JAMB Mathematics 2018 Paper

y = 23$_{\text{five}}$ + 101$_{\text{three}}$

23$_{\text{five}}$ = $2 \times 5^1 + 3 \times 5^0$

= 13$_{\text{ten}}$

101$_{\text{three}}$ = $1 \times 3^2 + 0 \times 3^1 + 1 \times 3^0$

= 10$_{\text{ten}}$

y$_{\text{ten}}$ = 13$_{\text{ten}}$ + 10$_{\text{ten}}$

= 23$_{\text{ten}}$

= 10111$_{\text{two}}$