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This problem trains for: AMC-10, AMC-12, AIME.

A positive integer ends with the digit 3. If this digit is moved and becomes the first digit of the number then the number increases three times. What are the first three digits of the smallest integer with this property?

Write the number in expanded form:

equation

where k represents the cluster of digits that precede the last one.

After moving the last digit in first position:

equation

calculate an expression for k, which must be a positive integer:

equation

and substitute in the second expression to find a Diophantine equation in N alone:

equation

equation

Solve for N:

equation

equation

and take the integers out of the fraction:

equation

Now the condition is for the fraction to be an integer. This will happen if and only if:

equation

or:

equation

Since 29 is prime we can apply Fermat's little theorem to find that:

equation

and therefore:

equation

The fraction has a numerator with 26 digits and the result of dividing it by 29 has 25 digits. As a result, the first two digits of the number we seek are 10 and the third digit is the first digit of the division:

equation

which is 3. Therefore, the required digits are 103.

Second solution

Let us derive the number using multiplication. We know that the last digit is 3. Multiplying by 3 we get 9 and this means that the next digit up is 9. We can apply this repeatedly to find the entire number:

equation