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Just Repeat

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A number of pleasing solutions were received.

Here is one offered by Andrei of School 205 Bucharest:

I observed that number 234234 is:

234234 = 234x1000 + 234 = 234(1000 + 1) = 1001 x 234


But 1001 is always divisible by 91:

1001 = 7 x 11 x 13 = 77 x 13 = 91 x 11 = 143 x 7

The same is true for 973973:

973973 = 1001 x 973

In fact, if you repeat any 3-digit number in that manner, you obtain a number divisible by 1001, so by 7, 11, 13, 77, 91 and 143:

Now, I look through other patters:

- if you repeat a 2-digit number twice,you obtain a number divisible by 101:

E.g. 65 ...6565

6565/101 = 65

- if you repeat a 2-digit numberthree times you obtain a number divisible by 3, 7, 13, 37 and all combinations obtained by multiplying them (21, 39, 111, 91, 259, 351, 273, and 3367).

- if you repeat a 4-digit number twice

Here, I obtained a prime number.

I did not continue further, as I would obtain too large numbers.


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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.

Big Powers

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.

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The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice.

NRICH is part of the family of activities in the Millennium Mathematics Project.

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