# data[].attributes.body
100x4=400
25X4=100
and 3 left over

# data[].attributes.body
1/2=100
1/4=50
1/5=40
40. 40. 40 40. 40 =200
40 Remainder 1

# data[].attributes.body
40 x 5 = 200 with 1 pencil left over

# data[].attributes.body
There are 125 pencils in each box.  There are 3 pencils left.

# data[].attributes.body
Plus 1 left over
200+1=201

# data[].attributes.attachment_link.title
I was still mixing up the story with this one!

# data[].attributes.body
Box 2: 100, 10, 10,5
Box 3: 100, 10, 10, 5
Box 4: 100, 10, 10, 5
3 left over

# data[].attributes.body
4 x 100 = 400
100 + 100 + 100 + 100 = 400
4 x 25 = 100
400 + 100 = 500
125 pencils in each box with 3 left over

# data[].attributes.body
box 2 125
box 3 125
box 4 125
125+125+125+125=500
3 leftover

# data[].attributes.body
Shared out 100 equally to each of 4 boxes.  Then, shared out the remaining 100 to each of 4 boxes which is 25.  100 + 25=125 in each box with 3 left over.

# data[].attributes.body
5x4=20
5x40=200
40 pencils in each box with one left over

# data[].attributes.body
plus 1 left over
66+1=67

# data[].attributes.body
4 equal groups of 125 = 125+125+125+125
Plus 3 left over

# data[].attributes.body
3+3=6
22 with 1 left over

# data[].attributes.body
4 groups of 20 + 2; 1 left over

# data[].attributes.body
I know 5x20=100, so 5x40=200.
40 and 1 left over
relate to nickels

# data[].attributes.body
40 in each box
5 boxes with 1 leftover

# data[].attributes.body
503-500 = 3
500 /4 = 100 R 100
100/ 4 = 25
125 pencils for each box R 3

# data[].attributes.body
box 1: 10, 10, 2
box2: 10, 10, 2
box3: 10, 10, 2
22 pencils in each box with 1 pencil left over

# data[].attributes.body
10       10         10
10      10          10
1           1            1
1           1         1
1

# data[].attributes.body
R 1

# data[].attributes.body
4 equal groups of 40=40+40+40+40+40
Plus 1 left over

# data[].attributes.body
2+2+2=6
22 in each box.
1 leftover.

# data[].attributes.body
Three groups of 22= 22+22+22=66 plus 1 left over

# data[].attributes.body
20         20.         20
//          //            //

Remainder : one
22 in each box