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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!

Have You Got It?

Can you explain the strategy for winning this game with any target?

Counting Factors

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

Gabriel's Problem

Age 11 to 14
Challenge Level

We received lots of great solutions, with some excellent explanations, from students from Nower Hill High School.

Abishek sent this clear explanation of how he went about solving the first problem, and here are Ben's approachSimran's approachPranavan's approachHolly's approachElisha's approach and Kania and Theivhikha's approach to the same problem. Rebecca also sent a clear explanation of her approach.

Another student from Nower Hill High School said: 

I tackled this problem by writing out the possible combinations of numbers whose products would equal the number on the side. After this, I would look  at the ones with the least amount of combinations or the least common numbers so that I can fill them in. Also, having the combinations written at the side will allow me to see which ones cross over with each other so that I could put them in the correct boxes for two calculations.

You can see her work on the first two questions here

Avipsa tackled questions 1 and 3 by splitting the numbers at the end of the rows and columns into the products of their prime factors. Here are her solutions.

Two other students from Nower Hill High School had a go at one of the more difficult questions, that required them to place all the numbers from 1 to 12. Both Mia and Hera discovered that there was more than one possible solution to the problem. 

Alice from Suton Valence School started with numbers she knew had few factors. She said:

The product of the middle column is 21, which is only divisible by 1, 3 and 7. Therefore the numbers in the middle column had to be 1, 3 and 7. 

The product of the left column and the middle row are multiples of 10 therefore they had to have 5 as one of their numbers. The only square they share is the middle square of the left column so that is where I put the 5.

Because 24 is not divisible by 7 I put 7 in the middle of the bottom row, leaving the 3 in the middle of the top row.

I then turned my attention to the middle row and calculated that 1 x 5 = 5 and  5 x 8 = 40, so I put the eight in the right-hand square of the middle row.

I then concluded that 9 was in the right-hand square of the bottom row because the products of the right-hand column and bottom row were high. 

Well done to the students of St Louis School in Milan. We got several solutions from students there. Tommasso explained how he worked out where 5 and 7 went.

First of all I have looked at which numbers have numbers 5 or 7 as factors. Because 5 and 7 can be written just once, only two numbers have these numbers as factors and you can place numbers 5 and 7.

For example, in the first problem just 60 and 40 are multiples of 5 and, in the second problem, just 210 and 126 are multiples of 7.

Dividing the numbers on the same row and column by 5 and 7, you can work out the other numbers in the row and column. Once you found these numbers you can more easily find the places of the other numbers. 

Davide explained how he used prime factors to work out where some of the numbers had to go. You can see his full solution here

Gianluca thought about how he could create his own problem. He said:

I've found out that the products at the end of each row multiply to 362880 and the products at the end of each column multiply too to 362880.

Why must this be true? Can you use this fact to create your own problem?