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Consider numbers whose fist digit is 1. Looking at each possible value for the second digit we find 9 such numbers.

110, 121, 132, 143, 154, 165, 176, 187, 198.

Similarly there are 8 numbers starting with 2; 7 numbers starting with 3...

Lastly there is only one number starting with 9: 990.

Hence the total is 9+8+7+6+5+4+3+2+1 = 45.

*This problem is taken from the UKMT Mathematical Challenges.*

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