ArithMind Mental Arithmetic Lessons
Stage 13 · Phase 3: Advanced Techniques advanced all

Squaring Numbers Ending in 5

Square any number ending in 5 mentally using a simple two-step formula.

The Pattern Behind a Five

To square a number is to multiply it by itself. When that number ends in 5, its square can be found by multiplying the part before the 5 by the next whole number, then appending 25. The earlier multiply-by-11 stage showed that a fixed multiplier creates a reliable digit pattern. This stage develops the same pattern awareness, but now the constant final digit of both factors guarantees a constant two-digit ending.

Write any positive whole number ending in 5 as (10n+5). Then ((10n+5)^2=100n^2+100n+25=100n(n+1)+25). Multiplication by 100 shifts (n(n+1)) two places left, leaving exactly enough room for 25.

A Useful Area Decision

Consider a square exhibition booth whose side is 45 feet. The organizer must order flooring by square foot. Instead of writing out (45×45), calculate (4×5=20) and attach 25: the area is 2,025 square feet. This shortcut is especially practical for square areas, perfect-square recognition, and checking computations that contain a side length ending in 5.

Method A: Prefix Times Successor

Remove the final 5 and call everything remaining the prefix. Multiply that prefix by the next integer. Attach “25” as a two-digit block, including its leading 2 and final 5.

Example one: 65²

The prefix is 6 and its successor is 7. Since (6×7=42), placing 25 after 42 gives 4,225. Notice that you do not calculate (6×6); the neighboring integer is essential.

Example two: 215²

The full prefix is 21. Its successor is 22, and (21×22=462). Append 25 to obtain 46,225. A magnitude check supports this: 215 is slightly above 200, whose square is 40,000.

Method B: Nearby-Base Verification

The difference-of-squares identity provides an independent check. Choose a nearby round number (a), write the target as (a\pm5), and use ((a\pm5)^2=a^2\pm10a+25). This method takes more steps but reveals why the final 25 survives.

Example three: 95²

Think (95=100-5). Then (100^2-2×100×5+5^2=10,000-1,000+25), which is 9,025. The prefix method agrees: (9×10=90), followed by 25.

Example four: 1,005²

Use (1,005=1,000+5). Its square is (1,000,000+10,000+25=\mathbf{1,010,025}). Alternatively, the prefix 100 times its successor 101 equals 10,100; appending 25 produces the same number.

Corrections for Common Slips

  • Multiplying the prefix by itself. For 75², (7×7) is not the leading block. Use consecutive numbers: (7×8=56), then append 25.
  • Reading only the tens digit as the prefix. In 135², the prefix is 13, not 3. The rule applies to every digit before the final 5.
  • Appending a single 25 without respecting place value. Think “multiply the leading block by 100, then add 25.” This prevents dropped zeros when the prefix product contains one.

Build a Perfect-Square Chain

For questions 1–6, write only “prefix × successor” before answering. For 7–12, calculate aloud without writing. For the final group, estimate between nearby hundreds before giving the exact square. Afterward, select four answers and verify them with the nearby-base identity. If an error occurs, identify whether the prefix, successor, multiplication, or attachment caused it.

Practice Set

  1. 15²
  2. 25²
  3. 45²
  4. 55²
  5. 75²
  6. 105²
  7. 115²
  8. 135²
  9. 165²
  10. 195²
  11. 225²
  12. 255²
  13. 315²
  14. 405²
  15. 515²
  16. 995²

Four Model Thought Processes

  • 55²: remove 5 to get prefix 5; use successor 6; (5×6=30); attach 25. Answer: 3,025.
  • 115²: prefix 11 meets 12; (11×12=132); shift 132 two places and add 25. Answer: 13,225.
  • 195²: prefix calculation (19×20=380); attaching the fixed ending creates 38,025, which is just below (200²=40,000).
  • 405²: prefix 40 times 41 is 1,640; preserve that entire block and follow it with 25. Answer: 164,025.

Check Your Results

  1. 225; 2. 625; 3. 2,025; 4. 3,025; 5. 5,625; 6. 11,025; 7. 13,225; 8. 18,225; 9. 27,225; 10. 38,025; 11. 50,625; 12. 65,025; 13. 99,225; 14. 164,025; 15. 265,225; 16. 990,025.

Learner Questions

Does this work for 5²?

Yes. The prefix is 0, and (0×1=0). Appending 25 yields 25. Usually you would simply recall this basic square.

What about negative numbers ending in 5?

Squaring removes the sign because a negative times a negative is positive. Thus ((-35)^2) is the same 1,225 as (35^2).

Can the rule handle a decimal such as 2.5?

Scale first: (2.5=25÷10). Since (25²=625), divide by (10²=100), giving 6.25. Tracking the scaling is necessary; simply appending 25 to 2 would be incorrect.

Practice Examples

Example 1: 35² = ?

  1. Remove the final 5, leaving the prefix 3.
  2. Multiply the prefix by its successor: 3 × 4 = 12.
  3. Attach the fixed two-digit ending 25.
  4. Thus 35² = 1,225.

Answer: 1,225

Example 2: 85² = ?

  1. The prefix before 5 is 8.
  2. Multiply 8 by the next integer, 9, to get 72.
  3. Place 25 after 72.
  4. The square is 7,225.

Answer: 7,225

Example 3: 125² = ?

  1. Treat 12 as the complete prefix, not as separate digits.
  2. Calculate 12 × 13 = 156.
  3. Append 25 to the prefix product.
  4. The result is 15,625.

Answer: 15,625

Example 4: 305² = ?

  1. Remove the ending 5 to identify the prefix 30.
  2. Multiply consecutive numbers: 30 × 31 = 930.
  3. Attach 25, preserving the zero in 930.
  4. Therefore 305² = 93,025.

Answer: 93,025