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| Mathematics Magazine for Grades 1-12 |
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Grade 2
Theory:Counting Even Numbers by Twos You can count by twos by either: Ø Adding 2 to the previous number or Ø Counting and skipping every other number The numbers that you would count if you started with
0 and counted by twos would be: 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24,
26, 28, 30 and so on. The numbers that you would count if you started with
1 and counted by twos would be: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25,
27, 29, 31 and so on. Solutions from the Previous Issue:Recover
the following operations by replacing the * 1.
* 6 + 2 * + * 1 = 96 Solution: We count the
units 6 + 1 + 9
= 16 The identity
became: *6 + 29
+ *1 = 96 We add the
tens. * + 2 + * + 1 =
9 or * + * = 6 There are
multiple combinations to add 2 numbers resulting 6 0 + 6 = 6
we have then: 06 + 29
+ 61 = 96 1 + 5 = 6
we have then: 16 + 29 + 51 = 96 2 + 4 = 6
we have then: 26 + 29 + 41 = 96 3 + 3 = 6
we have then: 36 + 29 + 31 = 96 4 + 2 = 6
we have then: 46 + 29 + 21 = 96 5 + 1 = 6
we have then: 56 + 29 + 11 = 96 6 + 0 = 6
we have then: 66 + 29 + 01 = 96 2.
4 * + * 2 + 1 * = 82 Solution: We count the
units * + 2 + * = 2; has the following solutions * =
0; and * = 0; We count the
tens: 4 + * + 1 = 8; The final
result is: 40 + 32
+ 10 = 82 Other
combinations for the units are: * + * = 10
1 + 9 = 10 41 + * 2 + 19
= 82;
2 + 8 = 10
42 + * 2 + 89 = 82;
3 + 7 = 10
43+ * 2 + 17 = 82;
4 + 6 = 10
44 + * 2 + 16 = 82;
5 + 5 = 10 45 + * 2 + 15 = 82;
6 + 4 = 10 46 + * 2 + 14 = 82;
7 + 3 = 10 47 + * 2 + 13 = 82; 8 + 2 = 10
48 + * 2 + 12 = 82; 9 + 1 = 10
49 + * 2 + 11 = 82; We are adding
the tens 4 + * + 1 + 1 = 8; * = 2 41 + 22
+ 19 = 82 42 + 22
+ 89 = 82 43 + 22
+ 17 = 82; 44 + 22
+ 16 = 82; 45 + 22
+ 15 = 82; 46 + 22
+ 14 = 82; 47 + 22
+ 13 = 82; 48 + 22
+ 12 = 82; 49 + 22
+ 11 = 82; 3.
** + 2* + * 5 = 71 Solution: We are count
the units: * + * + 5 = 11 * + * = 6 There are
multiple solutions for the unit’s addition: 0 + 6 = 6
*0 + 26 + * 5 = 71 1 + 5 = 6
*1 + 25 + * 5 = 71 2 + 4 = 6
*2 + 24 + * 5 = 71 3 + 3 = 6
*3 + 23 + * 5 = 71 4 + 2 = 6
*4 + 22 + * 5 = 71 5 + 1 = 6
*5 + 21 + * 5 = 71 6 + 0 = 6
*6 + 20 + * 5 = 71 We count the
tens: * + 2 + * + 1 =7 There are
multiple solutions of * + * = 4 0 + 4 = 4
00 + 26 + 45 = 71
01 + 25 + 45 = 71
02 + 24 + 45 = 71
03 + 23 + 45 = 71
04 + 22 + 45 = 71
05 + 21 + 45 = 71 06 + 20 + 45 = 71 1 + 3 = 4
10 + 26 + 35 = 71
11 + 25 + 35 = 71
12 + 24 + 35 = 71
13 + 23 + 35 = 71
14 + 22 + 35 = 71
15 + 21 + 35 = 71 16 + 20 + 35 = 71 2 + 2 = 4
20 + 26 + 25 = 71
21 + 25 + 25 = 71
22 + 24 + 25 = 71
23 + 23 + 25 = 71
24 + 22 + 25 = 71
25 + 21 + 25 = 71 26 + 20 + 25 = 71 3 + 1 = 4
30 + 26 + 15 = 71
31 + 25 + 15 = 71
32 + 24 + 15 = 71
33 + 23 + 15 = 71
34 + 22 + 15 = 71
35 + 21 + 15 = 71 36 + 20 + 15 = 71 4 + 0 = 4
40 + 26 + 05 = 71
41 + 25 + 05 = 71
42 + 24 + 05 = 71
43 + 23 + 05 = 71
44 + 22 + 05 = 71
45 + 21 + 05 = 71 46 + 20 + 05 = 71 Proposed Exercises:Replace the * in the followings identities:
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