Squaring Numbers Ending in 5: The n(n+1) Shortcut

Squaring any number ending in 5 takes two steps: multiply the prefix by itself plus one, then write 25 after the result.

That is the whole rule:

• 35²: prefix 3, so 3×4 = 12, append 25, answer 1,225.
• 125²: prefix 12, so 12×13 = 156, append 25, answer 15,625.

No grid method. No long multiplication. One small product and a fixed two-digit suffix.

This article derives the algebraic proof, builds a complete reference table for numbers 25 through 165, and shows where the shortcut applies in placement aptitude sections.

The Core Rule

Take any number whose last digit is 5. Call everything before the 5 the prefix (n). The square is computed in two steps:

1. Compute n × (n + 1)
2. Write that result, then immediately write 25 after it

Writing 25 after a result is equivalent to multiplying the result by 100 and adding 25. That is the algebraic content of the rule; the next section proves it.

Quick examples as a checklist:

• 25²: n = 2, 2×3 = 6, append 25. Answer: 625.
• 45²: n = 4, 4×5 = 20, append 25. Answer: 2,025.
• 75²: n = 7, 7×8 = 56, append 25. Answer: 5,625.
• 95²: n = 9, 9×10 = 90, append 25. Answer: 9,025.
• 105²: n = 10, 10×11 = 110, append 25. Answer: 11,025.
• 155²: n = 15, 15×16 = 240, append 25. Answer: 24,025.

Why the Rule Works: Algebraic Proof

Any number ending in 5 can be written as (10n + 5), where n is its prefix. Squaring:

(10n + 5)² = 100n² + 2 × 10n × 5 + 25
           = 100n² + 100n + 25
           = 100n(n + 1) + 25

The term 100n(n+1) is exactly n×(n+1) multiplied by 100. Multiplying by 100 shifts the digits two places to the left, which is the same operation as writing 25 in the two rightmost positions. The constant +25 fills those two places exactly.

Nothing is special about any particular example: every integer ending in 5 satisfies this identity. The proof works for 1-digit prefixes (n = 1 giving 15) through arbitrarily large prefixes.

Why the suffix is always exactly 25

The last two digits of (10n + 5)² come from 5² = 25. The cross-term 2 × 10n × 5 = 100n is always a multiple of 100, so it contributes zero to the units and tens positions. The prefix product 100n² is also a multiple of 100, contributing zero to units and tens. Only the 25 from 5² lands in the units and tens positions. This means the suffix 25 never changes regardless of n, and there is never any carry from the suffix into the prefix.

2-Digit Numbers: 25 Through 95

All eight 2-digit numbers from 25 to 95 that end in 5, using the two-step rule:

Number	Prefix n	n×(n+1)	Square
25	2	2×3 = 6	625
35	3	3×4 = 12	1,225
45	4	4×5 = 20	2,025
55	5	5×6 = 30	3,025
65	6	6×7 = 42	4,225
75	7	7×8 = 56	5,625
85	8	8×9 = 72	7,225
95	9	9×10 = 90	9,025

For 15, the same rule gives prefix 1, 1×2 = 2, answer 225. The pattern holds for every number ending in 5, starting from 5 itself (prefix 0: 0×1 = 0, answer 25).

The prefix products stay small for 2-digit numbers: the largest is 9×10 = 90. This means the full answer is always writable from memory for this range once the table is internalised.

3-Digit Numbers: 105 Through 165

The rule extends to 3-digit inputs without modification. The prefix is now a 2-digit number, and the prefix product n×(n+1) grows to 3 digits.

Number	Prefix n	n×(n+1)	Square
105	10	10×11 = 110	11,025
115	11	11×12 = 132	13,225
125	12	12×13 = 156	15,625
135	13	13×14 = 182	18,225
145	14	14×15 = 210	21,025
155	15	15×16 = 240	24,025
165	16	16×17 = 272	27,225

Computing the prefix product for 3-digit inputs

For 2-digit numbers, n×(n+1) fits within a single-digit or 2-digit result and is often immediate. For the 3-digit range above, the prefix product requires a short 2-digit multiplication.

A reliable decomposition for 13×14: break as 13×14 = 13×10 + 13×4 = 130 + 52 = 182. For 14×15: 14×15 = 14×10 + 14×5 = 140 + 70 = 210. Practise these prefix multiplications separately from the squaring shortcut. The two steps are independent.

The rule also extends to 4-digit inputs:

• 1,005²: prefix 100, 100×101 = 10,100, append 25, answer 1,010,025.
• 1,255²: prefix 125, 125×126 = 15,750, append 25, answer 1,575,025.

The two-step pattern has no upper bound.

Speed Practice for Aptitude Tests

The TCS NQT Numerical Ability section and AMCAT Quantitative Aptitude both include squaring and speed-multiplication problems. These sections typically run 25 to 35 questions in a 40 to 60-minute window. Identifying a squaring-ending-in-5 problem and applying the rule in 4 to 5 seconds, rather than running long multiplication for 15 to 20 seconds, compounds across a full section.

Keep the two steps distinct under time pressure:

• Step 1: Extract the prefix (everything before the 5).
• Step 2: Multiply n × (n+1). For single-digit n, this is a table fact. For two-digit n, use the break-into-tens decomposition above.
• Step 3: Write the prefix product, then write 25. Read the result.

Practice problems (work each before checking):

• 55²: n = 5, 5×6 = 30, answer = 3,025.
• 85²: n = 8, 8×9 = 72, answer = 7,225.
• 115²: n = 11, 11×12 = 132, answer = 13,225.
• 145²: n = 14, 14×15 = 210, answer = 21,025.
• 165²: n = 16, 16×17 = 272, answer = 27,225.

The only computation that changes between problems is the prefix product. The suffix 25 is a constant and costs zero mental effort once the rule is internalised.

For a related speed-multiplication technique that works by the same decompose-and-apply logic, see the guide on the shift-and-add method for multiplying by 111. FACE Prep’s article on calendar problems in aptitude tests applies the same derive-from-first-principles approach to odd-day calculations. For timed practice across all quantitative sections, the companion guide on solving clock problems for competitive exams covers the modular-arithmetic layer that clock questions test.

The n×(n+1) identity reduces a squaring operation to one prefix product and a fixed suffix. That same instinct to find the constant part of a problem and compute only what changes is what engineers apply when they test LLM behaviour systematically rather than guessing at prompt changes. TinkerLLM is where engineering students run those experiments at ₹499, applying one change at a time and reading the output exactly as the rule above applies one formula and reads the result.