Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, 134468.

Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, 66420.

We shall call a positive integer that is neither increasing nor decreasing a "bouncy" number; for example, 155349.

As n increases, the proportion of bouncy numbers below n increases such that there are only 12951 numbers below one-million that are not bouncy and only 277032 non-bouncy numbers below 10¹⁰.

How many numbers below a googol (10¹⁰⁰) are not bouncy?

Analysis

Whenever a problem asks for the number of elements in a set, as does this one, then the solution in typically a counting problem solved with combinometrics.

Our solution uses the general formula that counts the monotonically increasing/decreasing digits ignoring leading zeros.

Solution

Runs < 1 second in Python.

from Euler import binomial
 
n=100
print binomial(n+10,10) + binomial(n+9,9) - 10*n - 2

Comments

More information on the Euler module can be found on the tools page.

Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, 134468.

Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, 66420.

We shall call a positive integer that is neither increasing nor decreasing a "bouncy" number; for example, 155349.

Clearly there cannot be any bouncy numbers below one-hundred, but just over half of the numbers below one-thousand (525) are bouncy. In fact, the least number for which the proportion of bouncy numbers first reaches 50% is 538.

Surprisingly, bouncy numbers become more and more common and by the time we reach 21780 the proportion of bouncy numbers is equal to 90%.

Find the least number for which the proportion of bouncy numbers is exactly 99%.

Analysis

It was a surprise how long this program took to run. The is_bouncy() routine seemed to slow things down. We begin at the 90% point specified in the problem description and wait for the ratio to hit 99%.

Solution

Runs < 7 seconds in Python.

def is_bouncy(n):
  inc, dec, s = False, False, str(n)
  for i in range(len(s)-1):
    if s[i+1] > s[i]: inc = True
    elif s[i+1] < s[i]: dec = True
    if inc and dec: return True
  return False
 
n, p = 21780, 0.90
b = n * p
while p != 0.99:
  n += 1
  if is_bouncy(n): b += 1
  p = float(b)/n
 
print "Answer to PE112 = ", n

Comments

The float() function was required to convert to floating point.