<![CDATA[Module 5 Day 12 Challenge Part 5]]>https://forum.poshenloh.com/category/538RSS for NodeSun, 26 Sep 2021 02:42:28 GMTFri, 11 Dec 2020 21:19:25 GMT60<![CDATA[New way to do?]]>Actually I think there is a new way to do this; just let == be congruence, and p be the number of pennies. We know that

p == 3(mod 4)
p == 16(mod 25)

Subtracting 12 on 3(mod 4) gives us p == -9(mod 4), and subtracting 25 on 16(mod 25) gives us p == -9(mod 25), so that means that p == -9(mod 100) which is the same as p == 91(mod 100).

]]>https://forum.poshenloh.com/topic/664/new-way-to-dohttps://forum.poshenloh.com/topic/664/new-way-to-doFri, 11 Dec 2020 21:19:25 GMT