Problem
Bound an inequality at equality
Among 1 to 9, 9×box must be greater than 6×8. Find every number that can go in the box.
Operations
Your answer
Watch out It's tempting to round 48÷9 down to 5, but 9×5=45 is not greater than 48 — 5 doesn't work.
How to solve
Strategy Guess and Check — First compute the fixed left side, then test box values from small to large to find where 9×box first passes 48; listing the candidates keeps every value from 1 to 9 accounted for.
1STEP 1
Compute the left side
6×8 works out to a fixed 48.
6×8=48
6×8 means eight equal groups of 6, and that always comes out to the fixed number 48. Since the left side never changes, the whole problem becomes comparing 9×box against 48.
3.OA.C.7Multiplication Facts2STEP 2
Find where 9×box passes 48
9×5=45 isn't enough, but 9×6=54 passes 48 — the cutoff is 6.
9×5=45, 9×6=54
A box of 5 makes 9×box nine equal groups of 5, piling up to only 45 — short of 48. Raising the box to 6 adds one more to each of the nine groups, so the total climbs from 45 by another 9, reaching 54 and clearing 48.
Guess And Check3.OA.A.49 times the box first rises above 48 at box = 6, so the box has to be 6 or larger.
Why?
A box of 5 makes the right side 9 x 5 = 45, which is still under 48, so a box of 5 or anything smaller is too small.
Why?
9 x 5 is nine equal groups of 5, and nine groups of 5 pile up to only 45.
🧱Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.
Why?
A box of 6 makes the right side 9 x 6 = 54, and 54 is above 48, so a box of 6 already works.
Why?
9 x 6 is nine equal groups of 6, and nine groups of 6 pile up to 54.
🧱Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.
Why?
Raising the box from 5 to 6 adds one more to each of the nine groups, so the total climbs from 45 by another 9 up to 54, past 48.
🧱Distributive propertya groups of (b and c) is a groups of b plus a groups of c — you can split the group.
Why?
A box bigger than 6 puts even more into each of the nine groups, so the right side only grows and stays above 48 -- the box just needs to be 6 or more.
Why?
Every time the box goes up by one, all nine groups each gain one, so the total gains another 9 and can never shrink.
🧱Distributive propertya groups of (b and c) is a groups of b plus a groups of c — you can split the group.
3STEP 3
List every number that works
Anything at or above 6 only grows bigger, so 6, 7, 8, 9 all work.
□∈{6,7,8,9}
Once the box passes 6, each of the nine groups only gets bigger, so the right side keeps climbing and stays above 48. That means every value from 6 through 9 satisfies the condition.
Systematic List3.OA.A.4Answer
6, 7, 8, 9
□∈{6,7,8,9}
Check — 9×6=54, 9×7=63, 9×8=72, 9×9=81 are all greater than 48.
Takeaway
Turn one side into a single number, then use the times table to find the cutoff — everything above it works.
- Compute 6×8 → 48
- Check nine-times facts → 9×5=45, 9×6=54, cutoff is 6
- 6 through 9 all work → 6, 7, 8, 9
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