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:

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
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.

Ritika Mahajan

Learning Specialist- Whole School Numeracy