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Hiya there!
This is my solution to the question "Lesser Digits" in "The April Six". I sincerely hope that it is good enough to be published on the web page. By the way, that web page is very interesting.
Numbers that do not contain any of the digits 7, 8 and 9
Number that do not contain any of the digits 7, 8 or 9 | |
1 to 10 | 7 (1, 2, 3, 4, 5, 6, 10) |
11 to 20 | 7 |
21 to 30 | 7 |
31 to 40 | 7 |
41 to 50 | 7 |
51 to 60 | 7 |
61 to 70 | 6 |
71 to 80 | 0 |
81 to 90 | 0 |
91 to 100 | 1 |
Subtotal: (1 to 100) | 7 x 7 = 49 |
101 to 200 | 49 |
201 to 300 | 49 |
301 to 400 | 49 |
401 to 500 | 49 |
501 to 600 | 49 |
601 to 700 | 48 |
701 to 800 | 0 |
801 to 900 | 0 |
901 to 1000 | 1 |
Subtotal: (1 to 1000) | 49 x 7 =343 |
1001 to 2000 | 343 |
2001 to 3000 | 343 |
3001 to 4000 | 343 |
Total: (1 to 4000) |
343 x 4 = 1372
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Nisha Doshi of the Mount School, York, James Page, Jamie Arnall and Jack Adcock of Hethersett High School, and Astee Goh and Teo Yea Tian Raffles Girls' Primary School, Singapore sent good solutions to this and Ian Green , age 13 of Coopers Company and Coburn School, Upminster, Essex came very close. There were many incorrect solutions sent in for this one.
Choose two digits and arrange them to make two double-digit numbers. Now add your double-digit numbers. Now add your single digit numbers. Divide your double-digit answer by your single-digit answer. Try lots of examples. What happens? Can you explain it?
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.