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Properties of Integers: Practice Problems and Methods

Integer problems start with prime factorization. Learn the ideas specific to integers: the Euclidean algorithm, Diophantine equations and classification by remainders.

Basic, Standard: Grade 10 · Term 3 / Advanced: Grade 12 · Exam prep

Math problem generator

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Prime factorization, divisors and multiples

Number and sum of divisors

If N=paqbN = p^a q^b, it has (a+1)(b+1)(a + 1)(b + 1) positive divisors, with sum (1+p+⋯+pa)(1+q+⋯+qb)(1 + p + \cdots + p^a)(1 + q + \cdots + q^b).

Example: 72=23⋅3272 = 2^3 \cdot 3^2 has 4×3=124 \times 3 = 12 divisors with sum 15×13=19515 \times 13 = 195. The GCD uses the common primes with the smaller exponents; the LCM uses all primes with the larger exponents. For two numbers, ab=gcd⁡(a,b)×lcm⁡(a,b)ab = \gcd(a, b) \times \operatorname{lcm}(a, b).

The Euclidean algorithm and linear Diophantine equations

If a=bq+ra = bq + r, then gcd⁡(a,b)=gcd⁡(b,r)\gcd(a, b) = \gcd(b, r). Repeating this until the remainder is 0 is the Euclidean algorithm.

For ax+by=cax + by = c with aa, bb coprime, find one solution (x0, y0)(x_0,\ y_0) and subtract to get a(x−x0)=−b(y−y0)a(x - x_0) = -b(y - y_0). Since aa and bb are coprime, x−x0x - x_0 is a multiple of bb:

x=bk+x0,y=−ak+y0(k an integer)x = bk + x_0,\qquad y = -ak + y_0\quad (k \text{ an integer})

Remainders and base-n numbers

Every integer has the form 3k3k, 3k+13k + 1 or 3k+23k + 2. This classification proves facts such as "n2n^2 leaves remainder 0 or 1 when divided by 3". Congruences a≡b(modm)a \equiv b \pmod{m} make calculations like the remainder of 21002^{100} divided by 7 easy.

Base nn

1011(2)=1⋅23+0⋅22+1⋅2+1=111011_{(2)} = 1 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2 + 1 = 11. To convert from base 10 to base nn, divide by nn repeatedly and read the remainders from bottom to top.