수와 자리값

Ten of any unit bundles into one of the next-higher place, and the reverse on the way down. Children first carry ones into tens and tens into hundreds, then extend the same rule to thousands, to multiplying by ten (which shifts every digit up one place), and finally to tenths regrouping into a whole. The single idea — units convert at a fixed exchange rate — is what every later number topic rests on.

To maximize (or minimize) a number, product, quotient, or mixed number, place the biggest available digits in the places that carry the most weight. The reasoning starts with whole numbers, then governs products and quotients (where a small divisor enlarges the result), decimals, and mixed numbers. It is place-value comparison turned into an optimization strategy.

A hidden number is found by applying clues in turn — digit constraints, place-by-place comparison, and inequality bounds that fence the unknown into a finite range. Early versions narrow a four-digit number from digit clues; later versions bound an unknown digit or an unknown box value under a sum, product, or comparison constraint. The shared move is converting each condition into a restriction until one survivor remains.

One number can be rebuilt many ways — as sums of two addends or as different factor pairs giving the same product. Learners first see that the same quantity has many additive and multiplicative names, then use factor pieces to test divisibility. Flexible decomposition is the engine behind mental arithmetic and later divisibility work.

In a partly blanked addition, subtraction, multiplication, or long division, the missing digits are deduced column by column from the lowest place up, using how carries and borrows must propagate. The same cryptarithm logic deepens from three-digit sums to multi-digit products and division. It trains airtight place-value bookkeeping.

Decide or count the whole numbers that satisfy 이상/이하/초과/미만 (greater/less than or equal, strictly greater/less) boundary conditions, often as the overlap of two ranges on a number line.

Write each number as a product of prime powers and the question becomes a question about exponents. A perfect square has every exponent even, the divisor count multiplies (exponent + 1) over the primes, and a fraction terminates exactly when its reduced denominator has no prime but 2 and 5. Grade-school divisibility tested one divisor at a time; this reads the whole factor structure at once.

Powers multiply by adding exponents, nest by multiplying them, and distribute over a product -- each law read straight off the meaning of a power as repeated multiplication. The problems here rewrite to a common base or a common exponent, or factor a shared power out of a sum, and then compare what is left; pairing 2s with 5s to make powers of ten is the same rewrite, and some use a given inequality to bound the result.

With negative numbers in play, a product or quotient of nonzero numbers is negative exactly when an odd number of its factors are negative, and its size depends only on the absolute values. So a problem about unknown signed numbers starts by narrowing the sign candidates from a product or quotient, then brings in the other conditions -- a sum, an order, a quadrant, a reciprocal pair -- to decide which case holds, and only then compares sizes. Elementary ordering never needed this first step.

Every point on the number line (and each coordinate in the plane) is a signed position. On the line the distance between two points is the absolute value of their difference; in the plane the horizontal and vertical gaps are each the absolute difference of one coordinate. Directions become signs: reflecting in an axis flips the sign of one coordinate, reflecting in the origin flips both. |x - a| <= b is just 'within b of a'. Each absolute value hides a two-case split -- this side or that side -- and the extremes come from choosing the cases.

A situation with a threshold -- at least, cheaper than, before the deadline -- becomes a linear inequality in the unknown. The modelling is the same as for an equation; what changes is that the answer is an interval, so the boundary (included or not, rounded which way) decides it, and two thresholds on one situation give a range squeezed from both sides.