Difference between revisions of "2003 AMC 12B Problems/Problem 18"
(New page: suppose n=a*10^4+b*10^3+c*10^2+d*10+e, where a,b,c,d,e are integers between [0,9] then q=a*10^2+b*10+c, r=d*10+e, and (q+r)%11 = n%11, since 10000<n<99999 there are 9090-910+1=8181 n valu...) |
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where a,b,c,d,e are integers between [0,9] | where a,b,c,d,e are integers between [0,9] | ||
then q=a*10^2+b*10+c, r=d*10+e, and | then q=a*10^2+b*10+c, r=d*10+e, and | ||
| + | n%11=(a*100+b*10+c)%11 | ||
(q+r)%11 = n%11, since 10000<n<99999 | (q+r)%11 = n%11, since 10000<n<99999 | ||
there are 9090-910+1=8181 n values that are multiples of 11, | there are 9090-910+1=8181 n values that are multiples of 11, | ||
thus there are 8181 q+r values that are multiples of 11 | thus there are 8181 q+r values that are multiples of 11 | ||
| + | ''''Italic text'''' | ||
Revision as of 20:03, 12 June 2008
suppose n=a*10^4+b*10^3+c*10^2+d*10+e, where a,b,c,d,e are integers between [0,9] then q=a*10^2+b*10+c, r=d*10+e, and n%11=(a*100+b*10+c)%11 (q+r)%11 = n%11, since 10000<n<99999 there are 9090-910+1=8181 n values that are multiples of 11, thus there are 8181 q+r values that are multiples of 11 'Italic text'
