Numbers & Place Value

Problem

Round by looking at lower-place digits

The four-digit number 35□8 has an unknown tens digit shown as a box. We round this number to the nearest hundred in two ways: by always rounding up, and by rounding half up (the usual rounding). We must find every digit that can fill the box so that both methods give the same hundred.
Base-ten numbers
Your answer
How to solve
Strategy Make a Systematic List — Only ten digits can fill the box, so we can simply list 0 through 9 and, for each one, compute both roundings and compare. Listing the cases makes the rule jump out: a pattern appears at the tens digit 5, which we can state as the answer.
1STEP 1

Find what 'round up' gives

The ones digit 8 keeps the number off a whole hundred, so rounding up always lands on 3600.

3508 → 3600, 3518 → 3600, …, 3598 → 3600
2STEP 2

See when 'round half up' also gives 3600

Round half up looks at the tens digit instead: 5 or more goes to 3600, 4 or less to 3500.

□ ≥ 5 → 3600, □ ≤ 4 → 3500
3STEP 3

Match the two results

The two roundings agree only when the box holds 5, 6, 7, 8 or 9.

□ ∈ {5,6,7,8,9}
Answer
5, 6, 7, 8, 9
Check the boundary: 3548 rounds half up to 3500 (box 4) but rounds up to 3600 — they differ, correctly excluded. 3558 rounds half up to 3600 and rounds up to 3600 — they match, correctly included. The five digits 5–9 are exactly the ones at or above the halfway point, which makes sense.
Takeaway

Since the ones digit is 8, rounding up always reaches 3600 — so plain rounding matches only when the tens digit is big enough (5 or more) to round up too.

  • Find what 'round up' gives
  • See when 'round half up' also gives 3600
  • Match the two results
Where next?
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▶ Practice — 12 problems