2nd Grade · 2.NBT.A.2

Skip counting in 2nd grade

Skip counting means counting forward by the same jump each time, such as 5, 10, 15, 20. In 2nd grade, your child skip counts by 5s, 10s and 100s and keeps counting by ones all the way to 1000.

What the standard says

The official Common Core wording, quoted exactly. The plain version follows.

2.NBT.A.2 Count within 1000; skip-count by 5s, 10s, and 100s.

What it means, in plain words

In 1st grade, children count to 120. Counting to 1000 in 2nd grade does not mean saying every number from 1. Your child learns to start at any number and say what comes next. The tricky spots are where a new ten or a new hundred begins: 297, 298, 299, 300.

Skip counting in 2nd grade uses three jumps. Counting by 5s is the same as counting nickels: 5, 10, 15, 20. Counting by 10s adds one ten each time, so from 40 it goes 50, 60, 70. Counting by 100s adds one hundred each time, so from 200 it goes 300, 400, 500.

The harder part is starting from a number that does not end in 0. Counting by 10s from 47 gives 57, 67, 77. Only the tens digit changes, and the ones digit stays 7. This is the first step toward adding 10 or 100 in your head.

Your child should be able to:

Count by ones from any number up to 1000, including across a new hundred
Skip count by 5s, 10s and 100s
Count by 10s or 100s from a number that does not end in 0, such as 35 or 260
Find the missing number in a counting pattern

Example

Count by 10s starting at 176. What are the next four numbers?

Each jump of 10 adds 1 ten. The tens digit goes up by 1 and the ones digit stays 6.
176, 186, 196.
After 196, there are already 9 tens. One more ten makes 10 tens, which is a new hundred, so the next number is 206.
The next four numbers are 186, 196, 206 and 216.

Example

Count by 100s starting at 350. What are the next three numbers?

Each jump of 100 adds 1 hundred, so only the hundreds digit changes.
350, 450, 550, 650.
The next three numbers are 450, 550 and 650.

Where children usually get stuck

Counting by 10s from 85, your child says 95, then 100, then 105.
After the nineties, 100 is the number the child is used to hearing next. Counting by 10s from 85 never lands on 100, because the ones digit stays 5 the whole way.
What helps. Put 85 cents on the table and add one dime at a time. Say the new total out loud after each dime: 95, 105, 115.
Counting by ones, your child says "one hundred twenty-eight, one hundred twenty-nine, one hundred twenty-ten."
The child carries on the pattern of the ones (8, 9, 10) and does not yet see that ten ones roll over into a new ten.
What helps. Count out loud together and pause at every number that ends in 9. Ask what the next ten is called before you say it.
Your child can say 5, 10, 15, 20 quickly but cannot tell how much 6 nickels are worth.
The count was learned like a song. The child says the words but does not use them to count groups of 5.
What helps. Line up nickels and touch one each time you say a number. The last number you say is the total: 6 nickels are 30 cents.

Try four practice questions

Real 2nd grade questions from our question bank, easiest first. Answer them right here, check each one, and see a worked solution for any question.

1Warm-up
What number comes next in the pattern?
$\displaystyle 11, 16, 21, \underline{\phantom{000}}$
A
22
B
26
C
27
D
31
Show the answer and how to solve it
The answer is 26 because skip-counting by 5s means adding 5 each step, and 21 plus 5 is 26.
1. 1Look at the sequence. Each step adds 5: 11 to 16 is plus 5, and 16 to 21 is plus 5.
2. 2To find the next number, add 5 to 21.
3. 3The next number in the pattern is 26.
Common mistake. A common mistake is adding 10 instead of 5 (getting 31) or adding only 1 (getting 22). Skip-counting by 5s adds 5 each time.
The idea underneath. When you skip-count by 5s, the ones digit alternates between two values. The +5 step works the same way every time — just add 5 to the last number you see.
2On level
What number comes next in this pattern?
$\displaystyle 213, \; 313, \; 413, \; \underline{\phantom{000}}$
A
513
B
423
C
613
D
414
Show the answer and how to solve it
The next number is 513. Each number in the pattern is 100 more than the one before, so adding 100 to 413 gives 513.
1. 1Look at the pattern in the sequence.
2. 2From 213 to 313, the number grows by 100. From 313 to 413, it grows by 100 again. So each number is 100 more than the one before it.
3. 3To find the next number, add 100 to 413.
4. 4The next number in the pattern is 513.
Common mistake. A common mistake is adding 1 or 10 to 413 instead of 100. Always check by comparing two of the visible numbers — they tell you how much to add at each step.
The idea underneath. When skip-counting by 100 inside the same group of hundreds, only the hundreds digit grows by 1 — the tens and ones stay the same.
3On level
Find the missing number in the sequence below.
$\displaystyle 154, \ 155, \ \underline{\phantom{0000}}, \ 157, \ 158$
A
156
B
159
C
155
D
166
Show the answer and how to solve it
The missing number is 156. Counting up by 1 from 155 gives 156, and counting back by 1 from 157 also gives 156.
1. 1Look at the sequence. The numbers go up by 1 each step: 154, 155, then the blank, then 157, 158.
2. 2Count up by 1 from 155 to find the missing number.
3. 3Check by counting back by 1 from 157. You get the same answer.
Common mistake. A common mistake is picking 159, which comes after 158. But the blank is between 155 and 157 — counting up by 1 from 155 gives 156.
The idea underneath. When counting by 1, each step adds exactly 1 — the pattern continues without skipping.
4Stretch
What number completes the pattern?
$\displaystyle 493, \underline{\phantom{000}}, 513, 523$
A
503
B
502
C
403
D
513
Show the answer and how to solve it
The missing number is $503$ . Each number in the pattern is $10$ more than the previous one, so $493 + 10 = 503$ .
1. 1Look at the pattern below. Each number is $10$ more than the one before it.
2. 2The sequence shows $493$ , blank, $513$ , $523$ . From $513$ to $523$ we can see the step is $10$ , so the blank is $493 + 10$ .
3. 3The missing number is $503$ .
Common mistake. A common mistake is changing the ones digit when the sequence crosses a hundred boundary. Each step adds $10$ , so the ones digit always stays the same — $493$ , $503$ , $513$ , $523$ all end in $3$ .
The idea underneath. Counting by $10$ s changes only the tens digit — until the tens digit reaches $9$ . Then the tens rolls over to $0$ and the hundreds digit goes up by $1$ .