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For a positive integer n, E(n) is the summation of even digi

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For a positive integer n, E(n) is the summation of even digi

by Max@Math Revolution » Tue Oct 01, 2019 11:46 pm
[GMAT math practice question]

For a positive integer n, E(n) is the summation of even digit numbers of n. For example, E(148)=4+8=12 and E(3821)=8+2=10. What is the value of E(1)+E(2)+E(3)+...+E(2006)?

A. 12026
B. 12024
C. 12022
D. 12020
E. 12018
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by Max@Math Revolution » Thu Oct 03, 2019 11:23 pm
=>

When we consider numbers between 0 = 000 and 999, inclusive, the number of 3 digit numbers is 1000 if 0 is allowed for the largest digit.
Each of 0, 1, ..., 9 is used 3000/10 = 300 times.
Therefore, E(1) + E(2) + ... + E(999) = (2 + 4 + 6 + 8)*300 = 6000.
We have E(1001)+E(1002) + ... + E(1999) = 6000.
Then E(2000) + E(2001) + ... + E(2006)
= 2*7 + (2+4+6) = 26.
Hence, we have E(1) + E(2) + ... + E(2006) = 6000 + 6000 + 26 = 12026.

Therefore, A is the answer.
Answer: A
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