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Counting Strategies HNP Example Maths Recovery NZMaths Pupils are unable to consistently count a given number of objects because they lack the knowledge of counting sequences and/or the ability to match things one-to-one. I could not collect 7 counters. Emergent Stage 0 Emergent Stage 0 Early ** Pupils can count up to ten but cannot solve simple problems that involve joining and separating sets like 4 + 3. I collected 4 counters and collected another 3 counters but then I did not know what to do. ‘I have 4 and 3’. Or says - it makes 10 N/A One-to-one Counting Stage 1 Early *** When solving joining or separating of sets problems, pupils first rely heavily on counting physical materials, like their fingers. They count all the objects in both sets to find the answer. (They start their count at one). I used my fingers. 8 + 5 = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 Perceptual Counting Stage 1 Counting from One using Materials Stage 2 Early *** When solving joining or separating of sets problems, they can move away from using materials and are able to image visual patterns of the objects in their minds and count them. They count all the objects in both sets to find the answer. (They start their count at one). First * Pupils can solve number problems by counting on, or back. They keep track of the count using materials (e.g. fingers) or by imaging (in their heads). They understand that the end number in a counting sequence measures the whole set and can relate addition and subtraction of objects to the forward and backward number sequence by ones and tens etc. Early * Description of Stages of Thinking Pupils at this stage use a skip-counting strategy to solve multiplication problem. 1, 2, 4, 5, 10 I pictured the 8 and 5 counters in my head and counted: 8 + 5 = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 I thought 9 and then counted on: 9 + 4 = 9, 10,11, 12, 13 Or 32 + 21 = 32, 42, 52, 53 7 x 3 = I counted in 3s like this: 3, 6, 9, 12 …but I didn’t know the next numbers so I had to count like this: 13, 14, 15, 16, 17, 18,19, 20, 21 Figurative Counting Counting from One Stage 2 by Imaging Stage 3 Counting On Stage 3 Counting Down Stage 4 Advanced Counting (Counting on/back) Stage 4 Part-Part Whole Strategies HNP First ** First *** Second * Second ** Second *** Description of Stages of Thinking Pupils at this stage have begun to recognise that numbers can be split into parts and recombined in different ways. This is called part-whole thinking. Pupils use a limited range of mental strategies to estimate answers and solve addition and subtraction problems. Strategies used at this stage are most often based on a group of five or ten or use a known fact, such as a double. Pupils are familiar with a range of part-whole strategies and are learning to choose appropriately between these. They have well developed strategies for solving addition and subtraction problems. They see numbers as whole units but also understand that they can be ‘nested’ within these units. Pupils are able to solve multiplication answers from known facts. These pupils are also able to solve problems using a combination of multiplication and addition- based strategies. Pupils at this stage choose appropriately from a range of part-whole strategies to solve and estimate problems with whole numbers and decimals to one place. They are able to choose a multiplicative strategy to solve an addition problem where appropriate. Pupils are learning to manipulate factors mentally to solve multiplication and division problems. They move from partitioning additively to using a halving and doubling strategy. A key strategy of this stage is reversibility, solving division problems using multiplication. Example Maths Recovery NZMaths 48 + 7 as (48 + 2) + 5 Facile Stage 5 Early Additive Stage 5 N/A Advanced Additive Part Whole Stage 6 N/A Advanced Multiplicative Part Whole Stage 7 26 – 9 = (26 – 10) + 1 = 17 8 + 9 = (8 + 8) + 1 367 + 260 as (300 + 200) + (60 + 60) + 7 135 – 68 = 135 – 70 + 2 604 – 598 as 598 + ? = 604 6 x 3 = (5 x 3 ) + 3 = 15 + 3 = 18 6.5 + 7.7 = 6.5 + 8 = 14.5 – 0.3 = 14.2 55 + 44 + 33 = (5 x 11)+ (4 x 11) + (3 x 11) = 12 x 11 Understand that 5 x 12 can be partitioned into (5 x 10) + ( 5 x 2) Begins to understand how 5 x 12 is the same as 10 x 6