Problem
Build division expressions meeting conditions
Pick two of the cards 2, 4, 5, 6 and set them side by side as a two-digit number. Each card is used at most once in a number. Find the number divisible by both 7 and 8.
Reasoning · Grade 3-1 Multiplication and Division
Your answer
Watch out Testing every two-digit number is tempting, but combining the conditions leaves only multiples of 56.
How to solve
Strategy Make a Systematic List — A number that is divisible by both 7 and 8 must be a multiple of 7 x 8 = 56, because 7 and 8 share no common factor. So instead of testing every arrangement of the cards, I list the two-digit multiples of 56 and check which one can actually be built from the cards.
1STEP 1
Combine the two divisibility conditions
7 and 8 share no factor, so the number is a multiple of 56.
7 × 8 = 56
Being divisible by 7 means '7 fits in a whole number of times', and the same for 8; the smallest number both fit into is their product 56, so every common multiple is a multiple of 56.
3.OA.B.6Guess And CheckBecause 7 and 8 share no factor, a number divisible by both must be a multiple of their product 56.
Why?
A number's primes are fixed, so needing every prime of 7 and every prime of 8 means carrying both sets at once, which is 56.
🧱Every number has one prime factorizationKeep splitting a number into factors and you always land on the same bag of primes, no matter which split you tried first.
Why?
Dividing exactly means leaving no remainder, so any number that is not a whole count of 56 fails at least one of the two tests.
🧱Division with remainderAny number splits into a whole count of equal groups plus a leftover, and the leftover is always smaller than the group size.
2STEP 2
List two-digit multiples of 56
Only 56 is two digits; 112 is three.
56 × 1 = 56, 56 × 2 = 112
Skip-counting by 56 leaves only one stop inside the two-digit range, so the search is tiny.
3.OA.A.2Make A Systematic List3STEP 3
Check that 56 can be built from the cards
The cards hold 5 and 6, so 56 can be built.
56 = 5 6
The candidate is only useful if its two digits are actually on the cards, and here they are.
3.OA.A.2Make A Systematic ListAnswer
56
Check directly: 56 divided by 7 is 8 with no remainder, and 56 divided by 8 is 7 with no remainder. Both conditions hold, and 56 is a two-digit number made from the cards 5 and 6, so the answer is consistent.
Takeaway
If a number is divisible by 7 AND 8, it must be a multiple of 56 - that one idea instantly shrinks the search to a single number!
- Combine the two divisibility conditions
- List two-digit multiples of 56
- Check that 56 can be built from the cards