A $51 \mathrm{~cm}$ rod is built from $5 \mathrm{~cm}$ rods and $2 \mathrm{~cm}$ rods. All of the $5 \mathrm{~cm}$ rods must come first, and are followed by the $2 \mathrm{~cm}$ rods. For example, the rod could be made from seven $5 \mathrm{~cm}$ rods followed by eight $2 \mathrm{~cm}$ rods. How many ways are there to build the $51 \mathrm{~cm}$ rod?
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
Any number of $2 \mathrm{~cm}$ rods add to give a rod having an even length.

Since we need an odd length, $51 \mathrm{~cm}$, then we must combine an odd length from the $5 \mathrm{~cm}$ rods with the even length from the $2 \mathrm{~cm}$ rods to achieve this.

An odd length using $5 \mathrm{~cm}$ rods can only be obtained by taking an odd number of them. All possible combinations are shown in the table below.

| Number of 5 cm rods | Length in 5 cm rods | Length in 2 cm rods | Number of 2 cm rods |
| :---: | :---: | :---: | :---: |
| 1 | 5 | $51-5=46$ | $46 \div 2=23$ |
| 3 | 15 | $51-15=36$ | $36 \div 2=18$ |
| 5 | 25 | $51-25=26$ | $26 \div 2=13$ |
| 7 | 35 | $51-35=16$ | $16 \div 2=8$ |
| 9 | 45 | $51-45=6$ | $6 \div 2=3$ |

Note that attempting to use 11 (or more) $5 \mathrm{~cm}$ rods gives more than the $51 \mathrm{~cm}$ length required. Thus, there are exactly 5 possible combinations that add to $51 \mathrm{~cm}$ using $5 \mathrm{~cm}$ rods first followed by $2 \mathrm{~cm}$ rods.

ANswer: (A)

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