 --> 42/7 = 6 8² - 8 = 64 - 8 = 56 --> 56/8 = 7 9² - 9 = 81 - 9 = 72 --> 72/9 = 8 10² - 10 = 100 - 10 = 90 --> 90/10 = 9 You see? Or wasn't that the question? 0 Share this post Link to post Share on other sites

0 Posted 8 Oct 2011 (edited) · Report post no i think u have misunderstood the question... u answered it thinking that remainder means the quotient (the actual answer) we are asking for the remainder for instance lets say u have 5/2 the answer would = to 2 and the remainder would be 1 (speaking in long division way and without continuing to turn it into a decimal) so the question was what is the smallest number you can get that when u divide it by 2 your remainder is 1 and when u divide by 3 your remainder is 2 and so on and so on... the answer is 2519 no one could find anyth smaller.(but clever job on your answer were that the objective )hope that clears everything up Edited 8 Oct 2011 by guppy 0 Share this post Link to post Share on other sites

0 Posted 28 Oct 2011 (edited) · Report post i have the best method since all the remainders differ by 1 from the divisors so... let us assume the number to by x so adding one to x ie x+1 results another number of special property. ie it is divisible by all other numbers 2,3,4,5,6,7,8,9,10. to find the least number taking their LCM(lowest common multiple) results you 2520 yes x+1=2520 so x is equal to 2519 i can assure you no other method is better Edited 28 Oct 2011 by harikrishnan 0 Share this post Link to post Share on other sites

0 Posted 28 Oct 2011 · Report post i have the best method since all the remainders differ by 1 from the divisors so... let us assume the number to by x so adding one to x ie x+1 results another number of special property. ie it is divisible by all other numbers 2,3,4,5,6,7,8,9,10. to find the least number taking their LCM(lowest common multiple) results you 2520 yes x+1=2520 so x is equal to 2519 i can assure you no other method is better 0 Share this post Link to post Share on other sites

0 Posted 15 Nov 2011 · Report post Love it! Great problem 0 Share this post Link to post Share on other sites

0 Posted 15 Dec 2011 · Report post 2519 Lets say the number is X. SInce, it is divisible by the numbers 2,3,...,10, if we add 1 to it. SO, the number (X+1) is divisible by 2,3,4,...,10. Now, we need to find the smallest number divisible by 2,3,4,...,10. It should be the LCM(2,3,4,...,10) = 2520 So, X+1 = 2520 hence, X = 2519 0 Share this post Link to post Share on other sites

0 Posted 8 Jan 2012 · Report post the smallest we can find is 2519 0 Share this post Link to post Share on other sites

0 Posted 12 Jan 2012 · Report post surely the correct answer on my account would be ... NO 0 Share this post Link to post Share on other sites

0 Posted 25 May 2012 · Report post i think 839 is a smaller solution to the problem 0 Share this post Link to post Share on other sites

0 Posted 5 Jun 2012 · Report post if you check out each number out each number's posiblities and put it into a formula, you will obtain: for the number 2: (x is any natural number, and it's diferent for every equation) 1+ 2x Number 3 29+30x Number 4 19+ 20x Number 5 9+ 10x Number 6 29 + 30x Number 7 69 + 70x Number 8 39 + 40x Number 9 89+90x Number 10 9 + 10X So we need a number that is equivalent to all those equations. we have: 1+2X 29+30X 19+20X 9+10X 69+70X 39+40x 89+90X So, we go and analyse the equations who give bigger products for the smallest number of X. those are: 89+90x and 69 + 70x. Replacing one of the x for another letter (y) we get: 89+90y = 69 + 70x <=> 20 = 70x - 90y With that, we can now replace the "x" and the "y". for that formula to work, after some calculations, I notice that the x must go for 8 + 9z and the y must go 6+7z So, starting for the 1st, if z = 0, x=8 and y=6. we replace them in the equation 89 + 90 x 6 = 629. However, this number doesnt exist on the equation 39+40x, so its invalid. You keep going, z=1; x= 17 and y = 13, the final =1259, also doesnt exist on the equation 39+40x, so its invalid. z=2, the final is 1889, also invalid. Then, finally, z=43 y = 6+(7x3) =27........ 89 + 90x27 = 2519; which is valid for all the equations. 0 Share this post Link to post Share on other sites

0 Posted 6 Jun 2012 · Report post I actually believe I've found a smaller number, if negative numbers are allowed. -1 ! -1 = -1 * 10 + 9 -1 = -1 * 9 + 8 -1 = -1 * 8 + 7 -1 = -1 * 7 + 6 -1 = -1 * 6 + 5 -1 = -1 * 5 + 4 -1 = -1 * 4 + 3 -1 = -1 * 3 + 2 -1 = -1 * 2 + 1 0 Share this post Link to post Share on other sites

0 Posted 6 Nov 2012 (edited) · Report post For those of those familiar with SQL, these types of problems are easily solved by perfoming a query on a Numbers (or Tally) table. For example I have run the following selection on a numbers table (Numbers) containing 100,000,000 Rows (1,2,3....100,000,000) for the first 17 rows of the sequence (2 remainer 1 up to 18 remaining 17) The answer is 12,252,239 SELECT Min(Number) FroM Numbers WHERE Number%2 = 1 AND Number%3 = 2 AND Number%4 = 3 AND Number%5 = 4 AND Number%6 = 5 AND Number%7 = 6 AND Number%8 = 7 AND Number%9 = 8 AND Number%10 =9 AND Number%11 = 10 AND Number%12 = 11 AND Number%13 = 12 AND Number%14 = 13 AND Number%15 = 14 AND Number%16 = 15 AND Number%17 = 16 AND Number%18 = 17 (To get the solutions to next rows in the sequence: 19 remainder 18, 20 remainder 19...) requires a bigger Number table Edited 6 Nov 2012 by brifri238 0 Share this post Link to post Share on other sites

0 Posted 6 Nov 2012 (edited) · Report post FYI: If you extend the sequence from 2 remainder 1 .... to bigger numbers possible solutions are (I am not absoutely sure they are the smallest number that works for each but these work!) 20 remainder 19 ......................232,792,559 22 remainder 21.......................232,792,559 23 remainder 22....................5,354,228,879 24 remainder 23....................5,354,228,879 30 remainder 29.............2,329,089,562,799 31 remainder 30...........72,201,776,446,799 36 remainder 35.........144,403,552,893,599 40 remainder 39......5,342,931,457,063,199 Edited 6 Nov 2012 by brifri238 0 Share this post Link to post Share on other sites

0 Posted 18 Jan 2014 · Report post i think 839 is a smaller solution to the problem 839 / 9 has a remainder of 2, when it's supposed to be 8, so 839 isn't a solution. 0 Share this post Link to post Share on other sites

Posted · Report post

I just found a number with an interesting property:

When I divide it by 2, the remainder is 1.

When I divide it by 3, the remainder is 2.

When I divide it by 4, the remainder is 3.

When I divide it by 5, the remainder is 4.

When I divide it by 6, the remainder is 5.

When I divide it by 7, the remainder is 6.

When I divide it by 8, the remainder is 7.

When I divide it by 9, the remainder is 8.

When I divide it by 10, the remainder is 9.

It's not a small number, but it's not really big, either.

When I looked for a smaller number with this property I couldn't find one.

Can you find it?

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