Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
Make a set of numbers that use all the digits from 1 to 9, once and
once only. Add them up. The result is divisible by 9. Add each of
the digits in the new number. What is their sum? Now try some other
possibilities for yourself!
Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the number of noughts in
10 000! and 100 000! or even 1 000 000!
Because each of the five options given is a member of the sequence, each is also a multiple of 9. So we require the third factor of the number, that is the factor consisting of several 2s followed by a single 3, to be a multiple of 9 also. This will be true if and only if the sum of its digits is a multiple of 9.
For each number in the sequence, the number of 2s in its third factor equals the number of 0s in that number. The options given have 4, 6,8, 10 or 12 zeros, corresponding with their third factors having digit sums of 11, 15, 19,23 and 27 respectively. Of these, only 27 is a multiple of 9 so the correct answer is 20,000,000,000,007.
This problem is taken from the UKMT Mathematical Challenges.View the archive of all weekly problems grouped by curriculum topic