mathlove 10/03/2017 at 10:09
We see that since \(bcd\) is divisible by , \(d\) must equal either or , but it cannot equal , so \(d=5\) . We notice that since \(abc\) must be even, \(c\) must be either or . However, when \(c=2\), we see that \(e\equiv2\) \(\left(mod3\right)\), which cannot happen because and are already used up; so \(c=4\) . This gives \(e\equiv3\left(mod4\right)\), meaning \(e=3\) . Now, we see that \(b\) could be either or , but is not divisible by , but is. This means that \(c=4\) and \(a=1\) .Selected by MathYouLike
FA FIFA Club World Cup 2018 16/01/2018 at 22:05
We see that since bcd is divisible by [$5$] , d must equal either [$0$] or [$5$] , but it cannot equal [$0$] , so d=5 . We notice that since abc must be even, c must be either [$2$] or [$4$] . However, when c=2, we see that e≡2 (mod3), which cannot happen because [$2$] and [$5$] are already used up; so c=4 . This gives e≡3(mod4), meaning e=3 . Now, we see that b could be either [$1$] or [$2$] , but [$14$] is not divisible by [$4$] , but [$24$] is. This means that c=4 and a=1 .
tranthuydung 10/03/2017 at 12:22
Why did you asked this when you knew how to solve it?
Please help mathlove to solve this problem!