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This problem trains for: SAT-I, AMC-8, GMAT, AMC-10.

Sherlock Holmes found this sequence on a cigar stub that he was examining with his magnifying glass:

Because the brand of the cigar was "Havana-101" he assumed that the street number of the secret meeting place was the 101st term of the sequence. What street number did he go to? (character inspired from Arthur Conan Doyle's "Adventures of Sherlock Holmes")

The 'tens' place value increases by 1 after each group of 3 terms. How many groups of 3 are there in 101 and, if there is not an exact number of groups, how many terms are left over? This question can be answered using integer division (division with remainder):

There are 33 groups of 3 and 2 terms left over.

The 'tens' increase by 33 - one time after each group to 330 and the left over terms are 331 and, finally 332.