The multiplication $$abc\times de=7632$$ uses each of the digits $$1$$ to $$9$$ exactly once. What is the value of $$b$$? 
Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization
Note first that $$7632 =2\times2\times2\times2\times3\times3\times53$$. Therefore either the two-digit number $$de = 53$$ or the three-digit number $$abc$$ is a multiple of $$53$$. Since the multiplication uses each of the digits $$1$$ to $$9$$ once and $$7632$$ contains a $$3$$, the option $$de= 53$$ is not allowable. Hence we need to find a three-digit multiple of $$53$$ that does not share any digits with $$7632$$ and divides into $$7632$$ leaving an answer that also does not share any digits with $$7632$$. We can reject $$2 \times 53 = 106$$ since it contains a $$6$$ but $$3 \times 53 = 159$$ is a possibility. The value of $$7632\div159$$ is $$2\times2\times2\times2\times3 = 48$$ which does not have any digits in common with $$7632$$ nor with $$159$$. We can also check that no other multiple of $$53$$ will work. Therefore the required multiplication is $$159 \times 48 = 7632$$ and hence the value of $$b$$ is $$5$$. 