Problem
Count numbers built from digit cards
Using the four digit cards 8, 0, 5, and 3 each exactly once, build four-digit numbers and count how many of them are greater than 5000.
Base-ten numbers
Your answer
Watch out The count is only numbers greater than 5000, not every four-digit number you can build — a number led by 3 does not qualify.
How to solve
Strategy Make a Systematic List — Whether a number is greater than 5000 is decided almost entirely by the thousands digit, so I split the count by which card sits in the thousands place (a subproblem), then systematically list the arrangements of the remaining three cards.
1STEP 1
Decide which card can be in the thousands place
Only a thousands digit of 5 or more clears 5000, so of the cards 8, 0, 5, 3 just 5 and 8 can lead.
thousands digit ≥ 5 → 5 or 8
Comparing multi-digit numbers starts from the highest place, so the thousands digit alone tells you if you have cleared 5000.
4.NBT.A.2Identify SubproblemsA four-digit number built from the cards 8, 0, 5, and 3 is greater than 5000 exactly when the card in the thousands place is 5 or 8.
Why?
Whether a four-digit number clears 5000 is settled by its thousands digit alone, because that place is worth a whole thousand.
Why?
Moving the thousands digit up by one adds a full 1000, while the hundreds, tens, and ones places together can only ever reach 999, so they can never make up for a smaller thousands digit.
Why?
Each place holds ten of the place beneath it, so one thousand is bigger than any amount the three lower places can pile up.
🧱Place-value groupingTen ones make one ten and ten tens make one hundred — each place bundles ten of the last.
Why?
Of the four cards, only 5 and 8 are large enough in the thousands place to pass 5000, while 3 keeps the number down in the 3000s and 0 cannot lead a four-digit number at all.
Why?
Because the thousands digit decides the size, a 3 there gives a number below 4000, and a 0 there means there is no thousands at all, so the leading card has to be 5 or 8.
🧱Place-value groupingTen ones make one ten and ten tens make one hundred — each place bundles ten of the last.
2STEP 2
Count arrangements with 8 in the thousands place
Fix 8 in front; the other cards 0, 5, 3 fill three places in 6 orders, each at least 8035 — all beat 5000.
3 × 2 × 1 = 6
Three different cards into three slots gives 6 different orderings, and starting with 8 guarantees the number passes 5000.
4.NBT.A.2Make A Systematic List3STEP 3
Count arrangements with 5 in the thousands place
Fix 5 in front; 0, 8, 3 fill the rest in 6 orders, and even the smallest, 5038, still beats 5000.
3 × 2 × 1 = 6, 5038 ＞ 5000
Even the smallest number that starts with 5 here, 5038, is past 5000, so none of these 6 are lost.
4.NBT.A.2Make A Systematic List4STEP 4
Add the cases
Add the two leaders: 6 numbers starting with 8 plus 6 starting with 5 give 12.
6 + 6 = 12
The two starting digits give separate, non-overlapping lists, so we simply add the counts.
4.NBT.A.2Make A Systematic ListAnswer
12
6 + 6 = 12
There are 18 four-digit numbers in all using these cards (3 choices for a nonzero leading digit times 6 arrangements). Only those starting with 5 or 8 beat 5000, which is exactly 12 of them, a believable fraction.
Takeaway
Look at the biggest place first: only cards 5 and 8 can lead a number past 5000, then just count the orderings!
- Decide which card can be in the thousands place
- Count arrangements with 8 in the thousands place
- Count arrangements with 5 in the thousands place
- Add the cases
Where next?
Level upsuggested
Another one like thissuggested
Step back and solidifysuggested