Page 60 - ICSE Math 5

Table of 3 Table of 4
T O T O

3 4
6 8
9 1 2

1 2 1 6
1 5 2 0
1 8 2 4
2 1 2 8
2 4 3 2

2 7 3 6
3 0 4 0
Step 2: Add the numbers obtained in table of 3 and the respective digits at the tens place
in table of 4.
3 + 0 = 3, 6 + 0 = 6, 9 + 1 = 10, 12 + 1 = 13, 15 + 2 = 17, 18 + 2 = 20, 21 + 2 = 23,
24 + 3 = 27, 27 + 3 = 30 and 30 + 4 = 34.

Step 3: To form the table of 34, write the digit at the ones place in table of 4 as it is under
the ones place of the new table. Then, write the sums obtained in step 2 as the
digits at the tens and hundreds places of the new table as shown.

Table of 3 Table of 4 Table of 34
T O T O H T O
3 (3 + 0 = 3) 4  3 4
6 (6 + 0 = 6) 8  6 8
9 (9 + 1 = 10) 1 2  1 0 2

1 2 (12 + 1 = 13) 1 6  1 3 6
1 5 (15 + 2 = 17) 2 0  1 7 0
1 8 (18 + 2 = 20) 2 4  2 0 4

2 1 (21 + 2 = 23) 2 8  2 3 8
2 4 (24 + 3 = 27) 3 2  2 7 2
2 7 (27 + 3 = 30) 3 6  3 0 6
3 0 (30 + 4 = 34) 4 0  3 4 0
 In this way, every single time each group will write multiplication table of the number
given by the teacher.
 The group which forms the multiplication table first in this manner correctly will earn a
point.
 The group with maximum points at the end will be declared the winner.

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