Question 1012036: When a two-digit number is divided by 9, the remainder is 4. When it is divided by 6, the remainder is also 4, what is the smallest possible value of the number?
Answer by ValorousDawn(53)   (Show Source):
You can put this solution on YOUR website! 22.
You can first right out a list of numbers that satisfy this property, essentially multiplies of the numbers but plus 4. Your list is
4, 13, 22, 31, 40
4, 10, 16, 22, 28
If you want an analytical way, you'll have to take a twist on the Chinese remainder theorem. Split the the numbers into their prime factors (2*3),(3^2), and form your statements based off of that.
x=4 mod 2
x=4 mod 3
x=4 mod 6
x=4 mod 9
The case of 9 is better than 3, because it covers the three case, as well as more of it. Since 6 and 9 share a common factor, 2 already covers this. We are left with.
x=0 mod 2
x=4 mod 9
Through inspection the smallest case for this solution is 4, but that does not work. You then add the product of the modulo, to ensure congruence. So 4+2*9=4+18=22
Your answer is 22.
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