Use Square Remainders, Identities, and Sums
第20讲 完全平方数 · Lesson 20.6
Reduce enormous square expressions to short remainder cycles, compare equal sums of squares systematically, and turn visual square patterns into reusable identities.
Build a square-remainder cycle
When only a remainder is needed, replace a large number by its remainder first. Squaring the small remainder gives the same final remainder.
The replacement rule
If two numbers differ by a multiple of m, then their squares also differ by a multiple of m.
Explore one complete cycle
Checkpoint
Need a hint? Start here
Replace each number by its remainder, square that small number, then reduce again.
3² leaves 2 and6² leaves1. All whole numbers can leave0–6; squares leave only0,1,2,4. 2;1;seven remainder classes; unreduced sum14 (whose final remainder is0).
Reduce a very long sum of squares
A full remainder cycle can be bundled into blocks. Only the incomplete tail affects the final answer when each full block has remainder zero.
The first seven square remainders modulo 7 are 1,4,2,2,4,1,0. Predict the remainder of 8² before using the laboratory.
Your prediction is checked with this mission. Try it before reading the worked steps below.
original problem
Find the remainder when
is divided by 7.
One seven-term block
Every complete block of seven consecutive indices contributes remainder 0.
Split the target index
The first 287 blocks disappear modulo 7. The final five square residues are:
Square-sum remainder laboratory
Checkpoint
Need a hint? Start here
Split the number of terms into complete cycles and a tail. Multiply the full-block remainder before adding the tail.
2014=7×287+5. Each full block has remainder0; the tail residues sum to13. Prediction1. Checkpoint:287;5;13;final remainder6.
Combine a power cycle with a square remainder
Different parts of one expression may have different cycles. Reduce each part separately, then add and reduce once more.
original problem
Power part
So 2²⁰¹⁵ leaves the same remainder as the second entry of the cycle:
Square part and final sum
Power-plus-square laboratory
The base is a positive whole number from 1 through 99. Exponent 0 gives power value 1. The repeating part may start after some initial entries; the exact calculation includes these entries.
Checkpoint
Need a hint? Start here
Find the power cycle and the square remainder separately. Add the two remainders and reduce once more.
Power cycle2,4,1 has length3. Exponent2015 is2 beyond a multiple of3;2015 itself leaves6 modulo7. 3;4;1;final remainder5.
Scale a known square sum
When every term is multiplied by the same number, every square is multiplied by that number’s square.
Transfer problems
Given 1²+2²+⋯+25²=5525, calculate 3²+6²+⋯+75². Also find the least positive n for which 5 divides 1²+2²+⋯+n².
Scaled-square-sum laboratory
Least n for divisibility by 5
Checkpoint
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Multiplying every root by c multiplies every square by c².
9×5525=49725. The first partial sum1 is not divisible by5, but1+4=5 is. 9;49725;least n2.
Match two sums of three squares
Select six different digits from 1 through 9. A valid match uses three on each side, with no digit repeated anywhere, and gives equal sums of squares. Your selected match is checked together with the fixed checkpoint below.
Square-sum matcher
Left triple
Right triple
Systematic search size
There are
unordered groups of three distinct digits to examine. There are 9×8×7 ordered choices. Each group appears in six orders, so divide by 6. Group them by their square sum, then pair disjoint groups with the same total.
Checkpoint
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Choose all six digits from 1–9 and use each at most once. Add squares separately from digit sums.
16+9+64=4+49+36=89. Both digit sums are15. There are9×8×7÷6=84 unordered triples. 89;15;15;84.
Audit the worked example for completeness
The book displays two equal-square-sum pairings and concludes that the requested digit sum is 15. We preserve that original path, then separately test every assignment allowed by the printed conditions.
Printed condition
Six different symbols represent six different digits from 1 through 9. You can call them A, B, C, D, E, F: the condition is A²+B²+C²=D²+E²+F², and the requested sum is D+E+F. The original characters are shown below:
The worked example lists these two cases:
Both displayed cases give a digit sum of 15 on each side.
| Left triple | Right triple | Common square sum | Digit sums | Status |
|---|
Checkpoint
Need a hint? Start here
One counterexample is enough to disprove uniqueness. A complete search must check every disjoint pair of groups.
Six disjoint equal-square pairings exist. Either side may be the requested side; possible digit sums are12,13,14,15,16,18. Six;two displayed;no;minimum12;maximum18.
Recognize repunit-square patterns
A number made entirely of 1s is called a repunit. Squaring a short repunit creates a rising-and-falling digit ridge.
original question
Is 12,345,654,321 a perfect square?
Choose the number of 1s
The simple ridge pattern holds without carrying for lengths 1 through 9.
For six 1s, the peak digit is 6. Multiplication counts how many pairs of 1s contribute to each place: 1,2,3,4,5,6,5,4,3,2,1.
Checkpoint
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Count how many pairs of 1s contribute to each place. For up to nine 1s, those counts need no carrying.
The product digits rise to6 then fall:12345654321=111111². Six;111111;12345654321;peak6.
Build a square from odd layers—and evaluate the printed pattern
Each new border around a square has an odd number of cells. The same chapter also includes a large patterned multiplication that can be evaluated without rewriting the worked example.
original Exercise 13
Adding the next L-shaped border changes (n−1)² into n² because:
Evaluate the expression exactly as printed
Clarifying note: the upward side skips 4 in the printed expression. Its parenthesized sum is therefore 45. This page evaluates that printed expression as-is.
Checkpoint
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The nth L-shaped border has 2n−1 cells. For the printed product, first add exactly the terms shown, then use 45=50−5.
Ten odd numbers sum to100; the20th odd number is39. The parenthesis totals45. 100;39;45;product55555544444445.
Independent workshop
Correct all eight answers to complete the workshop.
Need a hint? Start here
Choose a method before calculating. Use the earlier local hint for that method, then verify the root, all stated conditions and the endpoints.
For1+3+⋯+29, there are15 terms, so the sum is15². 1;6;5;49725;6;111111;225;55555544444445.
Exit ticket
Complete all ten missions, including all five exit questions, to earn the certificate.
Fresh challenge: use the methods from this lesson on these new values. Check your calculations by squaring, listing a short cycle, or verifying every condition.
Optional reflection — not automatically graded
Square Remainder & Identity Architect
This certifies that Student completed all ten missions and can use square cycles, identities, and systematic searches.
Chapter 20 • Perfect Squares
Need a hint? Start here
Choose a method before calculating. Use the earlier local hint for that method, then verify the root, all stated conditions and the endpoints.
144 leaves4 modulo7;1²+⋯+10²=385 leaves0; scaling roots by5 multiplies the sum by25. 4;0;25;no;625.