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Problem of the Week
Problem B and Solution
Triple Trouble!
Problem
Ahmed wonders how many 3-digit whole numbers he can create which are not multiples of 10,
and which have no repeating digits.
a) How many such numbers are there with a 1 in the hundreds place?
b) How many such numbers are there with a 2 in the hundreds place?
c) How many such numbers are there in total?
Solution
a) Three-digit numbers with a 1 in the hundreds place can be organized
according to the tens digit as follows.
If the tens digit is 0, they are 102, 103, 104, 105, 106, 107, 108, 109.
If the tens digit is 1, there are none, since the 1 would be repeated.
If the tens digit is 2, they are 123, 124, 125, 126, 127, 128, 129.
If the tens digit is 3, they are 132, 134, 135, 136, 137, 138, 139.
If the tens digit is 4, they are 142, 143, 145, 146, 147, 148, 149.
Etcetera.
Observing the pattern we see there are 8 such numbers with tens digit 0,
since the ones digit can be anything except 1 or 0. But for the possible tens
digits 2, 3, 4, · · · , 9, there are only 7 possible numbers in each group, since
the ones digit can’t be 0 (no multiples of 10), nor 1, nor the specific tens
digit for that group.
Thus there are 8 + (8 × 7) = 8 + 56 = 64 such numbers with a 1 in the
hundreds place.
b) For numbers with a 2 in the hundreds place, the outcome would be the same,
namely one group of 8 (201, 203, 204, 205, 206, 207, 208, 209), and 8 groups
of 7 (e.g., 231, 234, 235, 236, 237, 238, 239), for a total of 64 such numbers.
c) The same pattern would occur for hundreds digits 3, 4, 5, · · · , 9, with the
greatest possible such number being 987. Thus the grand total of the 9
groups of such 3-digit numbers is 64 × 9 = 576 .