Reasoning · Grade 3-2 Completing Equations

Problem

Build division to extremize quotient or remainder

Pick 3 of the cards 3, 4, 6, 7, 8: two make a two-digit number and one divides it. No card twice in one division. A remainder is always smaller than the divisor. Build the largest quotient and the largest remainder.
Your answer
How to solve
Strategy Guess and Check — There are only a few sensible choices, so I reason about what makes a quotient or remainder big, then test the best candidates. A quotient grows when the top number is large and the divisor is small; a remainder can never reach the divisor, so I aim the divisor at 8 (giving a possible remainder up to 7) and look for a top number that leaves exactly 7.
1STEP 1

(1) Make the quotient as large as possible

Big number over small divisor gives 87 ÷ 3 = 29.

87 ÷ 3 = 29
2STEP 2

(2) Make the remainder as large as possible

A remainder caps at 7 under divisor 8, so 47 ÷ 8 it is.

47 ÷ 8 = 5 remainder 7
Answer
87 ÷ 3 = 29, 47 ÷ 8 = 5 remainder 7
Both answers respect the rules. In (1), 3 × 29 = 87 exactly, so 29 is a true quotient, and no smaller divisor than 3 exists among the cards. In (2), the remainder 7 is less than the divisor 8 as required, and a remainder of 7 is the largest a one-digit divisor can ever leave, so it cannot be beaten.
Takeaway

Want a big quotient? Divide a big number by the smallest card. Want a big remainder? Use the biggest divisor 8 and land just 7 past a multiple of it!

  • (1) Make the quotient as large as possible
  • (2) Make the remainder as large as possible