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How many positive integers less than or equal to 4000 can be written down without using the digits 7, 8 or 9?

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Nines and Tens

Explain why it is that when you throw two dice you are more likely to get a score of 9 than of 10. What about the case of 3 dice? Is a score of 9 more likely then a score of 10 with 3 dice?


Stage: 3 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Class 9Y-1 from Downend School sent us the correct solution and showed that it satisfied the criteria:

The five coins are 1p, 10p, 10p, 20p, 20p.
The thirteen amounts are 1p, 10p, 11p, 20p, 21p, 30p, 31p, 40p, 41p, 50p, 51p, 60p, 61p.
These amounts only end in the digits 0 or 1.
11p, 31p, 41p and 61p are all prime amounts of money.

Similar solutions were sent by Kate and Hannah from Marist College in New Zealand; students from Wingham Primary School; James from Wellington School; Can, Jack and Lee from Ashmount Primary School; Patrick and Natalie also from Ashmount Primary School; Thomas, Jack, Jim, Jenny and Timothy from Bay House School; Tasmin from St. Pauls; Kirsty from Herts and Essex High School; Levi from Greenwood Junior School; Tommy from St Mary's Hall, Lincoln and Jaehyung from ACS Egham International School, and Daniel from Ysgol Dyffryn Dulas Corris.

Jae explained his reasoning:

The first clue I used was: He can not make exactly 70p.
With that clue, I knew that for example it couldn't be 10, 10, 50, 5, 20 or 20, 20, 20, 10, 1

Then I looked at the clue The amounts of money he can make end with one of two possible digits.
That meant in the ones, it could be 5 and 0 or 1 and 0.

The last clue I used was He can make 13 different amounts of money with his coins.
I used trial and error and I got 1p, 10p, 10p, 20p and 20p.

It couldn't make exactly 70p and it was under £1 as well.
1, 10, 11, 20, 21, 30, 31, 40, 41, 50, 51, 60, 61 was the combinations he could make. That was 13 combinations all together.

And if you added all of the coins together, it would be 61p, which is a prime number.
Also there was only two digits in the ones place, 1 and 0. That is how I got my answer.

Alex from Bristol Grammar School also explained his reasoning:

A 1p coin, two10p coins, two 20p coins. A 1p coin is needed to make the prime number above 10 and the rest must be mutliples of 10 so that the final digit does not change. The 10 and 20 p coins were chosen through system of elimination.

Sammy & Lucy from Hove Park School also showed that their solution satisfied all the criteria:

A Man has 5 Coins and his coins are 10p, 10p, 20p, 20p, 1p

The specifications are:
He can make 13 different amounts of money with his coins.
The amounts he can make are.... 21p, 11p, 10p, 20p, 1p, 30p, 40p, 50p, 60p, 31p, 41p, 51p, 61p Which totals to 13 amounts.

The amounts of money he can make end with one of two possible digits.
All of the amounts above ended with either a 0 or a 1.

He cannot make up exactly 70 pence.
70p is not one of the totals seeing as 61 is the highest.

He cannot afford an item costing £1.
£1 cannot be one of the totals seeing as 61p is the highest.

He can make a prime number bigger than 10 with his coins.
61 is a prime number and over 10.

Well done to you all.