Squaring 2-Digit Numbers Ending in 1, 4, 5, and 6

Four trailing-digit patterns cover almost every 2-digit square you’ll face in a placement aptitude section: numbers ending in 5, 4, 6, and 1 each reduce to a single fast arithmetic step.

Three of the four share the same algebraic backbone. The difference-of-squares identity a² − b² = (a+b)(a−b) simplifies dramatically when the two numbers are consecutive (|a − b| = 1): the identity becomes a² − b² = a + b. That means you only need to know one anchor square and add or subtract a small two-term sum.

Why these four trailing digits specifically? Because 5 is the midpoint of the decimal system, and multiples of 5 are easy to square. Numbers ending in 4 or 6 sit exactly one step from a multiple of 5, keeping the difference-of-squares adjustment small. Numbers ending in 1 sit one step above a round multiple of 10, which are the easiest squares of all.

Numbers Ending in 5

The full derivation and reference table for ×5 squares lives in the squaring numbers ending in 5 guide. The short version: multiply the prefix digit by the next integer, then append 25.

These three anchor squares come up repeatedly in the sections below, so it is worth keeping them in working memory:

• 35²: prefix 3, so 3 × 4 = 12, append 25, answer 1225.
• 45²: prefix 4, so 4 × 5 = 20, append 25, answer 2025.
• 75²: prefix 7, so 7 × 8 = 56, append 25, answer 5625.

The ×5 squares at 25, 35, 45, 55, 65, 75, 85, and 95 are the anchors for the three shortcuts below. Once those eight values are memorised, all of ×4, ×6, and ×1 become one-step adjustments.

Numbers Ending in 4

Rule: square the neighbor ending in 5, then subtract the sum of the two numbers.

Why it works

Set a as the number ending in 5 and b as the number ending in 4. Since a − b = 1, the identity gives a² − b² = a + b, so b² = a² − (a + b). The subtracted term is always the sum of the two consecutive numbers.

Worked examples

• 24²: anchor = 25² = 625; subtract (25 + 24) = 49; answer 576.
• 34²: anchor = 35² = 1225; subtract (35 + 34) = 69; answer 1156.
• 44²: anchor = 45² = 2025; subtract (45 + 44) = 89; answer 1936.
• 54²: anchor = 55² = 3025; subtract (55 + 54) = 109; answer 2916.
• 74²: anchor = 75² = 5625; subtract (75 + 74) = 149; answer 5476.

Speed note: the anchor square is a memorised value, and the sum of two consecutive numbers is trivial to compute. Both steps together take under 3 seconds with practice.

Numbers Ending in 6

Rule: square the neighbor ending in 5, then add the sum of the two numbers.

Why it works

Set a as the number ending in 6 and b as the number ending in 5. Since a − b = 1, the identity gives a² − b² = a + b, so a² = b² + (a + b). The direction flips because the number ending in 6 is above the anchor, not below it.

Worked examples

• 26²: anchor = 25² = 625; add (25 + 26) = 51; answer 676.
• 36²: anchor = 35² = 1225; add (35 + 36) = 71; answer 1296.
• 46²: anchor = 45² = 2025; add (45 + 46) = 91; answer 2116.
• 56²: anchor = 55² = 3025; add (55 + 56) = 111; answer 3136.
• 76²: anchor = 75² = 5625; add (75 + 76) = 151; answer 5776.

Symmetry check: 34² + 36² = 1156 + 1296 = 2452, and 2 × 35² + 2 = 2 × 1225 + 2 = 2452. This confirms the identity (n−1)² + (n+1)² = 2n² + 2, a useful internal consistency check when drilling.

Numbers Ending in 1

Rule: square the round number directly below, then add the sum of the two numbers.

Why it works

Set a as the number ending in 1 and b as the round number below (b = a − 1). Since a − b = 1, the identity gives a² − b² = a + b, so a² = b² + (a + b). The anchor here is a multiple of 10, whose square is always easy: (N × 10)² = N² × 100.

Worked examples

• 21²: anchor = 20² = 400; add (20 + 21) = 41; answer 441.
• 31²: anchor = 30² = 900; add (30 + 31) = 61; answer 961.
• 41²: anchor = 40² = 1600; add (40 + 41) = 81; answer 1681.
• 61²: anchor = 60² = 3600; add (60 + 61) = 121; answer 3721.
• 81²: anchor = 80² = 6400; add (80 + 81) = 161; answer 6561.

The round-number anchor is the easiest of the three anchors to compute on the fly: recall a single-digit square, shift two decimal places. 80² = 64 × 100 = 6400. Done.

Reference Table

Number	Anchor square	Adjustment	Result
24²	25² = 625	−(25+24) = −49	576
34²	35² = 1225	−(35+34) = −69	1156
44²	45² = 2025	−(45+44) = −89	1936
54²	55² = 3025	−(55+54) = −109	2916
74²	75² = 5625	−(75+74) = −149	5476
26²	25² = 625	+(25+26) = +51	676
36²	35² = 1225	+(35+36) = +71	1296
46²	45² = 2025	+(45+46) = +91	2116
56²	55² = 3025	+(55+56) = +111	3136
76²	75² = 5625	+(75+76) = +151	5776
21²	20² = 400	+(20+21) = +41	441
41²	40² = 1600	+(40+41) = +81	1681
61²	60² = 3600	+(60+61) = +121	3721
81²	80² = 6400	+(80+81) = +161	6561

For numbers ending in 5, the reference table covering 15 through 95 is in the squaring numbers ending in 5 guide.

The adjustment column follows a pattern. For numbers ending in 4, the adjustment increases by 20 each decade: −49, −69, −89, −109, −129, −149. For numbers ending in 6, the adjustment is the same sequence but positive: +51, +71, +91, +111, +131, +151. For numbers ending in 1, the adjustment grows by 20 each decade starting at +41. If your computed adjustment for a given number does not fit this sequence, you have made an arithmetic error rather than a shortcut error.

Each of these four trailing-digit patterns is a one-step adjustment from a known anchor. That same instinct, finding the nearest easy value and computing a small gap, appears across quantitative aptitude. The shift-and-add trick for multiplying by 111 and the angle shortcuts in clock problems are built on the same structural move. For the quantitative sections of TCS NQT and AMCAT, fluency with these adjustments means fewer seconds on routine calculations and more attention for the harder reasoning questions. Once you’ve applied that fast-pattern instinct to aptitude drills, TinkerLLM puts the same approach to work on AI prompting at ₹299: find the structure, apply the rule, verify the output.