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Numeracy Challenge 

Numeracy Challenge

Numeracy Challenge 3: 

 

Hello everyone, 

Many congratulations to the budding mathematicians for having a go at the challenge and explaining your thinking with elegance.  

Names: 

Sanuj Alwis Year 7 

Anuk Alwis Year 7  

Jason   Hao Year 8 

Ethan Hao Year 10 

Chloe Guo Year 10 

Cloris Fang Year 8 

Aiden Kan Year 10 

Responses: 

  1. Each letter gets multiplied by 3 as you move down the alphabet. So that means the formula to evaluate this problem would be: (Let the position of letter = x) 3^(x-1)  Ex1, for C, it would be 3^(3-1) and 3^2 = 9 Ex2, if we are trying to figure out t, it would be 3^(20-1) and 3^19 = 1162261467  So Z would be equivalent to 3^(26-1) which is 847288609443 
  2. The answer is 6 and the pairs are 12 x 37 = 444 15 x 37 = 555 18 x 37 = 666 21 x 37 = 777 24 x 37 = 888 27 x 37 = 999  How I figured this out is because I found the factors for 111 which are 3 and 37. Because 37 is a prime number I just found multiplications of 3 that have 2 digits and from there I just multiplied it with 37. 

Brilliant work. Looking forward to the responses for this week’s challenge.  

 

https://forms.office.com/r/WTf67KQNRx 

 

 

 

 

 

Ritika Mahajan 

Learning Specialist- Whole School Numeracy