Difference between revisions of "2003 AMC 12B Problems/Problem 18"
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suppose n=a*10^4+b*10^3+c*10^2+d*10+e, | 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] | 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)%11, r=d*10+e, and |
n%11=(a*100+b*10+c)%11 | 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 |
| − | |||
Revision as of 20:05, 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)%11, 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
