Reasoning · Grade 2-2 Rules and Defined Operations

Problem

Recover the starting number

Multiply all the digits of a number to make a new one, and repeat until a single digit is left. The chain length counts how many multiplications that took. Build the chains for 76, 47 and 246, then find a two-digit number ending at 8 with length 4.
Your answer
How to solve
Strategy Look for a Pattern — Part (1) just follows the digit-multiply rule step by step. For part (2) the rule is hard to run backwards cleanly (many numbers share the same digit-product), so the surest method is to apply the rule forward to candidate two-digit numbers and list their chain lengths until one with length 4 ending in 8 appears.
1STEP 1

Build the chain for 76

76 goes through 42 to 8: length 2.

76 → (7 × 6=42) → (4 × 2=8) chain length 2
2STEP 2

Build the chain for 47

47 runs 28, 16, then 6: length 3.

47 → 28 → 16 → 6 chain length 3
3STEP 3

Build the chain for 246

246 multiplies all three to 48, then 32, then 6: length 3.

246 → (2 × 4 × 6=48) → 32 → 6 chain length 3
4STEP 4

Part (2): search two-digit numbers for chain length 4 ending in 8

Testing big digits gives 77 → 49 → 36 → 18 → 8, length 4.

77 → 49 → 36 → 18 → 8 chain length 4
Answer
76 has length 2, 47 and 246 length 3 / the two-digit number of length 4 is 77
Recount the arrows: 76 has 2, 47 has 3, 246 has 3 — matching the example style where each arrow is one step. For 77 the four products 49, 36, 18, 8 are each correct and the last value 8 is a single digit, so the chain length is 4 and the final digit is 8, as required.
Takeaway

Just multiply the digits and keep going — counting the arrows gives the chain length, and a little hunting shows 77 takes the most steps!

  • Build the chain for 76
  • Build the chain for 47
  • Build the chain for 246
  • Part (2): search two-digit numbers for chain length 4 ending in 8