Chapter 6.2 — Exercise 6.1 — Square Properties

Finding a Units Digit Without Squaring the Whole Number

Only the units digit of a number affects the units digit of its square, so there's never a need to square the entire number just to answer this kind of question.

• 39 → units digit 9, and 9² = 81 → units digit 1.
• 297 → units digit 7, and 7² = 49 → units digit 9.
• 5125 → units digit 5, and 5² = 25 → units digit 5.
• 7286 → units digit 6, and 6² = 36 → units digit 6.
• 8742 → units digit 2, and 2² = 4 → units digit 4.

Spotting Perfect Squares and Ruling Out Non-Squares

121 (= 11²) and 256 (= 16²) are perfect squares; 136, 321 and 600 are not — each fails the units-digit test or simply isn't the square of any whole number nearby. Giving a reason, not just an answer, is the more useful skill: 257 ends in 7 (not a valid square ending), 4592 ends in 2, 2433 ends in 3, and 6098 ends in 8 — all four ruled out instantly by the units-digit rule. 5050 needs a different reason: it ends in only one zero, and a perfect square that's a multiple of 10 must end in an even number of zeroes, so 5050 is ruled out too.

Even or Odd — Decided by the Original Number

Since squaring never changes whether a number is even or odd, this question never actually requires computing the square at all: 431 and 17779 are odd, so their squares are odd; 2826, 8204 and 99998 are even, so their squares are even. The entire question collapses to "is the original number even or odd?" — squaring a five-digit number is unnecessary work here, and the same shortcut scales to numbers of any size without becoming any harder.

Counting Integers Between Two Consecutive Squares

Using the 2n rule directly: between 25² and 26² there are 2×25 = 50 integers; between 56² and 57² there are 2×56 = 112; between 107² and 108² there are 2×107 = 214. None of these require computing either square in full — only recognising that the two given numbers are consecutive integers and applying 2n to the smaller one.

Summing Consecutive Odd Numbers Without Adding

1+3+5+7+9 = 5² = 25 (five terms, the 5th odd number is 9). Extending the same idea: 1+3+5+7+9+11+13+15+17 has nine terms (17 is the 9th odd number), so the sum is 9² = 81. Continuing to 1+3+...+25 has thirteen terms (25 is the 13th odd number), giving 13² = 169. The only work needed is counting how many odd numbers are in the list — the sum itself is immediate once that count is known.

Why the Zero-Count Rule Works

The rule that a perfect square multiple of 10 must end in an even number of zeroes follows directly from how factors of 10 split into 2's and 5's. A number ending in exactly one zero has exactly one factor of 2 and one factor of 5 that aren't matched by more of the same — and since a perfect square's prime factorisation groups every prime in pairs, a single unpaired 2 or 5 is exactly what a units-digit-zero count of one would leave behind. Two trailing zeroes, by contrast, correspond to a matched pair of both 2 and 5, which a perfect square's structure allows perfectly well. This is the same "group primes in pairs" idea from prime factorisation, just applied specifically to the factors of 10.

The Instinct This Exercise Builds

Every question above can be answered without fully computing a square — a units-digit check, an even/odd check, or a term count is always enough. That's the actual point of this exercise: recognising when a shortcut applies removes the need for arithmetic entirely, which becomes essential once the numbers involved grow far larger in the exercises ahead.

Combining Several Checks at Once

Real questions rarely test just one property in isolation — a number might need checking against the units-digit rule, the zero-count rule, and the even/odd rule all at once before a confident answer emerges. Take 6098 from this exercise: it ends in 8, which alone rules it out as a perfect square, no further checks needed. But a number like 4900 needs more care — it ends in acceptable digits (00), has an even count of trailing zeroes, and is itself even, so none of the quick rules rule it out, even though a further check (4900 = 70²) is needed to actually confirm it. Recognising when the fast rules are enough to rule a number out, versus when they merely fail to rule it out and a real calculation is still needed, is part of what this exercise is training — the rules are a filter, not a final verdict, and treating them as the latter is a common source of wrong confident answers — always be ready to fall back on prime factorisation or direct division once the quick checks have been exhausted without a clear result one way or the other.

Where a Quick Rule Gets Over-Trusted

• Assuming a units-digit match guarantees a perfect square. Ending in 4, 5, 6, 9, 0 or 1 is necessary but not sufficient — it rules nothing in, only rules impossible endings out.
• Miscounting zeroes on numbers ending in 0. A perfect square that's a multiple of 10 needs an even number of trailing zeroes — one zero alone is always a giveaway that it isn't one.
• Forgetting which number the 2n rule applies to. The formula 2n uses the smaller of the two consecutive numbers being squared, not their average or the larger one.

Practising the Shortcuts on Numbers of Your Own Choosing

A good way to test whether these shortcuts have genuinely sunk in is to pick a handful of unfamiliar numbers and work through all five question types on them without looking anything up. Choose a five- or six-digit number, find its units digit's square, decide whether it could possibly be a perfect square, check whether it's even or odd, and — if it happens to be one more than another number you've chosen — count the integers between their squares using 2n. Doing this with self-chosen numbers, rather than only ones already assigned in an exercise, is what turns these rules from "answers I can produce when asked" into "facts I reach for automatically" without ever needing a second conscious thought about it at all.

Faster Methods Ahead

Recognising perfect squares quickly here is what makes the systematic methods in Exercise 6.2 (prime factorisation) and Exercise 6.3 (the division method) faster to apply, since knowing in advance whether a number is even plausible as a perfect square saves wasted effort on numbers that can't be. For the underlying patterns this exercise draws on, revisit the Introduction to Square Numbers.