BECE Mathematics 2010 Paper

Convert the numbers from base 5 to base 10 so you can know the order of the numbers to find the missingnumbers in base ten and change back to base 5 $32_{\text{five}} = 3 \times 5^1 + 2 \times 5^0$Any number/letter raised to the power 0 is 1$\;32_{\text{five}} = 3 \times 5 + 2 \times 1$$\;32_{\text{five}} = 15 + 2$$\;32_{\text{five}} = 17$$\;33_{\text{five}} = 3 \times 5^1 + 3 \times 5^0$$\;33_{\text{five}} = 3 \times 5 + 3 \times 1$$\;33_{\text{five}} = 15 + 3$$\;33_{\text{five}} = 18$$\;34_{\text{five}} = 3 \times 5^1 + 4 \times 5^0$$\;34_{\text{five}} = 3 \times 5 + 4 \times 1$$\;34_{\text{five}} = 15 + 4$$\;34_{\text{five}} = 19$ The next sequence order would be 20 and 21 so you can convert the 20 and 21 to base 520 base 10 to base 5You list the remainders from the bottom to the top. $20 = 40_{\text{five}}$ 21 base 10 to base 5You list the remainders from the bottom to the top. $21 = 41_{\text{five}}$ The missing numbers are 40 and 41. Alternatively, when the number gets to 35 since there is no 5 in base 5, 5 is 10 in base 5 so you write the 0 andadd the 1 to 3 making 40, then you continue adding 1 so the next sequence would be 41.