The numbers $a,b,c$ are the digits of a three digit number which satisfy $49a+7b+c = 286$. What is the three digit number $(100a+10b+c)$?
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Qingchenl
26.01.2006 22:17
Actually, I got a different answer
486/49=5R41
41/7=5R6
6/1=6
So we get a=5, b=5, c=6
5(100)+5(10)+6(1)=556
JesusFreak197
27.01.2006 04:52
Shows me not to do things in my head so often... $286-245=41$, not $31$.
trimo
23.07.2017 16:12
49a + 7b is a multiple of 7. Now, 7 (7a+b) + c = 286 = 7×40+6. So, c = 6 & 7a+b = 40 = 7×5+5. So, a = 5 & b = 5. Hence the required number is 556.