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Adding All Nine

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!

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Counting Factors

Is there an efficient way to work out how many factors a large number has?

Product 100

Age 11 to 14 Short
Challenge Level

Answer: 18

100 = 50$\times$2$\times$1$\times$1     not all different
       = 10$\times$5$\times$2$\times$1     good! Sum = 18

Is there another way?
Factors of 100 not used: 4, 20, 25

100 = 4$\times$25$\times$1$\times$1     not all different
       = 4$\times$5$\times$5$\times$1       not all different. Cannot use 4
       = 25$\times$2$\times$2$\times$1     not all different. Cannot use 25
100 = 20$\times$5$\times$1$\times$1     not all different. Cannot use 20

This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.