Problem
A product of 0 only happens when one factor is 0; the visible 3 and 1 are not 0, so one hidden card must be 0.
Zero times any number is zero, so a product of 0 forces a 0 card to exist.
3.OA.B.5Eliminate PossibilitiesA product of 5 from single digits must be 5 x 1; card 4 is already 1, so the other hidden card must be 5.
5 is only 5 x 1 among one-digit products, and a 1 is already on a card, so the partner card is 5.
3.OA.C.7Eliminate PossibilitiesThe remaining hidden card must be 5.
A pair of cards multiplies to 5, and since the 3 and the 0 card cannot build a 5, that pair has to be the hidden card together with the 1 on card 4 — so the hidden card multiplied by 1 must equal 5.
Multiplying a number by 1 leaves it exactly as it is, so the only card that comes out as 5 after multiplying by 1 is 5 itself.
The four cards are 5, 3, 0, 1; multiplying the two largest gives 5 x 3 = 15, beating 5 x 1 = 5 and any pair using 0.
To make the biggest product, multiply the two largest cards, 5 and 3.
3.OA.C.7Make A Systematic ListZero makes 0 and one keeps a number the same -- those facts unmask the hidden cards, pure Grade 3 thinking!