i encountered the following trouble for homework- one "oddie" is a 3 digit number through all 3 number odd. How many "oddies" space divisible by 3?

There room 125 "oddies" and also 300 3-digit numbers divisible through 3, however I still have no idea just how to deal with this. Please provide me a hint or miscellaneous to start me off with.

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Hint: Enumerate the 25 number pairs that an oddie deserve to start with: 11, 13, 15, …, 99, and also group them right into columns follow to the sum of the 2 digits.

Each initial pair can be completed through a third digit in some number of ways; the variety of ways depends only on which pillar the early stage pair is in. Count the variety of pairs in every column, and also multiply by the variety of possible completions for that column. Add up the an outcome for every column.

I believe that this will take only a couple of minutes, due to the fact that there are only 25 initial pairs.

The variety of numbers which have k digits varieties from $10^k-1$ to $10^k-1$.The divisibility v 3is confirm by taking the amount of the digits and also checking that amount for divisibilitythrough 3. $10^k-1$ has a digit sum of 1; $10^k-1+1$ has actually a digit amount of 2;so the first number in the term divisible through 3 is $10^k-1+2$,and the last is (all-9"s) $10^k-1$, and also each 3rd is additionally divisble.

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The variety of numbers through all-odd number (1, 3, 5, 7, or 9) in the interval$<10^k-1,10^k-1>$ is $5^k$. (This is obvious, because we obtainall number in the interval with $k$ increasing by 1 by placingany of the 5 numbers 1, 3, 5,7 or 9 in former of the number of the previousinterval.)

The adhering to enumeration is recursive:To gain the set of numbers for $k$ think about the number in the interval for $k-1$with all-odd digits. Represent by $D_k,d$ the number of odd number in the expression $<10^k-1,10^k-1>$which have a remainder $\equiv d \pmod 3$, so $D_1,0=2$ native 3 and 9, $D_1,1=2$from 1 and also 7, $D_1,2=1$ native 5.and $D_k,0+D_k,1+D_k,2=5^k$.Placing a 1 in prior of a numbers move the collection of number in these remainder classesfrom $D_k,0$ come $D_k+1,1$, native $D_k,1$ come $D_k+1,2$and from $D_k,2$ to $D_k+1,0$. A similar argument holds for placing a 3,a 5, a 7 or a 9 in prior of this numbers:$$D_k+1,0 = 2D_k,0 + D_k,1 + 2D_k,2;$$$$D_k+1,1 = 2D_k,0 + 2D_k,1 + D_k,2;$$$$D_k+1,2 = 1D_k,0 + 2D_k,1 + 2D_k,2.$$Solvingthis direct system the equations offers $D_k,0 = 2,8,41,208,...$ through $3D_k,0=5^k+b(k)$, through $D_k,0 = 6D_k-1,0-6D_k-2,0+5D_k-3,0$and $b(k)=1,-1,-2,-1,1,2,...$ ($k\ge 1$) is one auxiliary simple sequence with period length 6.