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I decomposed 503 into groups of 100. This gave me  5 groups not 4 so i took a hundred.  I broke into 25s  and distributed  the pencils that were the 4 boxes with hundreds. This gave me 125 in each of the 4 boxes with 3 pencils left over.

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Count by 10s till you get to 200 making sure all groups have the same amount of pencils... 1 pencil left over

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Count by 10s till you get to 200 making sure all groups have the same amount of pencils... 1 pencil left over

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I divided into 201 by five boxes and partitioned 40 pencils into each box. When I worked in reverse this gave me the 5 groups of 40 per box and that is the same as 200 pencils when added.  This left me one pencil left over. 

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I divided into 201 by five boxes and partitioned 40 pencils into each box. When I worked in reverse this gave me the 5 groups of 40 per box and that is the same as 200 pencils when added.  This left me one pencil left over. 

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Divided 200 by 5 = 40 pencils in each box

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Divided 200 by 5 = 40 pencils in each box

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201-200 = 1 pencil left

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Used visuals and abstracts, base ten blocks. Pics unavailable.

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I drew boxes as an anchor for my multiples.

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4 hundreds in 400 and 4 quarters in a dollar

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4 hundreds in 400 and 4 quarters in a dollar

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I decomposed 500 into 400+100 and knew that there are 4 hundreds in 400 and the left over 100 I divided by 4 because I thought of it as 100 pennies (1 dollar) and I know that 4 quarters (25 cents) make a dollar. I added those numbers together and there were 3 pencils left over.

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When dividing 503 by 4, I saw there was 1 hundred (5/4). I then saw there was 2 tens (10/4). I then saw that the 20 of the 23 remaining could be divided evenly too (20/5=4).  So, 1 hundred + 2 tens + 5 ones. Answer: 125 with 1 left over.

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When dividing 503 by 4, I saw there was 1 hundred (5/4). I then saw there was 2 tens (10/4). I then saw that the 20 of the 23 remaining could be divided evenly too (20/5=4).  So, 1 hundred + 2 tens + 5 ones. Answer: 125 with 1 left over.

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4x100 = 400
4 x 25 = 100
400 + 100 = 500
503 - 500=3
100+25 = 125
So, 125 pencils, 3 left

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Divided 67 by three and broke it into 3 boxes of 20. Next, I had 7 pencils so I divided by three which gave me 2 pencils to be distributed into each box and I had 1 pencil left. This gave me 22 pencils left in each box and a remainder of 1 pencil. 

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Divided 67 by three and broke it into 3 boxes of 20. Next, I had 7 pencils so I divided by three which gave me 2 pencils to be distributed into each box and I had 1 pencil left. This gave me 22 pencils left in each box and a remainder of 1 pencil. 

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I drew boxes as an anchor for my multiples.

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1 left over

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There are 67 pencils and 3 boxes.

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20+2=22 r.1

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I know that 4x5=20 so 40x5=200 and there was 1 pencil left over.

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I drew boxes as an anchor for my multiples.

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Used base ten blocks to help me.

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I now 5x 4 =20, so 5 x 40 must = 200. So there will be one exra pencil

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by Elsa

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When dividing 201 bt 5, I noticed that if I set aside the 1 in the ones, I am left with a friendly number of 200. Since multiples of 5 always have either a 5 or a 0, I knew that I could divide 200 evenly by 5. I then looked to dividing the 20 tens by 5 to get 4 tens. Answer: 40 with 1 left over.

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When dividing 201 bt 5, I noticed that if I set aside the 1 in the ones, I am left with a friendly number of 200. Since multiples of 5 always have either a 5 or a 0, I knew that I could divide 200 evenly by 5. I then looked to dividing the 20 tens by 5 to get 4 tens. Answer: 40 with 1 left over.

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Drew 4 boxes, then used base ten blocks starting with 100 in each, then 10s, then 1s

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I made 4 containers. I put 100 in each container. That totaled 400. Then I placed 25 in each container to total 500. I had 3 left overs.

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Decomposed the 500 into 5 (100), then 100 into 4x25. Had 3 ones left over.

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Decomposed the 500 into 5 (100), then 100 into 4x25. Had 3 ones left over.

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Decomposed the 500 into 5 (100), then 100 into 4x25. Had 3 ones left over.

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I decomposed 67 into 60+7 and knew there were 3 groups of 20 in 60 (3x2=6).  Then I decomposed the 7 into 6+1 and knew there were 3 groups of 2 in 6 (3x2=6) and there was 1 pencil left over.

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I can divide 66 into 3 different groups and I know there will be one more left over. So Mr. Warren should put 22 pencils in three different boxes and have one left over.

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First I did 200 = A x 5. I know that 4 x 5 = 20, so 40 x 5 = 200 because I needed to multiply by a ten. To get to 201, I need one more pencil to the total. There are 40 pencils in each box and one extra pencil.

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When dividing 67 by 3, I knew each box would have 2 tens. since 7 is one more than 6 ones, I knew each box would have 2 ones with one left over. Answer: 22  with 1 left over.

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When dividing 67 by 3, I knew each box would have 2 tens. since 7 is one more than 6 ones, I knew each box would have 2 ones with one left over. Answer: 22  with 1 left over.