**1923**- Number 1 Volume 1**1924 Christmas**- Number 2 Volume 2**1924****1927 Christmas**- Number 8 Volume 4**1928 July**- Number 2 Volume 5**1930 January**- Number 1 Volume 7**1932 January**- Volume 9**1936 July**- Number 10 Volume 11**1937 July**- Number 12 Volume 11**1939 January**- Number 3 Volume 12**1939 December****1941 July****1942 July****1943 July****1944 July****1945 July****1946 July****1947 July****1948 July****1949 July****1950 July****1951 July****1952 July****1953 July****1954 July****1955 July**

*May we remind you that these are historical documents. Much of the content was written by pupils, and it naturally displays the languages and attitudes of the time.*

The Chronicle was originally identified by the edition number and volume. It isn't entirely clear what the system was behind the numbering, and perhaps the same was true at the time. However eventually the month and year became more prominent and the number and volume were dropped completely.

The answers to 'Problemania' in the July 1936 edition magazine aren't available (unless anyone has the next edition of the magazine). Graham had a bash at the problems and offers the following possible solutions:

- Scorcher should get the two half-crowns and one shilling. Roadhog should get one florin and one shilling.
Scorcher had given up two gallons and Roadhog one so it should be six shillings for one and three shillings for the other. For the younger readers, a half-crown was a coin worth two shillings and sixpence (i.e. two and a half shillings) and a florin was two shillings.

- The question seems rather ambiguous. £1 6s 0d or £1 10s 4d are possible answers.
If they all share the costs then it is simply £1 6s 0d. If it is assumed that the man on the ground floor doesn't use the stairs and only the others share the cost then £1 10s 4d would be the answer. One could argue that the theory should be that each flight costs £1 10s 4d and that the man on the top floor should pay all the cost of the sixth flight, plus half the next one down, plus one third of the next, and so on. However that doesn't seem to give a round number. It could then be argued that the first flight gets six times as dirty as the top flight and so on in which case the one sixth share solution applies again.

- 400 miles.
It is easy to over complicate this problem by assuming that when the men return to base they immediately set off again with new supplies. Forget that idea and assume they set off when they need to. It then becomes an easy problem. The key to solving it is to start from the end of the journey.

Hefty takes 40 days supplies on his last journey. None of them are left when he gets back. Therefore he must have met Heeman and Hardy when they had both run out of supplies and so must have met them after 10 days (10 days supplies for each of the 1 outbound and 3 inbound journeys gives the 40 days supplies carried).

Similarly Heeman took 40 days supplies on his last journey and again none were left. Therefore Hardy must have just run out of supplies when Heeman met him. That means the journey must have been 20 days (there was 1 outbound journey and 2 inbound to be covered by the 40 days supplies Heeman carried plus the 20 days of that Hefty had for them when they met).

The maximum journey Hardy could have made solo would be 40 days which is 20 outbound and 20 inbound. Is this possible? Indeed it is, as it is consistent with Heeman's turning after 10 days and Hefty after 20. So that means Hardy goes a total of 40 days before he turns around which is 400 miles.

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