<PRE><B>Question 744545</B>
The smallest number divisible by 2,3,4,5, or 6 with a remainder of 1 is  their LCM + 1.&lt;P&gt;
Find the LCM by Step 1:  Finding the prime factors of each number:&lt;P&gt;
2 is a prime, 3 is a prime, 4 factors to 2*2, 5 is prime, 6 factors to 2*3&lt;P&gt;
Step 2:  Look at each prime number represented.  Whichever factor group has the most of that number, use that many of that prime number in a product that will be the LCM:&lt;P&gt;
The most 2&#39;s are in the factorization of 4, so use 2*2.  There are at most one 3 (in the prime factorization of 3 or 6).  There are at most one 5.&lt;P&gt;
The LCM is 2*2*3*5 = 60.  LCM + 1 is 61.  But 61 isn&#39;t divisible by 7.&lt;P&gt;
Now increase by 60 (add 1 to multiples of 60) until you find one divisible by 7.&lt;P&gt;
61 NO&lt;P&gt;
121 NO &lt;P&gt;
181 NO&lt;P&gt;
241 NO &lt;P&gt;
301 Yes.  This is the answer.&lt;P&gt;
The other way to calculate this is using the Chinese Remainder Theorem.  This theorem only holds true for co-prime numbers.  That is numbers that only have 1 as a common factor.&lt;P&gt;
The list of numbers in this problem is 2,3,4,5,6, and 7.  They aren&#39;t co-prime.  But any odd number will have a remainder (mod) 1 when divided by 2, so we can keep that in mind but forget about 2 for the calculation.&lt;P&gt;
We need to find some x such that x = 1 mod 2 (meaning there is a remainder of 1 when x is divided by 2.)  That&#39;s any odd number.&lt;P&gt;
x = 1 mod 3&lt;P&gt;
x = 1 mod 4&lt;P&gt;
x = 1 mod 5&lt;P&gt;
x = 1 mod 6&lt;P&gt;
x = 0 mod 7&lt;P&gt;
Drop 6, because 6 isn&#39;t co-prime with 3.  That leaves 3,4,5,and 7 as the co-prime numbers.  Our answer has to be 1 mod 6, but we won&#39;t use 6 in the calculation.&lt;P&gt;
Step 1:  Multiply the co-prime numbers: 3*4*5*7 = 420 (the co-prime LCM)&lt;P&gt;
Step 2:  Divide 420 by each number.  Divide that result by the number to find the remainder.&lt;P&gt;
420/3 = 140 (call this the factor number) and 140/3 has a remainder of 2 (call this the mod number):  140=2 mod 3.  &lt;P&gt;
420/4 = 105 and 105 = 1 mod 4&lt;P&gt;
420/5 = 84 and 84 = 4 mod 5&lt;P&gt;
420/7 = 60 and 60 = 4 mod 7&lt;P&gt;
Step 3:  Add together (for each number): (desired remainder) * (factor number) * (mod number) =&lt;P&gt;
1*140*2 + 1*105*1 + 1*84*4 + 0*60*4 = 280 + 105 + 336 + 0 = 721&lt;P&gt;
Step 4:  Find the remainder of Step 3 result divided by the co-prime LCM.  &lt;P&gt;
721 / 420 = 1 remainder 301.&lt;P&gt;
This number (301, the remainder) satisfies the condition that it must be odd (remainder 1 when divided by 2.)  It also satisfies the remainder of 1 when divided by 6. 301 is the answer&lt;P&gt;
&lt;P&gt;Hope the solution helped.  Sometimes you need more than a solution.  Contact fcabanski@hotmail.com for online, private tutoring, or personalized problem solving (quick for groups of problems.)

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