Remainders
Number Theory
Level
2
A positive number has a remainder of 1 when divided by 4, a remainder of 2 when divided by 5, and a remainder of 3 when divided by 6.
What is the smallest number that has these properties?
The answer is 57.
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1 solution
This number must be in the forms: 4x+1, 5y+2 and 6z+3. A multiple of 5 always ends in 0 or 5 depending on whether the multiplier is even or odd so a number of the form 5n+2 must end in 2 or 7. 6z+3 is always odd as 6z is always even, being of the form 2(3z), and adding 3 to an even number produces as an odd number.
Therefore, the end of the number must be 7 rather than 2. The options (without considering 4x+1) are 27 57, 87... 27+30n.
For this number to be of the form 4x+1, 27+30n-1 must be divisible by 4. This equals 26+30n=2(15n+13). To produce a multiple of 4, 15n+13 must be even and therefore to be the lowest possible is 2(15 * 1 +13)=2 * 28 =56. We then add 1 to get our answer: 57.