# Divisibility

DIVISIBILITY BY 2

Materials:
…decimal system materials (wooden)
…blackboard or blank chart

Presentation:
Write a number and invite the child to bring that quantity with the materials. Try to make two equal groups from this quantity, making changes as necessary .If it is possible to make two equal groups, write “yes” next to the number and underline the last digit. If not, write “no”. Add or subtract one unit, and repeat the process. Examine many numbers in the same way. At a certain point the child will realize the rule: When the last digit is an even number or zero, the number is divisible by 2.
If the child does not reach this point of consciousness on her own, ask questions to call her attention to the relationship between the yes or no and the oddness or evenness of the last digit.

1126 yes
1125 no
1124 yes
1123 no
1122 yes
78 yes
79 no
80 yes
12 yes

DIVISIBILITY BY 4

Materials: Same as above

Presentation:
The procedure is the same as before, except that the last two digits are underlined. Now it is no longer a matter of oddness or eveness.
Rule: When the last two digits are divisible by 4, or they are both zeros, the number is divisible by 4.

816 yes
817 no
818 no
819 no
820 yes

DIVISIBILITY BY 5

Materials: Same as above

Presentation:
The procedure is the same as before, only the last digit is underlined.
Rule: When the last digit is 5 or 0, the number is divisible by 5.

125 yes
126 no
127 no
130 yes
45 yes
100 yes

DIVISIBILITY BY 25

Materials:
…ten bars and 40 or more golden unit beads
…small white square pieces of paper,
…pen

Presentation:
Put out one group of 25 with a little card over it. Place another group next to it in an inverse position to make it easy to visualize the group as 50. Place a little card over it (“50”). Continue placing groups of 25 (7 tens, 5 units for 75) to the left of the previous group, the respective card is placed over the top. Continue up to 8 groups of 25, substituting 100 squares after 100.

Invite the child to speculate on the next few multiples of 25.

Rule: When the last two digits of a number are 23, 50, 75 or two zeros, the number is divisible by 25.

DIVISIBILITY BY 9

Materials:
…Peg board
…Box of pegs in hierarchic colors
…Small white square pegs
…pen

Presentation:
9 is a very important number, Why? It is the last digit that can remain loose in our system. It is the square of 3, and 3 is a perfect number. Therefore in this work we will consider 9 as the square of 3.
On the peg board use 9 green pegs to construct a square of 3 by 3. This is a square of 3. Dissolve the square into a column at the top left corner of the board, labeling it 9.
Form two more squares of 3 side by side. Remove one unit from one square and add it to the other. Now one group has ten, change the ten green unit pegs for one blue ten peg. Dissolve the pegs into two columns next to the previous column. Label 18.
Repeat the procedure with 3 squares. Add one peg to each of the first two squares, taking the pegs from the last square. Change each group of 10 to a blue peg and dissolve. Continue in this way up to 10 squares = 90.
Observe that as the units decrease, the tens increase. At one extreme there are 9 units, at the other, 9 tens. This special pattern exists only in the table of 9. The sum of each pair of digits is 9. i.e. 27, 2 = 7 9.
Rule: A number is divisible by 9 when the sum of its digits equals 9 or a multiple of 9. If a number is divisible by 9, it is also divisible by 3.

Note: This characteristic may have been noticed in the multiplication booklet for memorization.

PROOF WITH MULTIPLICATION WITH RULE OF DIVISIBILITY BY 9

Materials: Decimal system materials

Presentation:
Take the thousand cube and try to make 9 equal groups. 1000 is not divisible by 9, but take away one unit and try again. 999 is divisible by 9. Write 1000 ­ 1 = 999. Repeat the procedure for 100 and 10. Write 100 ­ 1 = 99, 10 ­ 1 = 9. Choose a number: 5643. Convert it to expanded notation:

3……………………………………………………………………….r.3
40 = 4 x 10
…………10 ­ 1 = 9………..40 – 4 = 36…………..r.4
600 = 6 x 100
…….100 ­ 1 = 99 …….600 – 6 = 594………..r.6
5000 = 5 x 1000
…1000 ­ 1 = 999 ….5000 – 5 = 4995….+ r.5
………………………………………………………………………… 18

10 ­ 1 = 9, but we have 4 tens. We must multiply the whole equation by 4, which gives us 36. 36 from our original 40 leaves us 4 as a remainder. Continue for the others. Add all the remainders in the end. Since 18 is divisible by 9, the whole number is divisible by 9. (cont.)
Show an example of multiplication: 643 x 1527 = 981,861. Make sums of 9 in the digits – 643 x 1527 = 981,861. Total the remaining digits (4 from the multiplicand, 1 and 5 from the multiplicand) multiply the sums ( 6 x 4 = 24 ), and add the digits of the product
( 2 + 4 = 6 ). This number should correspond to the sum of the remaining digits in the original product – 6. 6 is also the remainder when the original product ( 981,861 ÷ 9 = 109,095 r. 6) is divided by 9. The “remaining sums” from above multiplicand and multiplier (4 and 6) will also be the remainder when divided by 9: ( 643 ÷ 9 = 71 r. 4 and 1527 ÷ 9 = 169 r. 6).

Age: Ten years