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#69 Backward drops

Set by Katie Steckles and Peter Rowlett

Imagine a number sequence, created from the normal sequence of whole numbers, where each term is that number written backwards, e.g. the 12th term is 21 (and any leading zeroes are ignored). This goes 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 11, 21, 31, 41, and so on. Each of these numbers is bigger than the last, except that after 9 we get a drop to the next number, which is 1.

When is the next drop?

What is the largest drop in the first 100 terms of the sequence?

What are all the different drop sizes in the first 1000 terms of the sequence?

Solution next week

#68 Another 2025 puzzle

Solution

1936 is 44², and the divisors of 44 (apart from 44 itself) are 1, 2, 4, 11 and 22. The product of these numbers is indeed 1936.

This only works with square roots that have six divisors: 1, the number itself (which we are excluding), and four others. The factors other than 1 can be paired up so that each pair multiplies to give the number itself (e.g. 45 = 5 脳 9 = 3 脳 15), so if there are four such factors and we multiply them all together, we will get the square of the original number. 144 is the smallest square with this property.