Squares Near 10, 100 and 1000: The Base-Distance Trick

Squaring any number within 5 of 10, 100, or 1000 takes two arithmetic steps once you know one identity.

The identity is (a + b)(a - b) = a² - b², rearranged as n² = (n − d)(n + d) + d². Choose d so that one factor equals the base (10, 100, or 1000) and the key multiplication collapses to appending zeros. Add d², and the answer is done.

This is not a trick requiring memorisation. It is one algebraic identity, applied systematically. That makes it reliable under exam pressure, where a half-remembered formula is worse than no formula at all.

The One Formula Behind All Three Bases

The starting identity: a² - b² = (a + b)(a - b). Rearrange with a = n and b = d:

n² = (n + d)(n − d) + d²

This is exact for any n and any d. The insight is choosing d = |base − n|, which forces one bracket term to equal exactly 10, 100, or 1000.

n relative to base B	d	Round factor produced	n² formula
Below B	B − n	n + d = B	(n − d) × B + d²
Above B	n − B	n − d = B	B × (n + d) + d²

Multiplying by 10 means appending one zero. By 100, two zeros. By 1000, three zeros. No column arithmetic for that step.

The binomial expansion teaches (B ± d)² = B² ± 2Bd + d² and is mathematically identical. The practical difference: the binomial method starts from B², which is one million for B = 1000, a six-digit anchor that is error-prone to manipulate under time pressure. The base-distance method starts from B × (B ± 2d), where B is an exact power of ten.

Squaring Numbers Close to 10

For d at most 3, d² stays single-digit (1, 4, or 9). For d = 4, d² = 16, still a simple two-number addition.

Sub-case, n below 10 (d = 10 − n): n + d = 10, so n² = (n − d) × 10 + d². Sub-case, n above 10 (d = n − 10): n − d = 10, so n² = 10 × (n + d) + d².

Worked examples:

• 9² (d = 1, below 10): (9 + 1)(9 − 1) + 1 = 10 × 8 + 1 = 81. ✓
• 11² (d = 1, above 10): 10 × (11 + 1) + 1 = 10 × 12 + 1 = 121. ✓
• 8² (d = 2, below 10): (8 + 2)(8 − 2) + 4 = 10 × 6 + 4 = 64. ✓
• 12² (d = 2, above 10): 10 × 14 + 4 = 144. ✓
• 7² (d = 3, below 10): (7 + 3)(7 − 3) + 9 = 10 × 4 + 9 = 49. ✓
• 13² (d = 3, above 10): 10 × 16 + 9 = 169. ✓
• 6² (d = 4, below 10): (6 + 4)(6 − 4) + 16 = 10 × 2 + 16 = 36. ✓
• 14² (d = 4, above 10): 10 × 18 + 16 = 196. ✓

Notice the structure: the product (n − d) × 10 or 10 × (n + d) gives a two-digit number ending in zero. Adding d² fills in the units (and, for d = 4, the tens) digit. No carry chain is possible in any of these near-10 examples.

Squaring Numbers Close to 100

The base is now 100. The first product is (100 ± 2d) × 100, a four-digit number ending in two zeros. Then add d².

Worked examples:

• 98² (d = 2, below 100): (98 + 2)(98 − 2) + 4 = 100 × 96 + 4 = 9,604. Verify: 9,600 + 4 = 9,604. ✓
• 102² (d = 2, above 100): 100 × 104 + 4 = 10,404. Verify: 10,400 + 4 = 10,404. ✓
• 97² (d = 3, below 100): 100 × 94 + 9 = 9,409. Verify: 9,400 + 9 = 9,409. ✓
• 103² (d = 3, above 100): 100 × 106 + 9 = 10,609. Verify: 10,600 + 9 = 10,609. ✓
• 95² (d = 5, below 100): 100 × 90 + 25 = 9,025. Verify: 9,000 + 25 = 9,025. ✓
• 105² (d = 5, above 100): 100 × 110 + 25 = 11,025. Verify: 11,000 + 25 = 11,025. ✓
• 93² (d = 7, below 100): 100 × 86 + 49 = 8,649. Verify: 8,600 + 49 = 8,649. ✓
• 107² (d = 7, above 100): 100 × 114 + 49 = 11,449. Verify: 11,400 + 49 = 11,449. ✓

The pattern across all near-100 examples: the product (100 ± 2d) × 100 ends in exactly two zeros. The trailing two digits of the final answer come entirely from d². For d at most 9, d² is at most 81, so you add at most two digits into those two trailing positions. The base factor and d² never overlap; no carry chain occurs.

For d = 5 onward, d² is two digits (25, 36, 49…). Adding a two-digit d² to a number ending in two zeros is still trivial: the units of d² fill the units position and the tens of d² fill the tens position, with no carry into the hundreds column.

Squaring Numbers Close to 1000

The base is 1000. The product (1000 ± 2d) × 1000 gives a six-digit number ending in three zeros. Then add d².

Worked examples:

• 999² (d = 1, below 1000): (999 + 1)(999 − 1) + 1 = 1000 × 998 + 1 = 998,001. ✓
• 1001² (d = 1, above 1000): 1000 × 1002 + 1 = 1,002,001. ✓
• 998² (d = 2, below 1000): (998 + 2)(998 − 2) + 4 = 1000 × 996 + 4 = 996,004. ✓
• 1002² (d = 2, above 1000): 1000 × 1004 + 4 = 1,004,004. ✓
• 997² (d = 3, below 1000): 1000 × 994 + 9 = 994,009. ✓
• 1003² (d = 3, above 1000): 1000 × 1006 + 9 = 1,006,009. ✓
• 995² (d = 5, below 1000): 1000 × 990 + 25 = 990,025. ✓
• 1005² (d = 5, above 1000): 1000 × 1010 + 25 = 1,010,025. ✓

Carry-chain analysis for d = 2 (998² and 1002²): d² = 4, a single digit. Adding 4 to a number ending in three zeros requires no carry propagation:

• 998²: 996,000 + 4 = 996,004. Units position only, no carry.
• 1002²: 1,004,000 + 4 = 1,004,004. Same.

For d = 5, d² = 25, two digits. Adding 25 to a number ending in three zeros fills two of the three trailing positions:

• 995²: 990,000 + 25 = 990,025. The 5 fills the units, the 2 fills the tens, the third trailing zero stays as hundreds. Still carry-free.

Carry chains through the base factor can only appear if d² reaches or exceeds 1000, meaning d must be 32 or above, far outside the range where nearness to the base makes the trick useful. For any number within 10 of 1000, d is at most 10 and d² is at most 100. The addition is always carry-free.

Reading the Pattern in Reverse

Placement multiple-choice questions occasionally give candidate answers and ask which is correct. The base-distance structure provides a fast elimination filter: for a number near a base, d² determines the trailing digits of the answer.

Near 100 (trailing two digits = d²):

• d = 1: answer ends in 01 (e.g., 99² = 9,801 or 101² = 10,201)
• d = 2: answer ends in 04 (e.g., 98² = 9,604 or 102² = 10,404)
• d = 3: answer ends in 09 (e.g., 97² = 9,409 or 103² = 10,609)
• d = 7: answer ends in 49 (e.g., 93² = 8,649 or 107² = 11,449)

A candidate answer ending in 12 is ruled out immediately: 12 is not a perfect square, so no number with integer distance from 100 can produce that trailing pattern.

Near 1000 (trailing three digits = d², zero-padded):

• d = 1: answer ends in 001 (e.g., 999² = 998,001)
• d = 3: answer ends in 009 (e.g., 997² = 994,009)
• d = 5: answer ends in 025 (e.g., 995² = 990,025)

An option ending in 017 is inconsistent with any single-digit d and can be ruled out without computing.

This reverse-read becomes faster than the forward calculation once the pattern is internalised. The same principle of structural elimination appears in FACE Prep’s guide to calendar problems in aptitude tests and in clock reasoning for competitive exams, where modular arithmetic constrains the answer space before any calculation begins.

Placement Practice Set

Speed-squaring questions appear in the quantitative aptitude sections of TCS NQT and AMCAT. Both tests include 25 to 35 quantitative questions in timed windows, and saving 10 seconds per problem adds meaningful buffer for the harder items at the end.

Work each problem using the base-distance method, then verify:

• 96² (d = 4, base 100): 100 × 92 + 16 = 9,216. Verify: 9,200 + 16 = 9,216. ✓
• 104² (d = 4, base 100): 100 × 108 + 16 = 10,816. Verify: 10,800 + 16 = 10,816. ✓
• 94² (d = 6, base 100): 100 × 88 + 36 = 8,836. Verify: 8,800 + 36 = 8,836. ✓
• 106² (d = 6, base 100): 100 × 112 + 36 = 11,236. Verify: 11,200 + 36 = 11,236. ✓
• 996² (d = 4, base 1000): 1000 × 992 + 16 = 992,016. Verify: 992,000 + 16 = 992,016. ✓
• 1004² (d = 4, base 1000): 1000 × 1008 + 16 = 1,008,016. Verify: 1,008,000 + 16 = 1,008,016. ✓
• 993² (d = 7, base 1000): 1000 × 986 + 49 = 986,049. Verify: 986,000 + 49 = 986,049. ✓
• 1007² (d = 7, base 1000): 1000 × 1014 + 49 = 1,014,049. Verify: 1,014,000 + 49 = 1,014,049. ✓

The 111-multiplication trick uses the same underlying structure: one algebraic identity (111 = 100 + 10 + 1) reduces an apparently complex multiplication to a pattern of column additions. The base-distance trick reduces squaring to one round-number multiplication and a small addition. Both reward students who understand why the shortcut works rather than those who memorise steps.

The base-distance identity reduces squaring to pattern recognition and one trivial multiplication. TinkerLLM applies the same logic to LLM experiments: instead of reading about what changes model behaviour, you run the test and observe the result directly. Entry is ₹299.