RIC-0563 Developing algebraic thinking
LOCKER NUMBERS Teachers notes Locker numbers 2 Problem 1 Problem 2 Problem 3 Problem 4 For Problem 1, if the locker number is xyz, then: y = 2z x = 2y = 2(2z) = 4z Since z ≠ 0 (otherwise x, y = 0), z = 1 or 2. Other values of z will make x or y a 2-digit number. If z = 1, then y = 2, and x = 4. The locker number is 421. If z = 2, then y = 4 and x = 8. The locker number is 842. For Problem 2, if the locker number is xyz, then: y = x + z With xyz > 700, x = 7 or 8. If x = 7, then z = 1 and y = 8, or z = 2 and y = 9. The locker numbers are 781 or 792. If x = 8, then z = 1 and y = 9. The locker number is 891. For Problem 3, if the locker number is xyz, then: z = 3, 6 or 9 y = x – 6 Since y must be non-negative, the only possible values for x are 6, 7, 8 or 9. If x = 9, then y = 3 and z = 6. If x = 8, then y = 2 and z = 6. If x = 7, then y = 1 and z = 6. If x = 6, then z = 6. The locker numbers are 936, 826 and 716. For Problem 4, if the locker number is xyz, then: x = y + 3 y = z + 3 so x = y + 3 = z + 3 + 3 = z + 6 Since x + y + z is odd, z = 1 or 3. If z = 1, then y = 4 and x = 7. The locker number is 741. If z = 3, then y = 6 and x = 9. The locker number is 963. 62 DEVELOPING ALGEBRAIC THINKING www.ricgroup.com.au R.I.C. Publications ® ISBN 978-1-74126-088-5
LOCKER NUMBERS Teachers notes Problem 5 For Problem 5, if the locker number is xyz, then: xyz < 500 z = y – 2 Since 2 is a factor of x, x = 2 or 4. If x = 2, then y = 6 and z = 4, or y = 8 and z = 6. The locker numbers are 264 and 286. If x = 4, then y = 2 and z = 0, or y = 8 and z = 6. The locker numbers are 420 and 486. Problem 6 For Problem 6, if the locker number is xyz, then: z = 0, 2, 4, 6 or 8 x = z – 1 y = z + 4 If z = 0, then x is a negative integer. If z = 6 or 8, then y is a 2-digit number. So, z = 2 or 4. If z = 2, then x = 1 and y = 6. If z = 4, then x = 3 and y = 8. The locker numbers are 162 and 384. R.I.C. Publications ® www.ricgroup.com.au DEVELOPING ALGEBRAIC THINKING 63 ISBN 978-1-74126-088-5
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LOCKER NUMBERS<br />
Teachers notes<br />
Problem 5<br />
For Problem 5, if the locker number is xyz, then:<br />
xyz < 500 z = y – 2<br />
Since 2 is a factor of x, x = 2 or 4.<br />
If x = 2, then y = 6 and z = 4, or y = 8 and z = 6.<br />
The locker numbers are 264 and 286.<br />
If x = 4, then y = 2 and z = 0, or y = 8 and z = 6.<br />
The locker numbers are 420 and 486.<br />
Problem 6<br />
For Problem 6, if the locker number is xyz, then:<br />
z = 0, 2, 4, 6 or 8 x = z – 1 y = z + 4<br />
If z = 0, then x is a negative integer.<br />
If z = 6 or 8, then y is a 2-digit number.<br />
So, z = 2 or 4. If z = 2, then x = 1 and y = 6.<br />
If z = 4, then x = 3 and y = 8.<br />
The locker numbers are 162 and 384.<br />
R.I.C. Publications ® www.ricgroup.com.au DEVELOPING ALGEBRAIC THINKING 63<br />
ISBN 978-1-74126-088-5
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