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Recognize prime structure, count divisors and reason with integer restrictions.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
Model an integer restriction with factors or a remainder equation and compute the requested quantity without guessing from a few examples.
Review multiplication, division, exponents and the distributive law. Divisibility of b by a means b = ak for an integer k, with a nonzero. This definition handles negative multiples without changing the factor relationship. A greatest common divisor is usually reported positive. If a question asks for positive divisors, do not double the answer to include negative ones. If it asks for all integer divisors, signs must be considered explicitly.
| Term | What it means |
|---|---|
| Prime factorization | An integer greater than one written as a product of primes. |
| Divisor | A nonzero integer that divides another integer without remainder. |
| Remainder | The amount left after complete groups, from zero up to one less than a positive divisor. |
| Parity | Whether an integer is even or odd. |
Integers include negative whole numbers, zero and positive whole numbers. A positive integer excludes zero; a nonnegative integer includes it. Prime numbers are positive integers greater than one with exactly two positive divisors. One is neither prime nor composite. Two is the only even prime, so an argument claiming that every prime is odd fails at a particularly important boundary. Before doing arithmetic, translate the question's words into a precise domain. A convenient decimal cannot serve as an example when the question specifies an integer.
Many apparently large calculations become short when you preserve structure. Divisibility asks whether a factor is present; a remainder asks which residue is left after removing complete groups; parity asks only whether a factor of two is present. These are different questions. Do not calculate a large power when only its last digit matters, and do not list every divisor when prime exponents already organize the possibilities.
Another way: steps
Identify the domain; choose factors or a remainder equation; calculate the requested feature; check boundaries and units.
Write 90 as 2 times 3 squared times 5. A positive divisor can use at most one factor of 2, two factors of 3 and one factor of 5. It cannot introduce a new prime. For two integers, the greatest common divisor uses the smaller exponent of each prime because that many copies occur in both. The least common multiple uses the larger exponent because it must contain enough copies for each integer to divide it. Multiplying the integers always produces a common multiple, but it is often larger than the least one because shared factors were counted twice.
If N = 2 cubed times 7 squared, each divisor has form 2 to the power a times 7 to the power b. The exponent a has four choices, zero through three; b has three choices, zero through two. Every pair gives one distinct divisor, so there are twelve. Adding the exponent counts would count separate decisions rather than combinations. Exponent zero matters because a divisor may omit a prime. This includes one, obtained by omitting every prime, and N itself, obtained by choosing all maximum exponents.
For a positive divisor d, write n = dq + r with integer q and 0 <= r < d. If n leaves remainder two on division by five, substitute 5q + 2 into the expression being tested. For example, 4n + 3 becomes 20q + 11, whose remainder is one, not eleven. Eleven is an intermediate residue that still contains two complete groups of five. This substitution proves the remainder for every allowed n. Testing n = 2 may suggest the result but does not show why adding any multiple of five leaves it unchanged.
Under the stated convention, -11 = 4(-3) + 1, so the remainder on division by four is one. Writing -11 = 4(-2) - 3 is a true equality but the final term is not the required remainder. Keep the allowed remainder interval visible. Programming languages and calculators can use different conventions for a remainder operation with negative inputs; a displayed result must still be interpreted using the mathematical convention in the question. In most practice, algebra is more transparent than relying on an unfamiliar calculator function.
Write an even integer as 2k and an odd integer as 2k + 1. The sum of two odd integers is 2(k + m + 1), hence even. Their product is 2(2km + k + m) + 1, hence odd. This is a general argument, whereas multiplying three selected pairs is only a sample. Consecutive integers always contain an even member, so n(n + 1) is even for every integer n. A consecutive pair need not contain a multiple of three. Separate the conclusion established by the structure from a stronger conclusion you merely expect.
The units digit of powers of seven cycles through 7, 9, 3 and 1, then repeats. Divide the exponent by four to locate its position. A remainder of zero means the fourth position, not a nonexistent zeroth entry. The cycle is justified because each next units digit comes from multiplying the previous units digit by seven and discarding complete tens. Similar reasoning finds a remainder modulo another small positive integer. Establish the cycle before extrapolating; the first two terms alone do not prove a period.
Squaring doubles every exponent in a prime factorization. Therefore a positive integer is a perfect square exactly when each prime exponent is even. To make 20k a square with the smallest positive integer k, factor 20 = 2 squared times 5 and supply one more factor of 5. Extra square factors would work but would not give the minimum k. This reasoning also distinguishes necessary conditions from sufficient ones: being divisible by four does not by itself make a number a square, as twelve demonstrates.
Suppose two maintenance routines repeat every twelve and eighteen hours from the same starting time. A simultaneous occurrence must be a positive multiple of both periods, so the first return is their least common multiple, thirty-six hours. A greatest common divisor of six hours describes a useful shared time unit but not the next simultaneous occurrence. The words in the situation decide which mathematical operation is relevant.
Now change the question: eighty-four red components and one hundred twenty-six blue components must be split into as many identical kits as possible, with no leftovers. The number of kits must divide both quantities, so use the greatest common divisor, forty-two. Each kit contains two red and three blue components. A correct raw number without the right interpretation would not solve the scheduling problem.
In the practice, enter only the requested numerical quantity. The explanation checks the model, but the automatic grade checks the answer, not your private argument. Review a model error separately from an arithmetic error and revisit a different case after a study gap.
One is not prime. A remainder is not the decimal part of a quotient. The least common multiple is not always the product, and the greatest common divisor is not always the smaller input. For divisor counting, add one to each prime exponent before multiplying. When a variable is an integer, boundary cases such as zero or two may defeat a rule that seems true for large positive odd numbers.
Factor the given integer.
48 = 2 to the fourth power times 3.
Only these primes can occur in a divisor.
List the exponent choices.
Five choices for 2; two choices for 3.
Include exponent zero for each prime.
Combine the independent choices.
5 times 2 = 10.
Each exponent pair creates a unique positive divisor.
Check the requested domain.
Ten positive divisors.
Negative divisors were not requested; both 1 and 48 are included.
Translate the stated condition.
n has remainder 3 modulo 8, so n = 8k + 3.
k is an integer; n itself is not fixed.
Substitute into the target.
5n + 4 = 40k + 19.
The same n must be used throughout.
Remove complete divisor groups.
19 = 8 times 2 + 3.
Nineteen is too large to be a remainder modulo eight.
Read the final remainder.
5n + 4 = 8(5k + 2) + 3.
The first term is divisible by eight.
Check the permitted interval.
The remainder is 3.
Three is at least zero and less than eight, as required.
Identify the requested set.
Integers from 1 through 80 divisible by 4 but not by 6.
The endpoints are included.
Count the initial multiples.
80 divided by 4 gives 20.
There are twenty multiples of four.
Find the overlap period.
The least common multiple of 4 and 6 is 12.
Numbers divisible by both are multiples of twelve.
Count the overlapping cases.
There are six multiples of 12 through 80.
The next, 84, is outside the interval.
Subtract the excluded cases.
20 - 6 = 14.
Every overlap was already in the initial count.
Verify the boundary behavior.
80 remains included; 72 is excluded.
The greatest multiple of four need not be a multiple of six.
Factor the original number.
54 = 2 times 3 cubed.
We want the smallest positive k making 54k a square.
Inspect each prime exponent.
The exponents one and three are both odd.
Choose what factors k must supply.
Complete and verify the product.
How many positive divisors does 45 have?
Answer:
What is the remainder when 73 is divided by 8?
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What is the greatest common divisor of 84 and 126?
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What is the least positive integer divisible by both 18 and 24?
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How many positive divisors does 72 have?
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An integer n has remainder 4 when divided by 7. What is the remainder when 3n + 2 is divided by 7?
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What is the units digit of 3 to the power 18?
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How many integers from 1 through 100 inclusive are divisible by 6 but not by 9?
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What is the smallest positive integer k for which 150k is the square of an integer?
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The sum of five consecutive integers is 35. What is the greatest of the five integers?
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Two laboratory alarms sound together at 09:00. One repeats every 16 minutes and the other every 24 minutes. How many minutes after 09:00 will they next sound together?
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Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A positive integer p leaves remainder 3 when divided by 5. What remainder does p squared leave when divided by 5?
Answer:
Tomorrow, recover the divisor-count rule from exponent choices rather than recalling it alone. Then explain why a shared scheduling period uses a least common multiple. Revisit Algebra I if translating restrictions into equations is difficult.
17. Complete a minimum-factor argument, step 3
k = 6; 54 times 6 = 324 = 18 squared.
One extra factor of each prime suffices, and additional factors cannot reduce k.